notes to point-set topology · 2020-04-30 · notes to point-set topology zhengyaowu abstract....

NOTES TO POINT-SET TOPOLOGY

ZHENGYAO WU

Abstract. Notes to the course “Basic Topology” that I taught during Fall 2019 at ShantouUniversity. Main reference: Bourbaki, General Topology, Chapters I, II, IV and IX.

Contents

1. September 9, Metric space, uniform structure, neighborhoods 32. September 16, Topological spaces, filters, bases of filters, Cauchy filters 103. September 23, Limit, completeness, interior, closure, cluster point, density 194. September 30, Minimal Cauchy filter, completion (1) 285. October 12, Continuity, Hausdorff, product space 386. October 14, Regular, extension of maps, homeomorphism 477. October 21, Completion(2), 558. October 28, Uniformizable, completely regular, compact(1), subspace 649. November 4, Tychonoff, compact(2) 7210. November 11, Compact(3), bounded, connected(1) 8111. November 18, Intervals, extreme / intermediate value theorems, metrizable, first /

second countable, basis of a topology 9112. November 25, Quotient space, open map, closed map 9913. December 2nd, Cone, suspension, non-Hausdorff, path connected 10814. December 9, Urysohn theorem, Tietze extension, Connected component, Cantor set 11715. December 16, Subbasis, isolated, perfect, Stone-Čech compactification 12416. December 23, Locally finite, refinement, paracompact, Lindelöf, Sorgenfrey 132

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NOTES TO POINT-SET TOPOLOGY 3

1. September 9, Metric space, uniform structure, neighborhoods

Topology studies properties of geometric objects preserved under continuous deformations suchas streching, twisting, crumping and bending, but not tearing or gluing1. Topology originates in1736 when Leonard Euler studied seven bridges of Königsberg.Point-set topology studies mainly continuity, compactness and connectedness.

Definition 1.1Let X be a set. A metric is a map d : X ×X → [0,+∞) such that(ECI) For all x, y ∈ X, d(x, y) = 0 iff x = y.(ECII) For all x, y ∈ X, d(x, y) = d(y, x).(ECIII) For all x, y, z ∈ X, d(x, z) ≤ d(x, y) + d(y, z).We call (X, d) a metric space.

Example 1.2

For all x = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn, define d(x, y) =√

n∑i=1

(xi − yi)2. Then d is a metric,

called the Euclidean metric.

Proof. The proof of (ECI) and (ECII) are left to the reader. We only prove (ECIII). For allx = (x1, . . . , xn), y = (y1, . . . , yn), z = (z1, . . . , zn) ∈ Rn, we need to show that√√√√ n∑

i=1(xi − zi)2 ≤

√√√√ n∑i=1

(xi − yi)2 +√√√√ n∑i=1

(yi − zi)2

Steps Statements Reasons

1. Define ai = xi−yi, bi = yi−zi. It is enough

to show√

n∑i=1

(ai + bi)2 ≤√

n∑i=1

a2i +√

n∑i=1

b2i .

ai + bi = xi − zi.

2. iffn∑i=1

aibi ≤√

n∑i=1

a2i

√n∑i=1

b2i . Square both sides of step 1.

3. iff(

n∑i=1

aibi

)2≤(

n∑i=1

a2i

)(n∑i=1

b2i

). Square both sides of step 2.

4.(

n∑i=1

a2i

)(n∑i=1

b2i

)−(

n∑i=1

aibi

)2≥ 0. = ∑

1≤i<j≤n(aibj − ajbi)2.

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Example 1.3Let X be a set. Let f : X → R be a function. For all x, y ∈ X, define d(x, y) = |f(x)− f(y)|. Themap d is a metric on X since | | is a metric on R.

1https://en.wikipedia.org/wiki/Topology

https://en.wikipedia.org/wiki/Topology

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Example 1.4Let X be the set of continuous functions [0, 1]→ R. For all f, g ∈ X, define

d(f, g) =∫ 1

0|f(t)− g(t)| dt.

The map d is a metric. The proof is left to the reader.

Example 1.5Let X be a set. For all x, y ∈ X, define

d(x, y) =

0, x = y

1, x 6= y

The map d is a metric. The proof is left to the reader.

Homework 1.6Let X be a set. Suppose d1 and d2 are metrics on X. Show that(1) For all x, y ∈ X, define d3(x, y) = d1(x, y) + d2(x, y). Show that d3 is a metric on X.(2) For all x, y ∈ X, define d4(x, y) = sup(d1(x, y), d2(x, y)). Show that d4 is a metric on X.

Definition 1.7Let (X, d) be a metric space. For each ε ∈ R, ε > 0, define

Uε = d−1([0, ε]) = {(x, y) ∈ X ×X : d(x, y) ≤ ε}.

We call U = {V ⊂ X ×X : ∃ε > 0, Uε ⊂ V } the uniform structure of (X, d).

Review 1.8Let X be a set, x ∈ X and Y ⊂ X ×X. We denote

• ∆X = U0 = {(x, x) : x ∈ X}.• Y −1 = {(x, y) ∈ X ×X : (y, x) ∈ Y }.• Y ◦ Y = {(x, z) ∈ X ×X : ∃y ∈ X, (x, y), (y, z) ∈ Y }.• Y (x) = {y ∈ X : (x, y) ∈ Y }.

Lemma 1.9Let (X, d) be a metric space.(BI) For all a, b > 0, there exists c > 0 such that Uc ⊂ Ua ∩ Ub.(UI’) For all a > 0, ∆X ⊂ Ua.(UII’) For all a > 0, there exists b > 0 such that Ub ⊂ U−1

a .(UIII’) For all a > 0, there exists b > 0 such that Ub ◦ Ub ⊂ Ua.

Proof. (BI) Take c = inf(a, b). If d(x, y) ≤ c, then d(x, y) ≤ a and d(x, y) ≤ b.(UI’) For all x ∈ X, d(x, x) = 0 ≤ a by (ECI).(UII’) Take b = a, Ub = Ua = U−1

a by (ECII).


(UIII’) Take b = a/2, if d(x, y) ≤ b and d(y, z) ≤ b, then d(x, z) ≤ d(x, y) + d(y, z) ≤ 2b = a by(ECIII). �

Definition 1.10The empty set ∅ has the uniform structure {∅}. Let X be a nonempty set. A uniformstructure of X is a set U of some subsets of X ×X such that(FI) If V ∈ U and W ⊃ V , then W ∈ U.(FII) If I is finite and Vi ∈ U for all i ∈ I, then ⋂

i∈IVi ∈ U.

(UI) If V ∈ U, then ∆X ⊂ V .(UII) If V ∈ U, then V −1 ∈ U.(UIII) For all V ∈ U, there exists W ∈ U such that W ◦W ⊂ V .If X is a set with a uniform structure U, then we call (X,U) a uniform space.

Example 1.11For ε > 0, define Vε = {(x, y) ∈ R ×R : |x− y| < ε}. We call U = {V ⊂ R ×R : ∃Vε ⊂ V } theadditive uniform structure of R.

Example 1.12Let X be a set. We call U = {V ⊂ X ×X : ∆X ⊂ V } the discrete uniform structure of X and(X,U) a discrete uniform space.

Example 1.13Let (X, d) be a metric space. For center x ∈ X and radius a ∈ R, a > 0, we call Va(x) ={z ∈ X : d(x, z) < a} the open ball; Wa(x) = {z ∈ X : d(x, z) ≤ a} the closed ball;Sa(x) = {z ∈ X : d(x, z) = a} the sphere.

Lemma 1.14Let (X, d) be a metric space. For x ∈ X, let V(x) = {V ⊂ X : ∃a > 0, Va(x) ⊂ V }.(VI) If V ∈ V(x) and V ⊂ Y ⊂ X, then Y ∈ V(x).(VII) If I is a finite set and Vi ∈ V(x) for all i ∈ I, then ⋂

i∈IVi ∈ V(x).

(VIII) For all V ∈ V(x), x ∈ V .(VIV) If V ∈ V(x), then there exists W ∈ V(x) such that for all y ∈ W , V ∈ V(y).

Proof. (VI)


1. ∃a > 0, Va(x) ⊂ V . V ∈ V(x).

2. Y ∈ V(x). Va(x) ⊂ V ⊂ Y

(VII)

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1. ∀i ∈ I, ∃ai > 0, Vai(x) ⊂ Vi. Vi ∈ V(x).

2. Let a = infi∈I

ai. I is finite.

3. Y ∈ V(x). Va(x) ⊂ ⋂i∈IVi

(VIII)


1. ∃a > 0, x ∈ Va(x) ⊂ V . V ∈ V(x).

(VIV)


1. ∃a > 0, Va(x) ⊂ V . V ∈ V(x).

2. For all y ∈ Va/2(x), Va/2(y) ⊂ Va(x). For all z ∈ Va/2(y), d(x, z) ≤ d(x, y) +d(y, z) < a/2 + a/2 = a by Defini-tion 1.1(ECIII).

3. Let W = {y ∈ V : ∃b > 0, Vb(y) ⊂ V }.Then W ∈ V(x).

Va/2(x) ⊂ W by step 2.

4. For all y ∈ W , V ∈ V(y). ∃b > 0, Vb(y) ⊂ V .

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Homework 1.15Let (X,U) be a uniform space (not necessarily a metric space). For each x ∈ X, define

V(x) = {V (x) : V ∈ U}, here V (x) = {y ∈ X : (x, y) ∈ V }.

Show that(VI) If V (x) ∈ V(x) and V (x) ⊂ Y ⊂ X, then Y ∈ V(x).(VII) If I is a finite set and Vi(x) ∈ V(x) for all i ∈ I, then ⋂

i∈IVi(x) ∈ V(x).

(VIII) For all V (x) ∈ V(x), x ∈ V (x).(VIV) If V (x) ∈ V(x), then there exists W (x) ∈ V(x) such that for all y ∈ W (x), V (x) ∈ V(y).


Definition 1.16Let X be a set with two metrics d1 and d2. We call d1 and d2 equivalent if (X, d1) and (X, d2)have the same uniform structure.

Example 1.17

For all x = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn, let d1(x, y) =√

n∑i=1

(xi − yi)2 be the Euclidean

metric. Define d2(x, y) = sup1≤i≤n

|xi − yi|. The map d2 is a metric on Rn (the proof is left to the

reader). The metrics d1, d2 are equivalent because

d2(x, y) ≤ d1(x, y) ≤√√√√ n∑i=1

(d2(x, y))2 =√nd2(x, y).

Homework 1.18Let (X, d) be a metric space. For all x, y ∈ X,(1) Define d1(x, y) =

√d(x, y). Show that d1 is a metric and d, d1 are equivalent.

(2) Define d2(x, y) = ln(1 + d(x, y)). Show that d2 is a metric and d, d2 are equivalent.(3) Define d3(x, y) = inf(1, d(x, y)). Show that d3 is a metric and d, d3 are equivalent.

Definition 1.19Let (X,U) and (X ′,U′) be uniform spaces. A map f : X → X ′ is uniformly continuous if forall V ′ ∈ U′, there exists V ∈ U such that (f × f)(V ) ⊂ V ′ (iff (f × f)−1(V ′) ∈ U).

Example 1.20(1) Let (X,U) be a uniform space. The identity map Id : X → X is uniformly continuous.(2) Let (X,U) and (X ′,U′) be uniform spaces. Every constant map c : X → X ′ is uniformlycontinuous.(3) Suppose (X,U) is a discrete uniform space. For all uniform space (X ′,U′), every map f : X →X ′ is uniformly continuous.

Definition 1.21Let X be a set with two uniform structures U1 and U2. If U1 ⊃ U2 (iff Id : (X,U1) → (X,U2) isuniformly continuous), then we say that U1 is finer than U2 and U2 is coarser than U1. It is anorder relation.

Example 1.22(1) The discrete uniform structure of X is the finest.(2) The uniform structure {X ×X} of X is the coarest.

Definition 1.23Let (X,UX) and (Y,UY ) be uniform spaces. Let UX×Y be the coarsest uniform structure of X×Ysuch that pr1 and pr2 are uniformly continuous, i.e. if U is a uniform structure of X × Y such

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that pr1 : (X × Y,U) → (X,UX) and pr2 : (X × Y,U) → (Y,UY ) are uniformly continuous, thenUX×Y ⊂ U and UX×Y is one of the U’s. We call UX×Y the product uniform structure; We call(X × Y,UX×Y ) the product uniform space of (X,UX) and (Y,UY ).

Lemma 1.24Let (X, d) be a metric space. Suppose [0,∞) have the additive uniform structure. Let UX be theuniform structure of (X, d). Let UX×X be the product uniform structure. The map d : X ×X →[0,∞) is uniformly continuous.Proof.


1. For all ε > 0, {(x1, x2, x3, x4) ∈X4 : d(x1, x3) < ε/2} ∈ UX×X ,

pr1 is uniformly continuous by Defini-tion 1.23.

and {(x1, x2, x3, x4) ∈ X4 : d(x2, x4) <ε/2} ∈ UX×X .

pr1 is uniformly continuous by Defini-tion 1.23.

2. W = {(x1, x2, x3, x4) ∈ X4 : d(x1, x3) <ε/2, d(x2, x4) < ε/2} ∈ UX×X .

UX×X is a uniform structure and Defini-tion 1.10(FII).

3. (d× d)(W ) ⊂ {a, b ≥ 0 : |a− b| < ε}. below

|d(x1, x2) − d(x3, x4)| = |d(x1, x2) −d(x2, x3) + d(x2, x3)− d(x3, x4)|

±d(x1, x3)

≤ |d(x1, x2) − d(x2, x3)| + |d(x2, x3) −d(x3, x4)|

Definition 1.1(ECIII)

≤ d(x1, x3) + d(x2, x4) Definition 1.1(ECII,III)

< ε/2 + ε/2 = ε.

4. d is uniformly continuous. Definition 1.19

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Lemma 1.25Let X be a set. Suppose for each x ∈ X, there exists a set V(x) whose elements are subsets of Xsuch that (VI)(VII)(VIII)(VIV) as in Lemma 1.14. Define T = {∅}∪{U ⊂ X : ∀x ∈ U, U ∈ V(x)}.Then we have(OI) If Ui ∈ T for all i ∈ I, then ⋃

i∈IUi ∈ T.

(OII) If I is a finite and Ui ∈ T for all i ∈ I, then ⋂i∈IUi ∈ T.


Proof. (OI)


1. If ⋃i∈IUi = ∅, then ⋃

i∈IUi ∈ T. (including

the case I = ∅)∅ ∈ T.

2. If ⋃i∈IUi 6= ∅, then for all x ∈ ⋃

i∈IUi, there

exists j ∈ I, x ∈ Uj.Union.

3. Uj ∈ V(x). Uj ∈ T.

4. ⋃i∈IUi ∈ V(x). Uj ⊂

⋃i∈IUi and (VI)

5. ⋃i∈IUi ∈ T. Def. of T.

(OII)


1. If I = ∅, then ⋂i∈IUi = X ∈ T. Def. of T and (VI)

2. If ⋂i∈IUi = ∅, then ⋂

i∈IUi ∈ T. ∅ ∈ T.

3. If I 6= ∅ and ⋂i∈IUi 6= ∅, for all x ∈ ⋂

i∈IUi,

x ∈ Ui for all i ∈ I.intersection

4. Ui ∈ V(x) for all i ∈ I. Ui ∈ T for all i ∈ I.

5. ⋂i∈IUi ∈ V(x). (VII).

6. ⋂i∈IUi ∈ T. Def. of T.

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2. September 16, Topological spaces, filters, bases of filters, Cauchy filters

Definition 2.1Let X be a set. Let T be a set of some subsets of X such that(OI) If I is a set and Ui ∈ T for all i ∈ I, then ⋃

i∈IUi ∈ T.

(OII) If I is a finite set and Ui ∈ T for all i ∈ I, then ⋂i∈IUi ∈ T.

We say that T is a topology, X (or (X,T)) is a topological space, elements of X are points,elements of T are open subsets of X.

Remark 2.2(1) implies that ∅ = ⋃

∅∈ T. (2) implies that X = ⋂

∅∈ T.

Some authors include {∅, X} ⊂ T in their definitions of “topology”.

Example 2.3Let X be a set. The trivial topology of X is T = {∅, X}.

Example 2.4Let X be a set. The discrete topology of X is the set T = {Y : Y ⊂ X}.

Homework 2.5(1) Find all possible topologies on {1, 2}.(2) Find all possible topologies on {1, 2, 3}.

Definition 2.6Let X be a topological space. Let A ⊂ X. Let V be a subset of X. We call V a neighborhoodof A if there exists an open set U in X such that A ⊂ U ⊂ V . Let x ∈ X. If V is a neighborhoodof {x}, then we say that V is a neighborhood of x.

Proposition 2.7Let X be a topological space. A subset U of X is open iff for all x ∈ U , U is a neighborhood of x.

Proof. Suppose U is open. By Definition 2.6, U is a neighborhood of x for all x ∈ U . Conversely,suppose for all x ∈ U , U is a neighborhood of x.


1. There exists an open subset Vx of X suchthat x ∈ Vx ⊂ U .

Definition 2.6

2. ⋃x∈U

Vx = U . U = ⋃x∈U{x} ⊂ ⋃

x∈UVx ⊂ U .

3. U is open in X. Definition 2.1(OI)


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Lemma 2.8Let X be a topological space and x ∈ X. Let V(x) be the set of all neighborhoods of x. Then(VI)(VII)(VIII)(VIV) as in Lemma 1.14 holds.

Proof. (VI) Suppose Y ⊂ X. If there exists V ∈ V(x) such that V ⊂ Y , then Y ∈ V(x).


1. There exists U open in X such that x ∈U ⊂ V .

V ∈ V(x) and Definition 2.6.

2. x ∈ U ⊂ Y . V ⊂ Y .

3. Y ∈ V(x). Definition 2.6.

(VII) Suppose I is a finite set. If Vi ∈ V(x) for all i ∈ I, then ⋂i∈IVi ∈ V(x).


1. For all i ∈ I, there exists Ui open in X

such that x ∈ Ui ⊂ Vi.Vi ∈ V(x) and Definition 2.6.

2. ⋂i∈IUi is open in X. Definition 2.1(OII)

3. x ∈ ⋂i∈IUi ⊂

⋂i∈IVi. Step 1.

3. ⋂i∈IVi ∈ V(x). Definition 2.6.

(VIII) For all V ∈ V(x), x ∈ V .


1. There exists U open in X such that x ∈U ⊂ V , so x ∈ V .

V ∈ V(x) and Definition 2.6.

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(VIV) If V ∈ V(x), then there exists W ∈ V(x) such that for all y ∈ W , V ∈ V(y).Let W = {y ∈ V : V ∈ V(y)}.


1. W 6= ∅. x ∈ W .

2. ∃U open in X such that x ∈ U ⊂ V . V ∈ V(x) and Definition 2.6.

3. W ∈ V(x). U ⊂ W by Proposition 2.7, and (VI)

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Example 2.9The set R has a topology TR whose open sets are unions of bounded open intervals (a, b). For eachx ∈ R, the set of all neighborhoods V(x) of x consists of {V ⊂ R : ∃a, b ∈ R, x ∈ (a, b) ⊂ V }.We call TR the Euclidean topology; We call (R,TR) the real line. Similarly, we define theEuclidean topology TQ on Q and we call (Q,TQ) the rational line.

Definition 2.10Let X be a set. Let F be a set of some subsets of X. We call F a filter of X if it satisfies (FI),(FII) as in Definition 1.10; and ∅ 6∈ F.

Example 2.11If X 6= ∅ and U is a uniform structure on X, then U is a filter. First, U satisfies (FI) and (FII) byDefinition 1.10. By Definition 1.10(UI), ∆X ⊂ V for all V ∈ U. Therefore ∅ 6∈ U.

Example 2.12Let X be a topological space. If ∅ 6= A ⊂ X, then the set V(A) of all neighborhoods of A is a filter,called the neighborhood filter of A. In particular, for x ∈ X, the set V(x) of all neighborhoodsof x is the neighborhood filter of x.

Example 2.13Let X be a set. Let (xj)j≥1 be an infinite sequence in X. Then F = {A ⊂ X : xj ∈ X −A for fintiely many j} is a filter.

Proof. (FI) If V ∈ F and W ⊃ V , then W ∈ F.



1. xj ∈ X − V for finitely many j. V ∈ F.

2. xj ∈ X −W for finitely many j. X −W ⊂ X − V since V ⊂ W .

3. W ∈ F. Def. of F.

(FII) If I is finite and Vi ∈ F for all i ∈ I, then ⋂i∈IVi ∈ F.


1. For all i ∈ I, xj ∈ X−Vi for finitely manyj.

Vi ∈ F.

2. xj ∈ X −⋂i∈IVi for finitely many j. X − ⋂

i∈IVi = ⋃

i∈I(X − Vi) and I is finite.

3. ⋂i∈IVi ∈ F. Def. of F.

∅ 6∈ F since X − ∅ = X contains every xj (infinitely many). �

Definition 2.14Let X be a set. Let F be a filter on X. We call B ⊂ F a basis of F if for all A ∈ F, there existsB ∈ B such that B ⊂ A. We also say that F is generated by B.

Lemma 2.15Let X be a set. A set B of some subsets of X is a basis of a filter F on X iff(BI) For all B1, B2 ∈ B, there exists B3 ∈ B such that B3 ⊂ B1 ∩B2 and(BII) ∅ 6∈ B and B 6= ∅.

Proof. Suppose B is a basis of a filter F on X. (BI)


1. B1 ∩B2 ∈ F. B ⊂ F and (FII)

2. ∃B3 ∈ B such that B3 ⊂ B1 ∩B2. B is a basis of U and Definition 2.14.

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(BII)


1. ∅ 6∈ B. B ⊂ F and ∅ 6∈ F by Definition 2.10.

2. X ∈ F. F 6= ∅ and Definition 2.10(FI).

3. ∃B ∈ B (such that B ⊂ X), i.e. B 6= ∅. B is a basis of F and Definition 2.14.

Conversely, suppose B satisfies (BI)(BII). Define F = {A ⊂ X : ∃B ∈ B, B ⊂ A}.(FI) If A ∈ F and A ⊂ A′, then A′ ∈ F.


1. There exists B ∈ B such that B ⊂ A. A ∈ F.

2. A′ ∈ F. B ⊂ A ⊂ A′ and def. of F.

(FII) If I is finite and Ai ∈ F for all i ∈ I, then ⋂i∈IAi ∈ F.


1.1. Suppose I 6= ∅. For all i ∈ I, there existsBi ∈ B such that Bi ⊂ Ai.

Ai ∈ F.

1.2. There exists B ∈ B such that B ⊂ ⋂i∈IBi. I is finite, (BI) and induction.

1.3. ⋂i∈IAi ∈ F. B ⊂ ⋂

i∈IBi ⊂

⋂i∈IAi and def. of F.

2.1. F 6= ∅. B ⊂ F; and B 6= ∅ by (BII).

2.2. Suppose I = ∅. ⋂∅

= X ∈ F. (FI)

Finally, ∅ 6∈ F since ∅ 6∈ B by (BII). Therefore F is a filter by Definition 2.10. �

Lemma 2.16Suppose X 6= ∅. A set B of some subsets of X ×X is a basis of a uniform structure U of X iff


(BI) For all B1, B2 ∈ B, there exists B3 ∈ B such that B3 ⊂ B1 ∩B2.(UI’) For all B ∈ B, ∆X ⊂ B.(UII’) For all B ∈ B, there exists B′ ∈ B such that B′−1 ⊂ B.(UIII’) For all B ∈ B, there exists B′ ∈ B such that B′ ◦B′ ⊂ B.

Proof. Suppose B is a basis of U. (BI) holds by Lemma 2.15.(UI’) follows from B ⊂ U and Definition 1.10(UI).(UII’)


1. B−1 ∈ U. B ⊂ U and Definition 1.10(UII).

2. There exists B′ ∈ B such that B′ ⊂ B−1. B is a basis of U and Definition 2.14.

3. B′−1 ⊂ B. Step 2.

(UIII’)


1. ∃W ∈ U such that W ◦W ⊂ B. B ⊂ U and Definition 1.10(UIII).

2. There exists B′ ∈ B such that B′ ⊂ W . B is a basis of U and Definition 2.14.

3. B′ ◦B′ ⊂ W ◦W ⊂ B. Steps 1 and 2.

Conversely, suppose (BI)(UI’)(UII’)(UIII’). Define U = {A ⊂ X : ∃B ∈ B, B ⊂ A}. We need toshow that U is a uniform structure of X. (FI)(FII) hold by Lemma 2.15.(UI)


1. ∀A ∈ U, ∃B ∈ B, B ⊂ A. Def. of U.

2. ∆X ⊂ B. (UI’)

3. ∆X ⊂ B ⊂ U . Steps 1 and 2.

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(UII)


1. ∀A ∈ U, ∃B ∈ B, B ⊂ A. Def. of U.

2. ∃B′ ∈ B, B′−1 ⊂ B. (UII’)

3. B′ ∈ U, B′−1 ⊂ B ⊂ A. B ⊂ U, steps 1 and 2.

4. A−1 ∈ U. B′ ⊂ A−1.

(UIII)


1. ∀A ∈ U, ∃B ∈ B, B ⊂ A. Def. of U.

2. ∃B′ ∈ B, B′ ◦B′ ⊂ B. (UIII’)

3. B′ ∈ U, B′ ◦B′ ⊂ B ⊂ A. B ⊂ U, steps 1 and 2.

Finally, U is a uniform structure on X by Definition 1.10. �

Corollary 2.17Let X 6= ∅ and (X, d) be a metric space. For all ε ∈ R and ε > 0, let Uε = {(x, y) ∈ X ×X : d(x, y) ≤ ε} as in Definition 1.7. The set B = {Uε : ε > 0} is a basis of the uniform structureof (X, d).

Proof. It follows from Lemma 2.16 and Lemma 1.9. �

Example 2.18Let X be a topological space and x ∈ X. Let V(x) be the neighborhood filter of x. A basis ofV(x) is also called a fundamental system of neighborhoods or a neighborhood basis or alocal basis of x.

Example 2.19For the real line R and x ∈ R, The set {[x− 1/n, x+ 1/n) : n = 1, 2, . . .} is a basis of V(x).


Example 2.20Let (X, d) be a metric space. The set of open balls {Va(x) : a ∈ R, a > 0} and the set of closedballs {Wa(x) : a ∈ R, a > 0} is a basis of V(x).

Definition 2.21Let X be a topological space. If Card(X) ≤ 1 or X satisfies the following axiom (H), then we callX a Hausdorff space or a separated space.(H) For all x 6= y in X, there exists U ∈ V(x) and V ∈ V(y) such that U ∩ V = ∅.

Lemma 2.22Every metric space (X, d) is Hausdorff.

Proof. Suppose Card(X) ≥ 2 and x 6= y in X.


1. d(x, y) > 0. x 6= y and Definition 1.1(ECI)

2. Suppose 0 < a < d(x, y)/2. ThenWa(x) ∈ V(x) and Wa(y) ∈ V(y).

Example 1.13

3. Wa(x) ∩Wa(y) = ∅. If z ∈ Wa(x) ∩ Wa(y), then d(x, y) ≤d(x, z) + d(z, y) ≤ a + a < d(x, y), a con-tradiction.

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Example 2.23Let (X,U) be a uniform space. Suppose X 6= ∅. By Example 2.11, U is a filter on X. LetS = {S ∈ U : S = S−1}. We have S is a basis of U. In fact, for all V ∈ U, V −1 ∈ U byDefinition 1.10(UII). Then S = V ∩ V −1 ∈ U by (FII). Also, S ∈ S and S ⊂ V .

Definition 2.24Let (X,U) be a uniform space. Let F be a filter of X. We call F a Cauchy filter of (X,U) if forall V ∈ U, there exists A ∈ F such that A× A ⊂ V .

Example 2.25Let (X, d) be a metric space. Let U be the uniform structure of (X, d). Let (xj)j≥1 be an infiniteCauchy sequence in X. Then F = {A ⊂ X : xj ∈ X − A for fintiely many j} is a Cauchy filter.

Proof. First, F is a filter on X by Example 2.13.

18 ZHENGYAO WU


1. For all V ∈ U, there exists δ > 0 such that{(x, y) ∈ X ×X : d(x, y) < ε} ⊂ V .

Definition 1.7

2. ∃n0, ∀m,n ≥ n0, d(xm, xn) < ε. (xn) is a Cauchy sequence in (X, d).

3. Define A = {xn : n ≥ n0}, A× A ⊂ V . If d(x, y) < ε, then (x, y) ∈ V .

4. A ∈ F. xj ∈ X − A iff 1 ≤ j < n0.

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Homework 2.26Let (X, d) be a metric space. Let U be the uniform structure of (X, d). Let F be a filter on X.Show that F is a Cauchy filter of (X,U) iff for all ε > 0 there exists A ∈ F such that for allx, y ∈ A, d(x, y) < ε.


3. September 23, Limit, completeness, interior, closure, cluster point, density

Definition 3.1Let X be a topological space and x ∈ X. Let F be a filter on X. We say that F converges to x(x is a limit point of F) if F ⊃ V(x).

Example 3.2Let X be a topological space and x ∈ X. Let (xn)n≥1 be an infinite sequence in X. By Exam-ple 2.13, F = {A ⊂ X : xn ∈ X − A for fintiely many n} is a filter. We have (xn) converges to xiff F converges to x.

Proof. The sequence (xn) converges to x, iff ∀V ∈ V(x), ∃n0, ∀n > n0, xn ∈ V , iff xn ∈ X − Vfor finitely many n for all V ∈ V(x), iff V(x) ⊂ F. �

Proposition 3.3Let (X,U) be a uniform space. Let F be a filter on X. If F is convergent, then F is a Cauchy filterof (X,U). In particular, for all x ∈ X, V(x) is a Cauchy filter.Proof.


1. X is a topological space. Homework 1.15 and Lemma 1.25

2. For all V ∈ U, there exists W ∈ U suchthat W ◦W ⊂ V .

Definition 1.10 (UIII).

3. There exists S ∈ U such that S = S−1

and S ⊂ W .Take S = W ∩ W−1 by Definition 1.10(UII) and (FI)

4. Suppose F converges to x. Then S(x) ∈ F. By Lemma 1.14, S(x) ∈ V(x) ⊂ F.

5. S(x)× S(x) ⊂ S ◦ S. For all y, z ∈ S(x), (y, x) ∈ S−1 = S and(x, z) ∈ S, so (y, z) ∈ S ◦ S.

6. F is a Cauchy filter of (X,U). S(x) × S(x) ⊂ S ◦ S ⊂ W ◦W ⊂ V bysteps 2-5.

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Example 3.4On the rational line Q, we have a sequence (xn) where xn =

n∑i=0

2−i(i+1)/2.

20 ZHENGYAO WU

We first show that (xn) is a Cauchy sequence. Suppose m > n,

|xm − xn| =m∑

i=n+12−i(i+1)/2 ≤

∞∑j=0

2−(n+1+j)(n+2+j)/2 = 2−n(n+3)/2∞∑j=0

2−(2jn+(1+j)(2+j))/2

≤ 2−n(n+3)/2∞∑j=0

2−1−2j = 2−n(n+3)/2 2−1

1− 2−2 < 2−n(n+3)/2.

Here we used 2jn+ (1 + j)(2 + j) ≥ (1 + j)(2 + j) = 2 + 3j + j2 ≥ 2 + 4j. For all 0 < ε < 1, taken0 = [

√−2 log2 ε] + 1 ≥

√−2 log2 ε, for all m > n > n0, |xm − xn| < 2−n0(n0+3)/2 < 2−n2

0/2 < ε.Next, we show that (xn) is divergent in Q. If lim

n→∞xn = a/b, (a, b ∈ Z, b > 0), then |a/b− xn| ≤

2−n(n+3)/2 for all n. Suppose xn = yn/2n(n+1)/2 where yn ∈ Z. Multiply b2n(n+1)/2 on both sides,we have |a2n(n+1)/2 − byn| ≤ b2−n. For all n > log2 b, b2−n < 1 and hence a2n(n+1)/2 − byn = 0.Therefore a/b = yn/2n(n+1)/2 = xn, a contradiction to the fact that xn is strictly increasing.

Homework 3.5Let (X, dX) and (Y, dY ) be metric spaces. Show that(1) A map f : X → Y is uniformly continuous iff for all ε > 0 there exists δ > 0 such that ifx1, x2 ∈ X and dX(x1, x2) < δ then dY (f(x1), f(x2)) < ε.(2) Suppose f : X → Y is a uniformly continuous map.If (x1, x2, . . .) is a Cauchy sequence in X, then (f(x1), f(x2), . . .) is a Cauchy sequence in Y .If F is a Cauchy filter of X, then f(F) = {V ⊂ Y : ∃U ∈ F, f(U) ⊂ V } is a Cauchy filter of Y .

Definition 3.6Let (X,U) be a uniform space. We call X complete if every Cauchy filter on X is convergent.By Example 3.2, if X is complete, then every Cauchy sequence in X is convergent.

Example 3.7Every discrete uniform space (X,U) is complete.

Proof. Suppose F is a Cauchy filter on X.


1. ∆X ∈ U. (X,U) is discrete and Example 1.12.

2. There exists A ∈ F such that A×A ⊂ ∆X . F is a Cauchy filter and Definition 2.24.

3. Suppose A = {x}. A 6= ∅ by Definition 2.10; If x 6= y in A,then (x, y) ∈ A× A−∆X .

4. For all V (x) ∈ V(x), A ⊂ V (x). Lemma 1.14(VIII)

5. V (x) ∈ F. F is a filter and Definition 2.10(FI).



6. V(x) ⊂ F, i.e. F converges to x. Definition 3.1.

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Definition 3.8Let X be a topological space and A ⊂ X. We call x ∈ X an interior point of A if A ∈ V(x).The interior of A is int(A) = {x ∈ A : A ∈ V(x)}.In particular, Proposition 2.7 can be rephrased as “A is open iff A = int(A)”.

Example 3.9Let R be the real line and a < b in R. Then int([a, b]) = (a, b).

Lemma 3.10Let X be a topological space and B ⊂ A ⊂ X. If B is open, then B ⊂ int(A).Proof.


1. For all b ∈ B, B ∈ V(b). Proposition 2.7.

2. A ∈ V(b). B ⊂ A and Lemma 2.8(VI).

3. b ∈ int(A). So B ⊂ int(A). Definition 3.8.

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Homework 3.11Let X be a topological space and A,B ⊂ X. Prove that(1) If B ⊂ A, then int(B) ⊂ int(A).(2) If A ∈ V(B), then B ⊂ int(A).(3) int(A ∩B) = int(A) ∩ int(B).

Definition 3.12Let X be a topological space and A ⊂ X.(1) If X − A is open, then we call A closed.(2) We call x a cluster point of A if for all V ∈ V(x), V ∩ A 6= ∅. The closure of A iscl(A) = {x ∈ X : ∀V ∈ V(x), V ∩ A 6= ∅}.

22 ZHENGYAO WU

Lemma 3.13Let X be a topological space and A ⊂ X.(1) If B is a closed subset of X such that A ⊂ B, then cl(A) ⊂ B.(2) A is closed iff A = cl(A).

Proof. (1)


1. X −B is open. B is closed and Definition 3.12(1)

2. For all x ∈ X −B, X −B ∈ V(x). Proposition 2.7

3. (X −B) ∩ A = ∅. A ⊂ B.

4. x ∈ X − cl(A). Definition 3.12(2).

5. cl(A) ⊂ B. X −B ⊂ X − cl(A) by steps 2-4.

(2) Suppose A is closed.


1. cl(A) ⊂ A. Take B = A in (1).

2. ∀a ∈ A, ∀V ∈ V(a), a ∈ V ∩ A. Lemma 2.8(VIII).

3. A ⊂ cl(A). Definition 3.12(2).

4. A = cl(A). Steps 1 and 3.

Conversely, suppose A = cl(A).


1. For all x ∈ X − A = X − cl(A), thereexists V ∈ V(x) such that V ∩ A = ∅.

Definition 3.12(2).

2. X − A is open. Proposition 2.7.

3. A is closed. Definition 3.12(1).


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Homework 3.14Let X be a topological space and A,B ⊂ X. Prove that(1) X − cl(A) = int(X − A), X − int(A) = cl(X − A).(2) If A ⊂ B, then cl(A) ⊂ cl(B).(3) cl(A ∪B) = cl(A) ∪ cl(B).(4) If A is open, then A ∩ cl(B) ⊂ cl(A ∩B).

Definition 3.15Let X be a topological space and A ⊂ X. We call A dense if cl(A) = X.

Lemma 3.16Let X be a topological space. A subset A ⊂ X is dense in X iff for all nonempty open subset Uof X, U ∩ A 6= ∅.

Proof. Suppose cl(A) = X and U is a nonempty open subset of X.


1. For all x ∈ U , U ∈ V(x). U is open and Proposition 2.7.

2. U ∩ A 6= ∅. x ∈ X = cl(A) and Definition 3.12(2).

Conversely, suppose for all nonempty open subset U of X, U ∩ A 6= ∅.


1. ∀x ∈ X, ∀V ∈ V(x), there exists an opensubset U of X such that x ∈ U ⊂ V .

Definition 2.6.

2. V ∩ A 6= ∅. V ∩ A ⊃ U ∩ A 6= ∅

3. X = cl(A). Definition 3.12(2)

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Homework 3.17The rational line Q is dense in the real line R.

24 ZHENGYAO WU

Example 3.18Let X 6= ∅ be a discrete topological space. The only dense subset of X is X itself because everysubset of X is closed.

Example 3.19Let X 6= ∅ be a trivial topological space. Every nonempty subset of X is dense in X because theonly nonempty closed subset of X is X itself.

Proposition 3.20Let (X,U) be a uniform space. Let S = {S ∈ U : S = S−1}. Let (X ×X,UX×X) be the productuniform space.(1) For all S ∈ S and for all M ⊂ X ×X, S ◦M ◦ S ∈ V(M) in X ×X.(2) cl(M) = ⋂

S∈SS ◦M ◦ S.

Proof. (1)


1. For all (x, y) ∈ X×X, (x, y) ∈ S◦M ◦S iffthere exists (p, q) ∈ M such that (x, p) ∈S and (q, y) ∈ S.

Review 1.8.

2. iff x ∈ S(p) and y ∈ S(q). Review 1.8 and S = S−1.

3. iff (x, y) ∈ S(p)× S(q) ∈ V((p, q)). S × S ∈ UX×X by Definition 1.23.

4. S ◦M ◦ S ∈ V(M). Definition 2.6 and (OI).

(2)


1. For all (x, y) ∈ X×X, (x, y) ∈ S◦M ◦S iffthere exists (p, q) ∈ M such that (x, p) ∈S and (q, y) ∈ S.

Review 1.8.

2. iff p ∈ S(x) and q ∈ S(y). Review 1.8 and S = S−1.

3. iff (p, q) ∈ S(x)× S(y) ∈ V((x, y)). S × S ∈ UX×X by Definition 1.23.

4. {S(x) × S(y) : S ∈ S} is a basis ofV((x, y)).

below



4.1. V(x)× V(y) is a basis of V((x, y)). Definition 1.23.

4.2. ∀U,U ′ ∈ U, ∃S ∈ S, S ⊂ U ∩ U ′. Definition 1.10(FII) and Example 2.23.

S(x)× S(y) ⊂ U(x)× U ′(y).

5. (x, y) ∈ cl(M) iff ∀S ∈ S, (S(x)×S(y))∩M 6= ∅ iff ∀S ∈ S, (x, y) ∈ S ◦M ◦ S.

Definition 3.12(2) and steps 1-3.

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Lemma 3.21Let (X,U) be a uniform space. The set {int(V ) : V ∈ U} is a basis of U as a filter.

Proof. Let S = {S ∈ U : S = S−1}.


1. ∀V ∈ U, ∃S ∈ S, S ⊂ V . Example 2.23.

∃S ′ ∈ S, S ′ ◦ S ′ ⊂ S. Definition 1.10(UIII).

∃W ∈ S, W ◦W ⊂ S ′. Definition 1.10(UIII).

2. W ◦W ◦W ⊂ V . W ◦W ◦W ⊂ W ◦W ◦W ◦W ⊂ S ′ ◦S ′ ⊂S ⊂ V .

3. W ◦W ◦W ∈ V(W ). Proposition 3.20(1).

4. W ⊂ int(V ). Step 2, 3 and Lemma 3.10.

5. {int(V ) : V ∈ U} is a basis of X. int(V ) ∈ U and int(V ) ⊂ V .

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Definition 3.22Let X be a topological space and x ∈ X. Let F be a set of some subsets of X. We say that x ∈ Xis a cluster point of F if x ∈ ⋂

A∈Fcl(A).

26 ZHENGYAO WU

Proposition 3.23Let X be a topological space. Let F be a filter on X. A point x ∈ X is a cluster point of F iffthere exists a filter F′ ⊃ F ∪ V(x).Proof.


1 x is a cluster point of F iff x ∈ ⋂A∈F

cl(A). Definition 3.22

2 iff ∀V ∈ V(x), ∀A ∈ F, V ∩ A 6= ∅. Definition 3.12

3 iff there exists a filter F′ ⊃ F ∪ V(x). F, V(x) are filters and Definition 2.14.

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Definition 3.24Let (X,U) be a uniform space. Let F be a Cauchy filter on X. We call F minimal if

{F′ is a Cauchy filter on X : F′ ⊂ F} = {F}.

Proposition 3.25Let (X,U) be a uniform space and S = {S ∈ U : S = S−1}. Let F be a Cauchy filter on X witha basis B. For S ∈ S and B ∈ B, let

S(B) = {y ∈ X : ∃x ∈ B ⊂ X, (x, y) ∈ S ⊂ X ×X}.

(1) The set {S(B) : S ∈ S, B ∈ B} is a basis of a filter F0.(2) F0 is the unique minimal Cauchy filter on X such that F0 ⊂ F.

Proof. (1) (BI) Suppose S1, S2 ∈ S and B1, B2 ∈ B.


1. ∃B3 ∈ B, B3 ⊂ B1 ∩B2. Lemma 2.15(BI) for B as basis of F.

2. ∃S3 ∈ S, S3 ⊂ S1 ∩ S2. Lemma 2.15(BI) for S as basis of U byExample 2.23.

3. S3(B3) ⊂ S1(B1) ∩ S2(B2). Def. of S(B).

(BII)



1.1. B 6= ∅. Lemma 2.15(BII) for B as basis of F.

1.2. S 6= ∅. Lemma 2.15(BII) for S as basis of U.

1.3. {S(B) : S ∈ S, B ∈ B} 6= ∅. Steps 1 and 2.

2.1 For all B ∈ B, B 6= ∅, ∃b ∈ B Lemma 2.15(BII) for B as basis of F.

2.2 For all S ∈ S, S 6= ∅. Lemma 2.15(BII) for S as basis of U.

2.3. Since ∆X ⊂ S, (b, b) ∈ S. S ∈ U and Definition 1.10(UI).

2.4 For all B ∈ B and S ∈ S, S(B) 6= ∅. b ∈ S(B) by steps 2.1 and 2.3.

Therefore {S(B) : S ∈ S, B ∈ B} is a basis of a filter F0 by Lemma 2.15.

28 ZHENGYAO WU

4. September 30, Minimal Cauchy filter, completion (1)

(2) F0 = {V ⊂ X : ∃S ∈ S, ∃B ∈ B, S(B) ⊂ V } by Lemma 2.15. It is a Cauchy filter.


1. ∀T ∈ S, ∃S ∈ S, S ◦ S ◦ S ⊂ T . Similar to steps 1, 2 of Lemma 3.21.

2. ∃A ∈ F, A× A ⊂ S. F is a Cauchy filter and Definition 2.24.

3. ∃B ∈ B, B ⊂ A. B is a basis of F and Definition 2.14.

4. B ×B ⊂ A× A ⊂ S. Steps 2 and 3.

5. S(B)× S(B) ⊂ S ◦ S ◦ S ⊂ T . Review 1.8.

6. F0 is a Cauchy filter. S is a basis of U and Definition 2.24.

Next, we show that F0 ⊂ F.


1. ∀V ∈ F0, ∃S ∈ S, ∃B ∈ B, S(B) ⊂ V . Def. of F0.


3. ∃B′ ∈ B, B′ ⊂ A. B is a basis of F and Definition 2.14.

4. ∃B′′ ∈ B, B′′ ⊂ B ∩B′. Lemma 2.15(BII)

5. B′′ ×B′′ ⊂ B′ ×B′ ⊂ A× A ⊂ S. Steps 2-4.

B′′ ⊂ S(B′′).

6. S(B′′) ⊂ S(B) ⊂ V . Steps 1 and 4.

7. B′′ ⊂ S(B′′) ⊂ V Steps 5 and 6.

8. V ∈ F. B′′ ∈ B and Definition 2.14.

Finally, we show that F0 is the smallest Cauchy filter such that F0 ⊂ F, i.e. if G is a Cauchy filtersuch that G ⊂ F, then F0 ⊂ G.



1. For all S ∈ S ⊂ U, there exists G ∈ Gsuch that G×G ⊂ S.

G is a Cauchy filter and Definition 2.24.

2. G ∈ F. G ⊂ F.

3. For all B ∈ B ⊂ F, G ∩B ∈ F. Definition 2.10(FII).

4. G ∩B 6= ∅. ∅ 6∈ F by Definition 2.10.

5. G ⊂ S(G ∩B) ⊂ S(B). (G ∩B)×G ⊂ G×G ⊂ S

6. S(B) ∈ G. Steps 2,5 and Definition 2.14.

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Homework 4.1Let (xn)n∈N be an infinite sequence in a setX. Let F = {A ⊂ X : xn ∈ X−A for finitely many n}.(1) For N ∈ N, let SN = {xn : n > N}. Show that B = {SN : N ∈ N} is a basis of the filter F.(2) Let (xnk

)k∈N be a subsequence of (xn)n∈N. LetG = {A ⊂ X : xnk∈ X−A for finitely many k}.

Show that G is finer than F.(3) Suppose X be a topological space and y ∈ X. Show that y is a cluster point of F iff thereexists a subsequence of (xn) whose limit is y.

Definition 4.2Let (X,U) be a uniform space and S = {S ∈ U : S = S−1}.

• Let X be the set of all minimal Cauchy filters on X.• For all S ∈ S, define S = {(F1,F2) ∈ X × X : ∃A ∈ F1 ∩ F2, A × A ⊂ S}. Let U be thefilter on X × X with basis B = {S : S ∈ S}.

We call (X, U) the completion of (X,U).

Proposition 4.3Let (X,U) be a uniform space. Its completion (X, U) is a uniform space.

Proof. (FI) and (FII) hold since U is a filter by Definition 4.2.(UI) For all U ∈ U, ∆

X⊂ U .


1. ∃S ∈ B, S ⊂ U B is a basis of U and Definition 2.14.

30 ZHENGYAO WU


2. ∆X

= {(F1,F1) : F1 ∈ X} ⊂ S. For all Cauchy filter F1, there exists A ∈F1 such that A×A ⊂ S by Definition 2.24.

3. ∆X⊂ S ⊂ U . Steps 1-2.

(UII) For all U ∈ U, U−1 ∈ U.


1. ∃S ∈ B, S ⊂ U B is a basis of U and Definition 2.14.

2. S = S−1 ⊂ U−1. Definition 4.2.

3. U−1 ∈ U. Definition 1.10(FI).

(UIII) For all U ∈ U, there exists Z ∈ U such that Z ◦ Z ⊂ U .


1. ∃S ∈ S, S ⊂ U U ∈ U.

2. There exists S ′ ∈ S such that S ′ ◦ S ′ ⊂ S. Lemma 2.16(UIII’) for S.

3. S ′ ◦ S ′ ⊂ S. below.

3.1. ∀(F1,F3) ∈ S ′ ◦ S ′, ∃F2 ∈ X such that(F1,F2) ∈ S ′ and (F2,F3) ∈ S ′.

Review 1.8

3.2. ∃M ∈ F1 ∩ F2 such that M ×M ⊂ S ′.

∃N ∈ F2 ∩ F3 such that N ×N ⊂ S ′. Definition 4.2.

3.3. M ∩N ∈ F2 and M ∩N 6= ∅. Definition 2.10(FII) and ∅ 6∈ F2.

3.4. (M ∪N)× (M ∪N) ∈ S ′ ◦ S ′ ⊂ S. For all x, y ∈ M ∪ N and z ∈ M ∩ N ,(x, z), (z, y) ∈ S ′ by step 3.2; and step 2.

3.5. M ∪N ∈ F1. M ⊂M ∪N , M ∈ F1 by step 3.2,

M ∪N ∈ F3. N ⊂M ∪N , N ∈ F3 by step 3.2,



M ∪N ∈ F1 ∩ F3. both by Definition 2.10(FI).

3.6. (F1,F3) ∈ S. Steps 3.4, 3.5 and Definition 4.2.

4. Take Z = S ′, Z ◦ Z ⊂ S ⊂ U . Steps 1 and 3.

Therefore, U is a uniform space on X by Definition 1.10. �

Proposition 4.4Let (X,U) be a uniform space. Its completion (X, U) is Hausdorff.

Proof. Suppose that (X, U) is not Hausdorff.


1. There exists F1 6= F2 ∈ X such that forall U1, U2 ∈ U, U1(F1) ∩ U2(F2) 6= ∅.

Definition 2.21(H).

2. ∀S ∈ B, ∃S ′ ∈ B, S ′ ◦ S ′ ⊂ S. B is a basis of U and Lemma 2.16(UIII’).

3. ∃F3 ∈ X, (F1,F3) ∈ S ′, (F3,F2) ∈ S ′. Steps 1, 2 and take U1 = U2 = S ′.

4. (F1,F2) ∈ S. (F1,F2) ∈ S ′ ◦ S ′ ⊂ S by steps 2 and 3.

5. ∃A ∈ F1 ∩ F2, A× A ⊂ S. Definition 4.2.

6. F4 = {P ⊂ X : ∃M ∈ F1, ∃N ∈ F2, M ∪N ⊂ P} is a filter on X.

Definition 2.10

7. F4 is a Cauchy filter on X. A ∈ F4 and Definition 2.24.

8. F1 ⊃ F4. M ∈ F1, M ⊂ P and Definition 2.10(FI).

F2 ⊃ F4. N ∈ F2, N ⊂ P and Definition 2.10(FI).

9. F1 = F4 = F2, a contradiction to step 1. Steps 8 and F1,F2 are minimal.

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32 ZHENGYAO WU

Lemma 4.5Let (X,U) be a uniform space. For x ∈ X, let V(x) = {V (x) : V ∈ U} be the neighborhood filterof x. Then V(x) is a minimal Cauchy filter.Proof.


1. F = {U ⊂ X : x ∈ U} is a filter on X. Definition 2.10.

2. F has basis B = {{x}}. Definition 2.14.

3. U has basis S = {S ∈ U : S = S−1}. Example 2.23.

4. The set {S(x) : S ∈ S} is a basis of V(x). Steps 2, 3 and Proposition 3.25(1).

5. V(x) ⊂ F is a minimal Cauchy filter. Proposition 3.25(2).

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Proposition 4.6Let (X,U) be a uniform space. Let (X, U) be its completion. Define i : X → X, x 7→ V(x).Suppose the filter generated by basis (i× i)−1(U) is F. Then U = F. In particular, i is uniformlycontinuous.

Proof. The map i is well-defined since V(x) is a minimal Cauchy filter by Lemma 4.5.


1. For all S ∈ S, (i× i)−1(S) ⊂ S. 1.1 and 1.2 below.

1.1. ∀(x, y) ∈ (i × i)−1(S), ∃A ∈ V(x) ∩V(y), A× A ⊂ S.

(i(x), i(y)) = (V(x),V(y)) ∈ S and Defi-nition 4.2.

1.2. (x, y) ∈ A× A ⊂ S. x ∈ A and y ∈ A by Homework 1.15(VIII).

2. For all S ∈ S, we have S ⊂ (i ×i)−1( S ◦ S ◦ S).

2.1. For all (x, y) ∈ S, (S(x)∪S(y))× (S(x)∪S(y)) ⊂ S ◦ S ◦ S.

Steps 2.1.1 to 2.1.4

2.1.1. For all z1, z2 ∈ S(x), (z1, z2) ∈ S ◦ S. (z1, x) ∈ S and (x, z2) ∈ S.

2.1.2. For all z1, z2 ∈ S(y), (z1, z2) ∈ S ◦ S. (z1, y) ∈ S and (y, z2) ∈ S.



2.1.3. For all z1 ∈ S(x), z2 ∈ S(y), (z1, z2) ∈S ◦ S ◦ S.

(z1, x) ∈ S, (x, y) ∈ S and (y, z2) ∈ S.

2.1.4. For all z1 ∈ S(y), z2 ∈ S(x), (z1, z2) ∈S ◦ S ◦ S.

(z1, y) ∈ S, (y, x) ∈ S and (x, z2) ∈ S.

2.2. S(x) ∪ S(y) ∈ V(x) ∩ V(y). Steps 2.2.1 and 2.2.2

2.2.1 S(x) ∪ S(y) ∈ V(x). S(x) ∈ V(x), S(x) ⊂ S(x)∪S(y) and (FI)

2.2.2 S(x) ∪ S(y) ∈ V(y). S(y) ∈ V(x), S(x) ⊂ S(x)∪S(y) and (FI)

2.3. (x, y) ∈ (i × i)−1( S ◦ S ◦ S),i.e. (i(x), i(y)) = (V(x),V(y)) ∈S ◦ S ◦ S.

Steps 2.1, 2.2 and Definition 4.2.

3. U ⊂ F. Step 1, B is a basis of U, S is a basis of U,and (FI) for F.

4.1. ∀T ∈ S, ∃S ∈ S, S ◦ S ◦ S ⊂ T . Similar to step 1, 2 of Lemma 3.21.

4.2. F ⊂ U. In particular, i is uniformly con-tinuous.

Step 2 with 4.1, B is a basis of U, S is abasis of U and (FI) for U.

5. F = U. Steps 3 and 4.2.

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Lemma 4.7Let (X,U) be a uniform space. Let F be a minimal Cauchy filter on X. For all A ∈ F, int(A) ∈ F.

Proof. Let S = {S ∈ U : S = S−1}. Let B be a basis of F.


1. For all S ∈ S ⊂ U, T = int(S) ∈ U. Lemma 3.21.

Furthermore, T ∈ S. T−1 = int(S−1) = int(S) = T .

2. For all B ∈ B and for all b ∈ B, T (b) ={x ∈ X : (b, x) ∈ T} is open.

For all x ∈ T (b), T (b) ∈ V(x) by Home-work 1.15 and T ∈ S. Then use Proposi-tion 2.7.

34 ZHENGYAO WU


3. For all B ∈ B, T (B) = {x ∈ X : ∃b ∈B, (b, x) ∈ T} is open in X.

Definition 2.1(OI).

4. For all A ∈ F, there exists S ∈ S andB ∈ B such that S(B) ⊂ A.

Proposition 3.25

5. T (B) ⊂ int(S(B)) ⊂ int(A). Steps 2 and S(B) ⊂ A and Lemma 3.10.

6. T (B) ∈ F. F is a minimal Cauchy filter and Proposi-tion 3.25.

7. int(A) ∈ F. Steps 5, 6 and Definition 2.10(FI).

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Proposition 4.8Let (X,U) be a uniform space with completion (X, U). Let i : X → X, x 7→ V(x). Then i(X) isdense in X.

Proof. Suppose F ∈ X, i.e. F is a minimal Cauchy filter. Suppose S ∈ B and S(F) = {F′ ∈X : ∃A ∈ F ∩ F′, A × A ⊂ S} is a neighborhood of F. We need to show that S(F) ∩ i(X) ={V(x) : x ∈ X, ∃A ∈ F ∩ V(x), A× A ⊂ S} 6= ∅.



2. int(A) ∈ F. Lemma 4.7.

3. int(A) 6= ∅. Definition 2.10.

4. S(F) ∩ i(X) 6= ∅. ∀x ∈ int(A), V(x) ∈ S(F) ∩ i(X).

5. i(X) is dense in X. Lemma 3.16.

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Lemma 4.9Let (X,U) be a uniform space. Let F be a Cauchy filter on X. If x ∈ X is a cluster point of F,then x is a limit point of F, i.e. F ⊃ V(x).


Proof.


1. There exists a filter F′ ⊃ F ∪ V(x). x is a cluster point of F and Proposi-tion 3.23.

2. F′ is a Cauchy filter. F is a Cauchy filter, F′ ⊃ F and Defini-tion 2.24.

3. ∃ a minimal Cauchy filter F0 ⊂ F. Proposition 3.25.

4. F0 = V(x) as minimal Cauchy filterscoarser than F′.

Uniqueness of Proposition 3.25.

5. F ⊃ V(x). Steps 3 and 4.

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Proposition 4.10Let (X,U) be a uniform space. Let Y ⊂ X be a dense subset. Suppose condition (?) is true.(?) “If F′ is a Cauchy filter on X and F′ has a basis consisting of some subsets of Y , then F′ isconvergent to an element of X. ”Then X is complete, i.e. every Cauchy filter on X is convergent to an element of X.Proof.


1. If every minimal Cauchy filter onX is con-vergent, then X is complete.

Proposition 3.25 and Definition 3.6.

2. Let F be a minimal Cauchy filter on X.For all A ∈ F, int(A) ∈ F.

Lemma 4.7.

3. int(A) 6= ∅. Definition 2.10.

4. Y ∩ A 6= ∅ for all A ∈ F. Y is dense in X and Lemma 3.16

5. FY = {Y ∩ A : A ∈ F} is a filter on Y . Definition 2.10.

6. FY is a Cauchy filter on Y . F is a Cauchy filter on X and Defini-tion 2.24.

36 ZHENGYAO WU


7. Let F′ be the filter of X with basis FY .Then F′ converges to some x0 ∈ X.

Condition (?).

8. F ⊂ F′. FY ⊂ F′ and Definition 2.10(FI).

9. x0 is a cluster point of F. Step 7, 8 and Proposition 3.23.

10. x0 is a limit point of F. Steps 9, F is Cauchy and Lemma 4.9.

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Proposition 4.11Let (X,U) be a uniform space. Its completion (X, U) is complete.

Proof. Let i : X → X, x 7→ V(x).


1. Let F be a Cauchy filter on i(X), let G bethe filter on X generated by basis i−1(F).Then G is a Cauchy filter.

below.

1.1. ∀S ∈ S, ∃A ∈ F, A× A ⊂ S. F is Cauchy and Definition 2.24.

1.2. (i× i)−1(S) ⊂ S. Step 1 of Proposition 4.6.

1.3. i−1(A)× i−1(A) ⊂ (i× i)−1(S) ⊂ S. Steps 1.1 and 1.2.

1.4. G is a Cauchy filter. i−1(A) ∈ G, step 1.3 and Definition 2.24.

2. There exists a unique minimal Cauchy fil-ter G0 on X such that G0 ⊂ G.

Proposition 3.25(2).

3. Let G0 be the filter on X whose basis isi(G0). Then G0 converges to G0 ∈ X.

below

3.1. For all S ∈ S, S(G0) ∩ i(X) = i(M) forsome M = ⋃{int(A) : A ∈ G0, A × A ⊂S} ∈ G0

int(A) ∈ G0 by Lemma 4.7, int(A) ⊂ M

and Definition 2.10(FI).

3.2. G0 ⊃ V(G0). ∀S ∈ S, S(G0) ⊃ i(M). So S(G0) ∈ G0.



4. i(G) = F. G ⊃ i−1(F) =⇒ i(G) ⊃ i(i−1(F)) = Fand Definition 2.10(FI) for F.

5. F converges to G0 in X. below

5.1. F = i(G) ⊃ i(G0) Step 2

5.2. F ⊃ G0 ⊃ V(G0) Step 3 and Definition 2.10(FI) for G0.

6. X is complete. Steps 5, Proposition 4.8 and Proposi-tion 4.10.

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38 ZHENGYAO WU

5. October 12, Continuity, Hausdorff, product space

Definition 5.1Let X and X ′ be topological spaces. A map f : X → X ′ is continuous at x0 ∈ X if for allneighborhood V ′ of f(x0), there exists a neighborhood V of x0 such that x ∈ V implies f(x) ∈ V ′.We call f continuous if f is continuous at every point of X.

Example 5.2The identity map IdX : X → X, x 7→ x is continuous.

Example 5.3Every constant map X → X ′, x 7→ c is continuous.

Example 5.4If X is a discrete topological space, then every map X → X ′ is continuous.

Proposition 5.5Let (X,U) and (X ′,U′) be uniform spaces. If f : X → X ′ is uniformly continuous, then it iscontinuous.

Proof. Suppose x ∈ X and y = f(x) ∈ X ′.


1. For all V ′ ∈ V(f(x)), V ′ = U ′(y), U ′ ∈ U′. Lemma 1.14.

2. There exists U ∈ U, (f × f)(U) ⊂ U ′. f is uniformly continuous and Defini-tion 1.19.

3. Let V = U(x). Then f(V ) ⊂ V ′. For all z ∈ V , i.e. (x, z) ∈ U . We have(f(x), f(z)) = (y, f(z)) ∈ U ′, i.e. f(z) ∈U ′(y) = V ′.

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Example 5.6Let R be the real line with the additive uniform structure. The map f : R → R, f(x) = x3 iscontinuous but not uniformly continuous.


First, we show that f is continuous. For all x0 ∈ R and for all ε > 0, take δ = inf(

1, ε

3x20 + 3|x0|+ 2

)

|x3 − x30| ≤ sup(|(x0 − δ)3 − x3

0|, |(x0 + δ)3 − x30|)

≤ sup(δ| − 3x20 + 3x0 − δ2|, δ|3x2

0 + 3x0 + δ2|)

≤ δ(3x20 + 3|x0|+ δ2) ≤ δ(3x2

0 + 3|x0|+ 1) < ε.

Next, we show that f is not uniformly continuous. For all δ > 0, take y = sup(δ, 16δ2 ) and x = y+δ,

we have |x3 − y3| = |(y + δ)3 − y3| = 3y2δ + 3yδ2 + δ3 > 3yδ(y + δ) ≥ 6δ2y > 1.

Homework 5.7Let f : X → X ′ be a map between topological spaces.(1) Prove that the following are equivalent:(1a) f is continuous at x0, i.e. ∀V ′ ∈ V(f(x0)), ∃V ∈ V(x0), f(V ) ⊂ V ′.(1b) ∀V ′ ∈ V(f(x0)), f−1(V ′) ∈ V(x0).(1c) Suppose S is a basis of V(f(x0)). ∀V ′ ∈ S, f−1(V ′) ∈ V(x0).(2) Prove that the following are equivalent:(2a) f is continuous, i.e. ∀x ∈ X, ∀V ′ ∈ V(f(x)), ∃V ∈ V(x), f(V ) ⊂ V ′.(2b) For all A ⊂ X, f(cl(A)) ⊂ cl(f(A)).(2c) For all closed subset F ′ of X ′, F = f−1(F ′) is a closed subset of X.(2d) For all open subset U ′ of X ′, U = f−1(U ′) is a open subset of X.

Proposition 5.8Let X be a topological space. The following are equivalent:(H) X is Hausdorff, i.e. ∀x 6= y ∈ X, ∃U ∈ V(x), ∃V ∈ V(y), U ∩ V = ∅.(H1) For all x ∈ X, {x} = C, where C = ⋂{U ∈ V(x) : U is closed}.(H4) If a filter F on X converges to x, then x is the only limit point of F.(H5) If a filter F on X converges to x, then x is the only cluster point of F.

Proof. The structure of the proof is (H) =⇒ (H1) =⇒ (H5) =⇒ (H4) =⇒ (H).(H) implies (H1).


1. x ∈ C. x ∈ U by Homework 1.15(VIII).

2. If there exists y ∈ C and x 6= y, ∃U ∈V(x), ∃V ∈ V(y), U ∩ V = ∅.

(H)

3. cl(U) ∈ V(x). U ∈ V(x) and Homework 1.15(VI).

40 ZHENGYAO WU


4. C ⊂ cl(U). cl(U) is closed by Lemma 3.13(2).

5. U ∩ V 6= ∅. y ∈ cl(U) and Definition 3.12(2).

6. A contradiction. Steps 2 and 5.

(H1) implies (H5).


1. x is a cluster point of F. F ⊃ V(x) and Proposition 3.23.

2. ∀y ∈ X−{x} = X−C, ∃U ∈ V(x) closedsuch that y 6∈ U .

(H1).

3. X − U ∈ V(y). y ∈ X − U is open and Proposition 2.7

4. y 6∈ cl(U). U ∩ (X − U) = ∅.

5. y is not a cluster point of F . U ∈ V(x) ⊂ F and Definition 3.22

(H5) implies (H4) since every limit point of F is a cluster point by Proposition 3.23.(H4) implies (H). Assume the opposite ∀U ∈ V(x), ∀V ∈ V(y), U ∩ V 6= ∅.


1. B = {U ∩ V : U ∈ V(x), V ∈ V(y)} is abasis of a filter F.

Lemma 2.15(BI)(BII)

2. F converges to both x and y. V(x) ∪ V(y) ⊂ B ⊂ F.

3. A contradiction. to (H4)

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Definition 5.9Let f : X → Y be a map between topological spaces. Let a ∈ X. Let F be the filter on Y generatedby basis f(V(a)) = {f(V ) : V ∈ V(a)}. If F ⊃ V(b) for some b ∈ Y , we say that lim

x→af(x) = b.

Proposition 5.10A map between topological spaces f : X → Y is continuous at a ∈ X iff lim

x→af(x) = f(a).

Proof. Let F be the filter on Y generated by basis f(V(a)) = {f(V ) : V ∈ V(a)}.


1. f is continuous at a ∈ X iff ∀V ′ ∈V(f(a)), ∃V ∈ V(a), f(V ) ⊂ V ′.

Definition 5.1

2. iff ∀V ′ ∈ V(f(a)), V ′ ∈ F. F has basis f(V(a)) and (FI).

3. iff limx→a

f(x) = f(a). V(f(a)) ⊂ F and Definition 5.9

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Lemma 5.11Let f : X → Y be a map between topological spaces. If Y is Hausdorff and lim

x→af(x) exists, then

limx→a

f(x) is unique.

Proof. Let F be the filter on Y generated by basis f(V(a)) = {f(V ) : V ∈ V(a)}.


1. The limit point of F is unique. Y is Hausdorff and Proposition 5.8(H4).

2. limx→a

f(x) is unique. Definition 5.9.

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Lemma 5.12Let (X,U) be a uniform space with completion (X, U). Let i : X → X, x 7→ V(x). Suppose(Y,UY ) is a complete Hausdorff uniform space and f : (X,U)→ (Y,UY ) is a uniformly continuousmap. There exists a unique map g0 : i(X)→ Y such that f = g0 ◦ i and g0 is uniformly continuous.Proof.

42 ZHENGYAO WU


1. f : X → Y is continuous. f is uniformly continuous and Proposi-tion 5.5.

2. For all a ∈ X, f(a) = limx→a

f(x). Proposition 5.10.

3. The map g0 : i(X) → Y , g0(i(a)) = f(a)is well-defined.

below.

3.1. If i(a) = i(b), then f(V(a)) = f(V(b)). V(a) = V(b).

3.2. limx→a

f(x) = limx→b

f(x). Y is Hausdorff and Lemma 5.11.

3.3. f(a) = f(b). Steps 2 and 3.2.

4. The map g0 is uniformly continuous. below.

4.1. ∀U ∈ UY , ∃S ∈ S, (f × f)(S) ⊂ U . f is uniformly continuous and Defini-tion 1.19.

4.2. (i× i)−1(S) ⊂ S. Step 1 of Proposition 4.6.

4.3. (g0 × g0)(S) = (f × f)((i × i)−1(S)) ⊂(f × f)(S) ⊂ U .

Steps 3, 4.1 and 4.2.

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Definition 5.13Let X be a set with two topologies T1 and T2. If T1 ⊃ T2, then we say that T1 is finer than T2

and T2 is coarser than T1. (Equivalently, Id : (X,T1)→ (X,T2) is continuous. )

Definition 5.14Let (Xi,Ti)i∈I be a family of topological spaces and X = ∏

i∈IXi. Suppose a topology TX on X

satisfies(1) ∀i ∈ I, pri : (X,TX)→ (Xi,Ti) is continuous.(2) If T′ is a topology on X and ∀i ∈ I, pri : (X,T′)→ (Xi,Ti) is continuous, then TX ⊂ T′.We call TX the product topology and (X,TX) the product space.

Lemma 5.15Let (X1,T1) and (X2,T2) be topological spaces. Let (X1 × X2,T) be their product space. ThenT = {⋃

i∈I(Ui × Vi) : Ui ∈ T1, Vi ∈ T2}.


Proof. Let T′ = {⋃i∈I

(Ui × Vi) : Ui ∈ T1, Vi ∈ T2}.(OI): If for all j ∈ J ,

⋃i∈I

(Uij × Vij) ∈ T′ where Uij ∈ T1 and Vij ∈ T2, then

⋃j∈J

(⋃i∈I

(Uij × Vij))

=⋃

(i,j)∈I×J(Uij × Vij) ∈ T′.

(OII): If J is finite and for all j ∈ J , ⋃i∈I

(Uij × Vij) ∈ T′ where Uij ∈ T1 and Vij ∈ T2, then

⋂j∈J

(⋃i∈I

(Uij × Vij))

=⋃i∈I

(⋂j∈J

Uij ×⋂j∈J

Vij) ∈ T′.


1. T′ is a topology on X1 ×X2. Definition 2.1.

2. The projection pr1 : (X1 × X2,T′) →(X1,T1) is continuous.

For all U ∈ T1, pr−11 (U) = U × X2 ∈ T′

and Homework 5.7(2d).

The projection pr2 : (X1 × X2,T′) →(X2,T2) is continuous.

For all V ∈ T2, pr−12 (V ) = X1 × V ∈ T′

and Homework 5.7(2d).

3. T ⊂ T′. T is the coarsest topology such that pr1

and pr2 are continuous.

4. For all U ∈ T1, pr−11 (U) = U ×X2 ∈ T. The projection pr1 : (X1 × X2,T) →

(X1,T1) is continuous and Home-work 5.7(2d).

For all V ∈ T2, pr−12 (V ) = X1 × V ∈ T. The projection pr2 : (X1 × X2,T′) →

(X2,T2) is continuous and Home-work 5.7(2d).

5. U × V = (U ×X2) ∩ (X1 × V ) ∈ T. Definition 2.1(OII) for T.

6. T′ ⊂ T. Definition 2.1(OI) for T.

7. T′ = T. Steps 3 and 6.

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Lemma 5.16Let (X1,T1) and (X2,T2) be topological spaces. Let (X1 ×X2,T) be their product space. For allA1] ⊂ X1 and A2 ⊂ X2, cl(A1 × A2) = cl(A1)× cl(A2).

44 ZHENGYAO WU

Proof.


1. pr1(cl(A1 × A2)) ⊂ cl(A1). pr1 is continuous and Homework 5.7(2b)

pr2(cl(A1 × A2)) ⊂ cl(A2). pr2 is continuous and Homework 5.7(2b)

2. cl(A1 × A2) ⊂ cl(A1)× cl(A1). Step 1.

3. For all (x1, x2) ∈ cl(A1) × cl(A2), thereexists V1 ∈ V(x1) and V2 ∈ V(x2) suchthat V1 ∩ A1 6= ∅ and V2 ∩ A2 6= ∅.

Definition 3.12(2).

4. (x1, x2) ∈ cl(A1 × A2). V1 × V2 ∈ V((x1, x2)), (V1 × V2) ∩ (A1 ×A2) 6= ∅ and Definition 3.12(2)

5. cl(A1)× cl(A1) ⊂ cl(A1 × A2). Steps 3 and 4.

6. cl(A1)× cl(A1) = cl(A1 × A2). Steps 2 and 5.

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Proposition 5.17Let X be a topological space. The following are equivalent:(H) X is Hausdorff, i.e. ∀x 6= y ∈ X, ∃U ∈ V(x), ∃V ∈ V(y), U ∩ V = ∅.(H2) The diagonal ∆X is closed in the product space X ×X.(H3) For all I, the diagonal ∆ = {(x)i∈I : x ∈ X} is closed in the product space XI = ∏

i∈IX.

Proof. The structure of the proof is (H) =⇒ (H3) =⇒ (H2) =⇒ (H).(H) implies (H3).


1. If x = (xi)i∈I ∈ XI −∆, then j 6= k in Isuch that xj 6= xk.

x 6∈ ∆.

2. ∃Uj ∈ V(xj), ∃Vk ∈ V(xk), Uj ∩ Vk = ∅. (H)

3. pr−1j (Uj) = Uj ×

∏i∈I, i6=j

X ∈ V(x). prj is continuous and Homework 5.7(2d).

pr−1j (Vk) = Vk ×

∏i∈I, i6=k

X ∈ V(x). prk is continuous and Homework 5.7(2d).



U = Uj × Vk ×∏

i∈I, i6=j,kX ∈ V(x). U = pr−1

j (Uj) ∩ pr−1k (Vk) is open by Defi-

nition 2.1(OII).

4. XI −∆ is open. U ∩∆ = ∅ by step 2; and Proposition 2.7.

5. ∆ is closed in XI . Definition 3.12(1).

(H3) implies (H2). Take I = {1, 2}.(H2) implies (H).


1. For all x 6= y in X, there exists U ∈V((x, y)) in X×X such that U ∩∆X = ∅.

(H2)

2. There exist V ∈ V(x) and W ∈ V(y) suchthat V ×W ⊂ U .

Lemma 5.15.

3. V ∩W = ∅. (V ×W ) ∩∆X = ∅ by steps 1 and 2.

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Corollary 5.18Let X be a topological space. Let Y be a Hausdorff space. Let f, g : X → Y be continuous maps.(1) The set {x ∈ X : f(x) = g(x)} is closed in X.(2) If {x ∈ X : f(x) = g(x)} is dense in X, then f = g.(3) The graph of f is closed in X × Y .

Proof. Let Z = {x ∈ X : f(x) = g(x)}.(1) Define h : X → Y × Y , h(x) = (f(x), g(x)).


1. For all open U ⊂ Y , f−1(U) is open in X. f is continuous and Homework 5.7(2d).

For all open V ⊂ Y , g−1(V ) is open in X. g is continuous and Homework 5.7(2d).

46 ZHENGYAO WU


2. h−1(U × V ) = f−1(U)∩ g−1(V ) is open inX.

Definition 2.1(OII).

3. h is continuous. Step 2 and Lemma 5.15.

4. ∆Y is closed in Y × Y . Y is Hausdorff and Proposition 5.17(H2).

5. Z = h−1(∆Y ) is closed in X. Step 3, 4 and Homework 5.7(2c).

(2)


1. Z is closed in X. (1).

2. Z = cl(Z). Lemma 3.13(2).

3. Z = X. cl(Z) = X by Definition 3.15.

(3) The graph of f : X → Y is G = {(x, f(x)) ∈ X × Y }.


1. pr2 : X × Y → Y is continuous. Definition 5.14.

2. f ◦ pr1 : X × Y pr1−−→ Xf−→ Y is continuous. f is continuous; and pr1 is continuous by

Definition 5.14.

3. G = {z ∈ X × Y : pr2(z) = f(pr1(z))}. If z = (x, y), then pr2(z) = f(pr1(z)) iffy = f(x).

4. G is closed in X × Y . (1).

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6. October 14, Regular, extension of maps, homeomorphism

Definition 6.1A topological space X is regular if it is Hausdorff and(OIII

′) For all F closed in X and for all x 6∈ F , there exists U ∈ V(F ) and V ∈ V(x) such thatU ∩ V = ∅.

Example 6.2Every metric space is regular.

Proof. For all z ∈ (X, d), let Va(z) = {y ∈ X : d(y, z) < a} denote the open ball with center zand radius a > 0. Suppose F is closed in X, x 6∈ F and d(x, F ) = inf(d(x, y) : y ∈ F ) = δ.


1. δ > 0. Otherwise below

1.1. x is a cluster point of F . ∀ε > 0, ∃y ∈ F, δ ≤ d(x, y) < δ + ε.

1.2. x ∈ F , a contradiction. F is closed and Definition 3.12(2).

2. Let U = ⋃z∈F

Vδ/2(z). Then U ∈ V(F ). U is open by Definition 2.1(OI).

3. Vδ/2(x) ∈ V(x). Vδ/2(x) is open.

4. U ∩ Vδ/2 = ∅. Otherwise d(x, z) ≤ d(x, y) + d(y, z) <

δ/2 + δ/2 = δ for some y ∈ U ∩ Vδ/2.

5. X is regular. Definition 6.1.

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Proposition 6.3A topological space X is (OIII

′) iff(OIII) For all x ∈ X, {Closed neighborhoods of x} is a basis of V(x).

Proof. Suppose X is regular. Let W be an open neighborhood of x ∈ X.


1. X −W is closed. W is open and Definition 3.12(1).

2. x 6∈ X −W . x ∈ W by Lemma 2.8(VIII).

48 ZHENGYAO WU


3. ∃U ∈ V(X −W ), ∃V ∈ V(x), U ∩V = ∅. X is regular and Definition 6.1.

4. cl(V ) ∈ V(x). V ∈ V(x) and Lemma 2.8(VI).

5. V ⊂ X − int(U). int(U) ⊂ U ⊂ X − V .

6. cl(V ) ⊂ X − int(U). X − int(U) is closed and Lemma 3.13(1)

7. X − int(U) ⊂ W . X −W ⊂ int(U) by U ∈ V(X −W ).

8. {Closed neighborhoods of x} is a basis ofV(x).

cl(V ) is a closed neighborhood of x con-tained in W by steps 4, 6 and 7.

Conversely, suppose {Closed neighborhoods of x} is a basis of V(x). Suppose F is closed in X andx 6∈ F .


1. X − F is a neighborhood of x. X − F is open and Proposition 2.7

2. There exists a closed neighborhood V of xsuch that V ⊂ X − F .

Given.

3. U = X − V ∈ V(F ). U is open and F ⊂ X − V

4. X is regular. Steps 2,3 and U ∩ V = ∅

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Theorem 6.4Let X be a topological space. Let A be a dense subset of X. Let Y be a regular space. A mapf : A→ Y extends uniquely to a continuous map f : X → Y iff for all x ∈ X, lim

y→xy∈A

f(y) exists in Y .

Proof. Suppose f : X → Y is continuous and extends f : A→ Y .


1. f(x) = limy→x

f(y). Proposition 5.10.



2. f(x) = limy→xy∈A

f(y) = limy→xy∈A

f(y). f extends f .

Conversely, suppose limy→xy∈A

f(y) exists. Define f(x) = limy→xy∈A

f(y).


1. f is well-defined. Y is Hausdorff by Definition 6.1, andProposition 5.8(H4).

2. ∀U ∈ V(f(x)), ∃V ∈ V(f(x)), V ⊂ U

and V is closed.Y satisfies (OIII) by Definition 6.1, andProposition 6.3

3. ∃W ∈ V(x), f(W ∩ A) ⊂ V . f(x) = limy→xy∈A

f(y) and Definition 5.9.

4. clY (f(W ∩ A)) ⊂ V . V is closed and Lemma 3.13(1)

5. W ∩ A is dense in W . A is dense in X.

6. For all z ∈ W , f(z) = limy→z

y∈W∩Af(y). Given.

7. f(z) ∈ cl(f(W ∩ A)). Definition 3.22

8. f(W ) ⊂ cl(f(W ∩ A)) ⊂ V ⊂ U . steps 2, 4 and 7.

9. f is continuous. Definition 5.1.

10. f is unique. A is dense in X and Corollary 5.18(2).

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Example 6.5By Definition 6.1, every regular space is Hausdorff. Conversely, not every Hausdorff space isregular. Let K = { 1

n: n ∈ Z>0}. Let T be a topology on R whose open sets are unions of open

intervals and elements of {(a, b)−K : a < b}. It is Hausdorff.Suppose K is closed in (R,T) and 0 6∈ K. Since lim

n→∞

1n

= 0, for all U ∈ V(0) and V ∈ V(K),U ∩ V 6= ∅. Hence (R,T) is not regular.

50 ZHENGYAO WU

Homework 6.6(1) Let X be a Hausdorff topological space. Show that for all x ∈ X, {x} is closed.(2) Let X, Y be a Hausdorff topological spaces. Show that the product space X ×Y is Hausdorff.

Lemma 6.7Let (X,U) be a uniform space. The set {cl(W ) : W ∈ S} is a basis of U as a filter.Proof.


1. For all U ∈ U, there exists W ∈ S suchthat W ◦W ◦W ⊂ U .

Similar to steps 1, 2 of Lemma 3.21.

2. cl(W ) = ⋂S∈S

S ◦W ◦S ⊂ W ◦W ◦W ⊂ U . Proposition 3.20.

3. {cl(W ) : W ∈ S} is a basis of U. Definition 2.10.

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Lemma 6.8Let (X,U) be a uniform space. Let A be a dense subset of X. The set {cl((A×A)∩U) : U ∈ U}is a basis of U as a filter.Proof.


1. A× A is dense in X ×X. Lemma 5.16

2. For an open U ∈ U, let V = U ∩ (A×A).We have U ⊂ cl(V ).

U = U∩(X×X) = U∩cl(A×A) ⊂ cl(U∩(A× A)) = cl(V ) by Homework 3.14(4).

3. cl(U) ⊂ cl(V ). Lemma 3.13.

4. cl(V ) ⊂ cl(U). V ⊂ U and Homework 3.14(2).

5. cl(V ) = cl(U). Steps 3 and 4.

6. {cl(U) : U ∈ U open} is a basis of U. Lemma 3.21 and Lemma 6.7.

7. {cl((A×A)∩U) : U ∈ U open} is a basisof U.

Steps 5 and 6.


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Definition 6.9Let X,X ′ be topological spaces. Let ϕ : X → X ′ be a map. We call ϕ a homeomorphism if ϕis a bijection and ϕ, ϕ−1 are continuous. We say that X and X ′ are homeomorphic and writeX ∼= X ′ (some authors use other symbols).

Example 6.10Let ϕ : X → X ′ be a bijection. IfX,X ′ are discrete topological spaces, then ϕ is a homeomorphism.

Example 6.11Let X be the discrete space Q. Let X ′ be the rational line Q. Then IdX : X → X ′ is a continuousbijection but not a homeomorphism. In fact, {0} is open in X but not in X ′.

Example 6.12A plane is homeomorphic to a punctured sphere.

Proof. Consider the plane R2 and the sphere S2 = {(x, y, z) ∈ R3 : x2 + y2 + z2 = 1}. The stere-ographic projection is ϕ : R2 → S2−{(0, 0, 1)} such that ϕ(x1, x2) is the intersection of S2 with

the line through (0, 0, 1) and (x1, x2, 0). The map ϕ(x1, x2) =(

2x1

x21 + x2

2 + 1 ,2x2

x21 + x2

2 + 1 ,x2

1 + x22 − 1

x21 + x2

2 + 1

)is continuous. Its inverse ϕ−1(x, y, z) =

(x

1− z ,y

1− z

), (z 6= 1) is also continuous. Therefore, ϕ

is a homeomorphism. �

Example 6.13A line segment is not homeomorphic to a circle.

Proof. Consider the line segment [0, 1] and the circle S1 = {(x, y) ∈ R2 : x2 + y2 = 1}.(1) For all continuous f : [0, 1]→ [0, 1], f has a fixed point x0. In fact,

• If f(0) = 0, the fixed point is x0 = 0;• If f(1) = 1, the fixed point is x0 = 1;• If f(0) > 0 and f(1) < 1, we have g(x) = f(x) − x is continuous on [0, 1], g(0) > 0 andg(1) < 0. By the intermediate value theorem, there exists x0 ∈ (0, 1) such that g(x0) = 0,i.e. f(x0) = x0.

(2) Suppose a : S1 → S1, a((x, y)) = (−x,−y). Then a does not have a fixed point. Otherewise−x = x, −y = y and hence x = y = 0, a contradiction to x2 + y2 = 1.(3) Suppose there is a homeomorphism ϕ : [0, 1]→ S1. By (1), f = ϕ−1 ◦ a ◦ ϕ is continuous witha fixed point x0. So a has a fixed point ϕ(x0), a contradiction to (2). �

Homework 6.14In R3 with the Euclidean topology, prove that the following are homeomorphic to each other:X1 is a punctured plane; X2 is cysubcoverr with height; X3 is a hyperboloid of one sheet.

52 ZHENGYAO WU

(Hint: We may assume that X1 = {(x1, x2) ∈ R2 : x1 6= 0 or x2 6= 0}; X2 = {(y1, y2, y3) ∈R3 : y2

1 + y22 = 1}; X3 = {(z1, z2, z3) ∈ R3 : z2

1 + z22 − z2

3 = 1}. )

Homework 6.15Suppose a < b in the real line R. Show that the open interval (a, b) is not homeomorphic to theclosed interval [a, b].

Lemma 6.16Every uniform space (X,U) satisfies(OIII): For all x ∈ X, closed neighborhoods of x form a basis of V(x).

Proof. Let B = {B ∈ U : B is closed}.


1. B is a basis of U. Lemma 6.7.

2. {B(x) : B ∈ B} is a basis of V(x). Definition 2.14.

3. j : {x}×X ↪→ X×X, B is closed inX×X,j−1(B) = {x}×B(x) is closed in {x}×X.

j is continuous and Homework 5.7(2c).

4. B(x) is closed in X. B(x) ∼= {x} ×B(x) and X ∼= {x} ×X.

5. (OIII) holds. Steps 2, 3.

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Corollary 6.17Every Hausdorff uniform space (X,U) is regular.

Proof. By Lemma 6.16, X satisfies (OIII). Since X is also Hausdorff, it is regular by Definition 6.1.�

Proposition 6.18Let (X,U) and (X ′,U′) be uniform spaces. Let f : X → X ′ be a uniformly continuous map. If Fis a Cauchy filter on X, then f(F) is a basis of a Cauchy filter on X ′.

Proof. Suppose F′ is the filter on generated by the basis f(F).



1. ∀V ′ ∈ U′, (f × f)−1(V ′) ∈ U. f is uniformly continuous and Defini-tion 1.19.

2. ∃A ∈ F, A× A ⊂ (f × f)−1(V ′). F is Cauchy and Definition 2.24.

3. f(A) ∈ F′. A ∈ F and f(F) ⊂ F′.

4. f(A)× f(A) ⊂ V ′. So F′ is Cauchy. Step 2, 3 and Definition 2.24.

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Proposition 6.19Let X be a topological space. Let A be a dense subset of X. Let X ′ be a complete, Hausdorffuniform space. A map f : A → X ′ extends to a continuous map f : X → X ′ iff for all x ∈ X,f(A ∩ V(x)) = {f(A ∩ V ) : V ∈ V(x)} is a basis of a Cauchy filter on X ′.Proof.


1. X ′ is a regular space. X ′ is a Hausdorff, uniform space andCorollary 6.17.

2. f extends to a continuous f iff for all x ∈X, lim

y→xy∈A

f(y) exists in X ′.Theorem 6.4.

3. iff f(A ∩ V(a)) is a basis of a convergentfilter F on X ′.

Definition 5.9.

4. If F is convergent, then it is Cauchy. Proposition 3.3.

5. If F is Cauchy, then it is convergent. X ′ is complete and Definition 3.6.

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Theorem 6.20Let (X,U) be a uniform space. Let A be a dense subset of X. Let (X ′,U′) be a complete Hausdorffuniform space. If f : A → X ′ is uniformly continuous, then it extends to a uniformly continuousmap f : X → X ′.

54 ZHENGYAO WU

Proof. First, we show that f extends to a continuous map f : X → X ′.


1. ∀x ∈ X, V(x) is a Cauchy filter on X. Proposition 3.3.

2. A ∩ V 6= ∅ for all V ∈ V(x). A is dense in X and Definition 3.15.

3. A ∩ V(x) is a Cauchy filter on A. Definition 2.10 and Definition 2.24.

4. f(A ∩ V(x)) is a basis of a Cauchy filteron X ′.

f is continuous and Proposition 6.18.

5. f extends to f . Proposition 6.19.

Next, we show that f is uniformly continuous. Let S′ = {V ′ ∈ U′ : V ′ = V ′−1}.


1. For all closed V ′ ∈ S, there exists V ∈ Usuch that (f × f)((A× A) ∩ V ) ⊂ V ′.

f is uniformly continuous and Defini-tion 1.19.

2. We may assume that V = cl(W ) inX×X,where W ⊂ A× A.

Lemma 6.8.

3. (f × f)(W ) ⊂ V ′. Steps 1, 2.

4. f × f is continuous. f is continuous and Homework 5.7(2d)

5 (f × f)(V ) = (f × f)(cl(W )) ⊂ cl((f ×f)(W )) = cl((f × f)(W )) ⊂ cl(V ′) = V ′.

V ′ is closed and Homework 5.7(2b).

6. f is uniformly continuous. Lemma 6.7 and Definition 1.19.

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7. October 21, Completion(2),

Proposition 7.1 The universal property for completionLet (X,U) be a uniform space with completion (X, U). Let i : X → X, x 7→ V(x). We have(P) If (Y,UY ) is a complete Hausdorff uniform space and f : (X,U) → (Y,UY ) is a uniformlycontinuous map, then there exists a unique map g : X → Y such that f = g ◦ i and g is uniformlycontinuous.Proof.


1. There exists a unique uniformly continu-ous g0 : i(X)→ Y such that f = g0 ◦ i.

Lemma 5.12.

2. i(X) is dense in X. Proposition 4.8

3. g0 extends to a uniformly continuousg : X → Y .

Step 3 and Theorem 6.20.

3. f = g ◦ i. f = g0 ◦ i and g extends g0.

4. g is unique. g extends g0, i(X) is dense in X by Propo-sition 4.8 and Corollary 5.18(2).

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Definition 7.2Let (X,U) and (X ′,U′) be uniform spaces. A map f : X → X ′ is called an isomorphism ofuniform spaces if f is a bijection and (f × f)(U) = U′ (iff f and f−1 are uniformly continuous).

Proposition 7.3Let (X,U) be a uniform space with completion (X, U). Let i : X → X, x 7→ V(x). Then (i, X, U)is unique up to isomorphism of uniform spaces.

Proof. Suppose (i1, X1,U1) is another triple.


1. There exists a unique uniformly continu-ous ϕ : X → X1 such that ϕ ◦ i = i1.

(P) for (i, X, U) by Theorem 6.20.

56 ZHENGYAO WU


2. There exists a unique uniformly continu-ous ψ : X1 → X such that ψ ◦ i1 = i.

(P) for (i1, X1,U1) by Theorem 6.20.

3. ψ ◦ ϕ ◦ i = i. Steps 1 and 2.

4. ψ ◦ ϕ = IdX . Similarly, ϕ ◦ ψ = IdX1 Corollary 5.18(2).

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Proposition 7.4A uniform space (X,U) is Hausdorff iff ⋂

V ∈UV = ∆X .

Proof. Let B = {B ∈ U : B is closed}.


1. ⋂B∈B

B = ⋂V ∈U

V . below.

1.1. ⋂B∈B

B ⊃ ⋂V ∈U

V . B ⊂ U.

1.2. ⋂B∈B

B ⊂ ⋂V ∈U

V . ∀V ∈ U, ∃B ∈ B, B ⊂ V .

Suppose ⋂V ∈U

V = ∆X .


2. ⋂B∈B

B is closed. B is closed for all B ∈ B, Defini-tion 3.12(1) and Definition 2.1(OI).

3. ∆X is closed. ∆X = ⋂V ∈U

V = ⋂B∈B

B.

4. X is Hausdorff. Proposition 5.17 (H)(H2)

Conversely, suppose X is Hausdorff.



5. ∆X is closed. Proposition 5.17(H)(H2)

6. ⋂B∈B

B ⊂ ∆X . ∆X ∈ B.

7. ∆X ⊂⋂B∈U

B. Definition 1.10(UI).

8. ∆X = ⋂B∈U

B = ⋂V ∈U

V . Steps 1, 6 and 7.

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Lemma 7.5Let X1, X2 be complete Hausdorff uniform spaces. Let Y1 ⊂ X1, Y2 ⊂ X2 be dense subsets.Every isomorphism of uniform spaces f : Y1 → Y2 extends to an isomorphism of uniform spacesf : X1 → X2.

Proof. Suppose g : Y2 → Y1 is the inverse of f .


1. f extends to a uniformly continuousf : X1 → X2.

f is uniformly continuous and Theo-rem 6.20.

g extends to a uniformly continuousg : X2 → X1.

g is uniformly continuous and Theo-rem 6.20.

2. f ◦ g = IdX2 . (f ◦ g)|Y2 = f ◦ g = IdY2 , Y2 is dense in X2

and Corollary 5.18(2)

g ◦ f = IdX1 . (g ◦ f)|Y1 = g ◦ f = IdY1 , Y1 is dense in X1

and Corollary 5.18(2)

3. f is an isomorphism of uniform spaces. f−1 = g by steps 1 and 2.

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Proposition 7.6Let (X,U) be a complete Hausdorff uniform space. Let A be a dense subset of X. If U′ is auniform structure on X, U′ is coarser than U and U′ ∩ (A× A) = U ∩ (A× A), then U′ = U.

58 ZHENGYAO WU

Proof. Let (X ′, U′) be the completion of (X,U′). Let i′ : (X,U′)→ (X ′, U′) be the canonial map.


1. The composition ϕ = i′ ◦ IdX : (X,U) →(X ′, U′) is uniformly continuous.

Proposition 4.6 and Example 1.20(1).

2. (A,U′ ∩ (A × A)) = (A,U ∩ (A × A)) isHausdorff.

(X,U) is Hausdorff and U′ ∩ (A × A) =U ∩ (A× A).

3. ϕ|A and i′|A : A → ϕ(A) are isomor-phisms.

ϕ|A, i′|A is injective by step 2.

4. ϕ : (X,U) → (X ′, U′) and i′ : (X,U′) →(X,U′) are isomorphisms.

A is dense in X, ϕ(A) = i′(X) is dense inX ′ by Proposition 4.8, and Lemma 7.5.

5. IdX : (X,U)→ (X,U′) is an isomorphism.So U = U′.

i′ : (X,U′) → (X ′, U′) is an isomorphismby ϕ = i′ ◦ IdX and step 4.

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Theorem 7.7Let (X, d) be a metric space with uniform structure U. Let (X, U) be the completion of (X,U).Then d : X ×X → [0,+∞) extends uniquely to a metric d : X × X → [0,+∞) such that U is theuniform structure of the metric space (X, d).Proof.


1. d : X ×X → [0,+∞) is uniformly contin-uous.

Lemma 1.24.

2. i : X → X is injective. X is Hausdorff by Lemma 2.22.

3. X is dense in X. i(X) is dense in X by Proposition 4.8.

4. X ×X is dense in X × X. Lemma 5.16.

5. d extends to d : X × X → [0,+∞) and dis uniformly continuous.

Theorem 6.20.



6. d(a, b) = lim(x,y)→(a,b)(x,y)∈X×X

d(x, y) for all (a, b) ∈

X × X.

Theorem 6.4.

Next, we show that d is a metric.(ECI). Suppose d(a, b) = lim

x→a, y→bx∈X, y∈X

d(x, y) = 0.


1. For all U ∈ U, there exists ε > 0 suchthat Sε ◦ Sε ◦ Sε ⊂ U . Here Sε = {(x, y) ∈X ×X : d(x, y) < ε}.

U is the uniform structure of (X, d),{S : S ∈ U, S = S−1} is a basis of Uand similar to steps 1, 2 of Lemma 3.21.

2. ∃S ∈ U, ∀x ∈ S(a) ∩ X, ∀y ∈ S(b) ∩X, d(x, y) < ε.

Definition 5.9 and d(a, b) =lim

x→a, y→bx∈X, y∈X

d(x, y) = 0

3. Replace S with T = Sε ∩ S. Then T 6= ∅. Definition 1.10(FII).

4. (a, b) ∈ U for all U ∈ U. (a, b) ∈ T ◦ Sε ◦ T ⊂ Sε ◦ Sε ◦ Sε ⊂ U .

5. X is Hausdorff. Proposition 4.4.

6. (a, b) ∈ ∆X. So a = b. Proposition 7.4.

Next, we show that d(a, a) = limx→a, y→ax∈X, y∈X

d(x, y) = 0. In fact, for all Sε and x, y ∈ Sε(a) ∩ X,

(i(x), a) ∈ S, (a, i(y)) ∈ S, so (i(x), i(y)) ∈ Sε ◦ Sε ⊂ S2ε by Proposition 4.3(UIII), step 3. Finally(x, y) ∈ S2ε by Proposition 4.6 step 1. Therefore d(x, y) < 2ε.(ECII). d(a, b) = lim

(x,y)→(a,b)(x,y)∈X×X

d(x, y) = lim(x,y)→(a,b)(x,y)∈X×X

d(y, x) = d(b, a) for all a, b ∈ X.

(ECIII). d(a, b) + d(b, c) = limx→a, y→bx∈X, y∈X

d(x, y) + limy′→b, z→cy′∈X, z∈X

d(y′, z).

Since limy→b, y′→by∈X, y′∈X

d(y, y′) = d(b, b) = 0, the above

= limx→a, y,y′→b, z→cx∈X, y,y′∈X, z∈X

d(x, y) + d(y, y′) + d(y′, z) ≥ limx→a, z→cx∈X, z∈X

d(x, z) = d(a, c) for all a, b, c ∈ X.

So, d is a metric by Definition 1.1.

60 ZHENGYAO WU

Let U′ be the uniform structure of (X, d). We show that U′ = U.


1. d : X×X → [0,+∞) is uniformly continu-ous, where X×X has the product uniformstructure of U and U.

Step 5 of the first part.

2. For all ε > 0, there exists V, V ′ ∈ U suchthat (d× d)(V ×V ′) ⊂ {(a, b) ∈ [0,+∞)×[0,+∞) : |a− b| < ε}.

Definition 1.19.

3. Take (x, y) ∈ V and (x, x) ∈ V ′, |d(x, y)−d(x, x)| < ε.

∆X⊂ V ′ by Definition 1.10(UI).

4. |d(x, y)| < ε. So U ⊂ U′. d(x, x) = 0 by (ECI).

5. U has basis (i× i)−1(U). Proposition 4.6.

6. U ∩ (X ×X) = U. i is injective since X is Hausdorff byLemma 2.22.

7. U′ ∩ (X ×X) = U. d extends d.

8. U′ = U. Proposition 7.6.

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Homework 7.8Let Q be the set of rational numbers. Let U be the uniform structure on Q of the Euclideanmetric. Define R = (Q, U). Let T be the topology of R provided by U.(1) Show that:(1a) The addition map f : Q×Q→ Q, f(x, y) = x+y extends uniquely to a uniformly continuousmap f : R ×R → R.(1b) The map g : Q→ Q, g(x) = −x extends uniquely to a uniformly continuous map g : R → R.(1c) The map h : Q → Q, h(x) = sup(x, 0) extends uniquely to a uniformly continuous maph : R → R.(2) Let Q+ = {x ∈ Q : x ≥ 0} and cl(Q+) its closure in (R,T). For all x ∈ R and y ∈ R, wedefine x ≤ y if y − x ∈ cl(Q+). In R, show that(2a) If x ≤ y and y ≤ z, then x ≤ z. (Hint: use (1a))(2b) If x ≤ y and y ≤ x, then x = y. (Hint: use (1c))


(2c) Either x ≤ y or y ≤ x.(2d) x ≤ y iff x+ z ≤ y + z.(2e) Suppose x, y ∈ Q. If x ≤ y in R, then x ≤ y in Q.(3) Show that(3a) For all a ∈ R, (−∞, a] and [a,+∞) are closed in (R,T).(3b) For all a ∈ R, (−∞, a) and (a,+∞) are open in (R,T).(3c) For all a ≤ b in R, [a, b] is closed in (R,T).(3d) For all a < b in R, (a, b) is open in (R,T).(4) Show that(4a) For all q ∈ Q+, [−q, q] = cl([−q, q] ∩Q) in (R,T).(4b) T = TE, where TE is the Euclidean topology on R.

Homework 7.9Let X be a nonempty set and R ⊂ X×X. If ∆X ⊂ R, R−1 ⊂ R and R◦R ⊂ R, then we call R anequivalence relation on X. Suppose A ⊂ X ×X such that ∆X ⊂ A and A−1 ⊂ A. Constructan equivalence relation R on X such that A ⊂ R.

Proposition 7.10Let (X,U) be a Hausdorff uniform space. If U as a filter has a countable basis, then there existsa metric d on X such that U is the uniform structure of the metric space (X, d).

Proof. Suppose (Vn)n=1,2,... is a countable basis of U.


1. There exists (Un)n=1,2,... such that Un ∈ S,U1 ⊂ V1 and Un+1 ◦Un+1 ◦Un+1 ⊂ Un ∩Vnfor all n ≥ 1.

below.

1.1. There exists U1 ∈ S such that U1 ⊂ V1. S is a basis of U by Example 2.23.

1.2. Suppose Un is constructed. There existsUn+1 ∈ S such that Un+1 ◦ Un+1 ◦ Un+1 ⊂Un ∩ Vn.

Un ∩ Vn ∈ U by Definition 1.10(FII), andsimilar to steps 1, 2 of Lemma 3.21.

2. (Un)n=1,2,... is a basis of U such that Un+1◦Un+1 ◦ Un+1 ⊂ Un for all n ≥ 1.

Un ⊂ Vn for all n ≥ 1, (Vn)n=1,2,... is abasis of U and Definition 2.14.

62 ZHENGYAO WU

Define g : X ×X → [0,+∞) such that

g(x, y) =

0, (x, y) ∈∞⋂n=1

Un;

2−k, (x, y) ∈k⋂

n=1Un − Uk+1;

1, (x, y) ∈ X ×X − U1.


3. g(x, y) = 0 iff x = y.∞⋂n=1

Un = ⋂U∈U

U = ∆X since (X,U) isHausdorff and Proposition 7.4.

4. g(x, y) = g(y, x). Un ∈ S for all n ≥ 1.

Define d : X ×X → [0,+∞) such that

d(x, y) = inf(p−1∑i=0

g(zi, zi+1) : x = z0, z1, . . . , zp−1, zp = y).


5. d(x, y) ≤ g(x, y). Def. of d.

6. d(x, x) = 0 for all x ∈ X. d(x, x) ≤ g(x, x) = 0.

7. For all x = z0, z1, . . . , zp−1, zp = y,p−1∑i=0

g(zi, zi+1) ≥ 12g(x, y).

below.

7.1. If p = 1, then g(z0, z1) = g(x, y) ≥12g(x, y).

z0 = x, z1 = y and g(x, y) ≥ 0.

7.2. Let a =p−1∑i=0

g(zi, zi+1). If a ≥ 12 , then the

inequality holds.g(x, y) ≤ 1.

7.3. When a <12 , there exists k ≥ 2, k ∈ Z

such that 2−k ≤ a < 2−(k−1).2−k is decreasing.

7.3.1 For h = sup(q :q−1∑i=0

g(zi, zi+1) ≤ a

2),h−1∑i=0

g(zi, zi+1) ≤ a

2 .

Def. of h.

7.3.2.p−1∑i=h+1

g(zi, zi+1) ≤ a

2 .h∑i=0

g(zi, zi+1) > a

2 by def. of h.



7.3.3. g(x, zh) ≤ a. So (x, zh) ∈ Uk. Step 7.3.1 and inductive hypothesis.

7.3.4. g(zh+1, y) ≤ a. So (zh+1, y) ∈ Uk. Step 7.3.2 and inductive hypothesis.

7.3.5. g(zh, zh+1) ≤ a. So (zh, zh+1) ∈ Uk. Step 7.2 and g ≥ 0.

7.3.6. (x, y) ∈ Uk ◦ Uk ◦ Uk ⊂ Uk−1. Steps 2, 7.3.3, 7.3.4 and 7.3.5

7.3.7. g(x, y) ≤ 2−(k−1) ≤ 2a. Step 7.3 and def. of g.

8. d(x, y) ≥ 12g(x, y). Step 7 and def. of d.

To be continued. �

64 ZHENGYAO WU

8. October 28, Uniformizable, completely regular, compact(1), subspace


9. If d(x, y) = 0, then g(x, y) = 0 and x = y. Steps 3 and 8.

10. d is a metric. Definition 1.1 and below.

10.1. (ECI) d(x, y) = 0 iff x = y. Steps 6 and 9.

10.2. (ECII) d(x, y) = d(y, x). Step 4.

10.3. (ECIII) d(x, z) ≤ d(x, y) + d(y, z). For all x = z′0, z′1, . . . , z

′s = y and y =

z′′0 , z′′1 , . . . , z

′′t = z, let zi = z′i for all 0 ≤

i ≤ s; let zs+j = z′′j for all 1 ≤ j ≤ t andp = s+ t, we have x = z0, z1, . . . , zp = z.

11. For all a > 0, Uk ⊂ d−1([0, a]) for all 2−k <a.

Step 5.

12. For all k ≥ 1, k ∈ Z, d−1([0, 2−(k+1))) ⊂Uk.

Step 8.

13. {d−1([0, a)) : a > 0} is a basis of U. Steps 2, 11 and 12.

Theorem 8.1Let (X,U) be a Hausdorff uniform space. There exists a family of metrics (di)i∈I on X such that

∀V ∈ U, ∃i ∈ I, ∃ε > 0, {(x, y) ∈ X ×X : di(x, y) < ε} ⊂ V.

Proof.


1. For all V ∈ U, there exists (Un)n=1,2,...

such that Un ∈ S, U1 ⊂ V and Un+1 ◦Un+1 ⊂ Un for all n ≥ 1.

Similar to step 1 of Proposition 7.10.

2. There exists a filter UV with basis(Un)n=1,2,..., and UV is a uniform structure.

Definition 1.10.

3. U = ⋃V ∈U

UV . UV ⊂ U and V ∈ UV .



4. UV is the uniform structure of a metricspace (X, dV ).

Proposition 7.10.

5. ∀V ∈ U, ∃ε > 0, {(x, y) ∈ X ×X : dV (x, y) < ε} ⊂ V .

Steps 3, 4 and Definition 1.7.

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Definition 8.2We call a topological space (X,T) uniformizable if there exists a uniform structure U on X suchthat for all x ∈ X, and for all A ⊂ X, A ∈ V(x) iff there exists U ∈ U such that A = U(x), hereU(x) = {y ∈ X : (x, y) ∈ U}.

Example 8.3Every uniform space is uniformizable. Conversely, if a topological space does not satisfy (OIII),then it is not uniformizable by Lemma 6.16.

Theorem 8.4A topological space (X,T) is uniformizable iff(OIV) For all x0 ∈ X and for all V ∈ V(x0), there exists a continuous function f : X → [0, 1] suchthat f(x0) = 0 and f(X − V ) = {1}.

Proof. Suppose X is uniformizable and U is a uniform structure on X compatible with T.


1. There exists U ∈ U such that U(x0) = V .

1. There exists a family of metrics (di)i∈I onX such that ∃i ∈ I, ∃ε > 0, {(x, y) ∈X ×X : di(x, y) < ε} ⊂ U .

Theorem 8.1.

2. X − V ⊂ {x ∈ X : di(x0, x) ≥ ε}. X×X−U ⊂ {(x, y) ∈ X×X : di(x, y) ≥ε}.

3. Define f(x) = inf(1, 1εdi(x0, x)).

3.1. f(x0) = inf(1, 1εdi(x0, x0)) = inf(1, 0) = 0. Definition 1.1(ECI).

3.2. ∀x ∈ X−V, f(x) = inf(1, 1εdi(x0, x)) = 1. Step 2.

66 ZHENGYAO WU

Conversely, suppose (OIV) holds. Let Φ be the set of all continuous functions X → [0, 1]. For allf ∈ Φ and for all ε > 0, let Uf,ε = {(x, y) ∈ X ×X : |f(x) − f(y)| < ε}. Let U be the uniformstructure on X with a basis consisting of finite intersections of Uf,ε. Let T′ be the topology of U.


1. For all V ∈ T and for all x0 ∈ V , V ∈V(x0).

Proposition 2.7.

2. There exists f ∈ Φ such that f(x0) = 0and f(X − V ) = {1}.

(OIV).

3. For all z ∈ Uf,1/2(x0), |f(z)| < 12 . (x0, z) ∈ Uf,1/2 so that |f(z)| = |f(x0) −

f(z)| < 12 by f(x0) = 0.

4. Uf,1/2(x0) ⊂ V . Uf,1/2(x0) ∩ (X − V ) = ∅ by step 3 andf(X − V ) = {1}.

5. T ⊂ T′. V ∈ T′ by Proposition 2.7.

6. Conversely, suppose U =n⋂k=1

Ufk,εk. For

all x ∈ X, U(x) is open in (X,T′).Homework 1.15 and Lemma 1.25.

7. U(x) =n⋂k=1{y ∈ X : |fk(x) − fk(y)| <

εk} =n⋂k=1

f−1k ((fk(x) − εk, fk(x) + εk)) is

open in (X,T).

(OII) and Homework 5.7(2d).

8. T′ ⊂ T. Steps 6 and 7.

9. T′ = T. Steps 5 and 8.

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Definition 8.5A topological space is completely regular if it is Hausdorff and uniformizable.

Proposition 8.6(OIV) implies (OIII): For all x0 ∈ X, {Closed neighborhoods of x0} is a basis of V(x0).Proof.



1. There exists a continuous f : X → [0, 1]such that f(x0) = 0 and f(X −V ) = {1}.

(OIV)

2. Let F = f−1([0, 12 ]).

2.1. F ∈ V(x0). 0 ∈ f(F ).

2.2. F ⊂ V . 1 6∈ f(F ).

2.3. F is closed. f is continuous and Homework 5.7(2c).

3. (OIII) holds. Steps 2.2, 2.3 and Definition 2.14.

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Corollary 8.7Every completely regular space X is regular.Proof.


1. X is Hausdorff and uniformizable. Definition 8.5.

2. X is Hausdorff and (OIV). Uniformizable iff (OIV) by Theorem 8.4.

3. X is Hausdorff and (OIII). (OIV) implies (OIII) by Proposition 8.6.

4. X is Hausdorff and (OIII′). (OIII) iff (OIII

′) by Proposition 6.3.

5. X is regular. Definition 6.1.

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Remark 8.8The “Deleted Tychonoff Corkscrew” is a regular topological space which is not completely regular.Search it if you like.

Definition 8.9Let X be a topological space.(C) We call X quasi-compact if every filter on X has at least one cluster point.

68 ZHENGYAO WU

We call X compact if it is Hausdorff and quasi-compact.

Proposition 8.10Let X be a topological space. The following are equivalent:(C) X is quasi-compact, i.e. every filter on X has at least one cluster point.(C1) Every maximal (up to inclusion) filter on X is convergent.(C2) Let (Fi)i∈I be a family of closed subsets of X. If ⋂

i∈IFi = ∅, then there exists a finite J ⊂ I

such that ⋂i∈J

Fi = ∅.(C3) (Borel-Lebesgue) For every open cover (Ui)i∈I of X, there exists a finite J ⊂ I such that(Ui)i∈J is a cover of X.

Proof. (C) implies (C1). Let U be an maximal filter on X.


1. U has a cluster point x. (C)

2. There exists a filter U′ ⊃ U ∪ V(x). Proposition 3.23

3. U′ = U ⊃ V(x). U is maximal.

(C1) implies (C). Let F be a filter on X.


1. There exists a maximal filter U on X suchthat U ⊃ F.

Zorn’s lemma.

2. U has a limit point x, i.e. U ⊃ V(x). (C1)

3. x is a cluster point of F. U ⊃ F ∪ V(x) and Proposition 3.23.

(C) implies (C2).


1. If not (C2), then for all finite J ⊂ I,⋂i∈J

Fi 6= ∅.



2. { ⋂i∈J

Fi : J ⊂ I finite} is a basis of a filterF on X.

Lemma 2.15

3. F has a cluster point x. (C)

4. x ∈ cl(Fi) for all i. Take J = {i} in step 2 and Definition 3.22.

5. x ∈ ⋂i∈IFi, a contradiction to ⋂

i∈IFi = ∅. cl(Fi) = Fi since Fi is closed and

Lemma 3.13(2).

(C2) implies (C).


1. If not, then there exists a filter F on X

without cluster point.(C)

2. ⋂M∈F

cl(M) = ∅. Definition 3.22

3. There exists a finite E ⊂ F such that⋂M∈E

cl(M) = ∅.(C2)

4. ∅ 6= ⋂M∈E

M ∈ F, a contradiction to step 3. Definition 2.10(FII) and ∅ 6∈ F.

(C2) iff (C3). ⋂i∈I(resp.J)

Fi = ∅ iff ⋃i∈I(resp.J)

Ui = X for Ui = X − Fi. �

Proposition 8.11Let f : X → X ′ be a continuous map between topological spaces. If X is quasi-compact, thenf(X) is quasi-compact.Proof.


1. Suppose (Ui)i∈I is an open cover of f(X).Then (f−1(Ui))i∈I is an open cover of X.

below.

70 ZHENGYAO WU


1.1. f−1(Ui) is open in X for all i ∈ I. Ui is open in X ′, f is continuous andHomework 5.7(2d).

1.2. X = f−1(⋃i∈IUi) = ⋃

i∈If−1(Ui). f(X) ⊂ ⋃

i∈IUi.

2. There exists J ⊂ I finite such that(f−1(Ui))i∈J is an open cover of X.

X is quasi-compact and Proposi-tion 8.10(C3).

3. (Ui)i∈J is an open cover of f(X). f(X) = f( ⋃i∈J

Ui) = ⋃i∈J

f(Ui).

4. f(X) is quasi-compact. Steps 1, 2 and Proposition 8.10(C3).

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Proposition 8.12If X is a Hausdorff space and A ⊂ X is quasi-compact, then A is closed.Proof.


1. ∀x ∈ cl(A), ∀V ∈ V(x), V ∩ A 6= ∅. Definition 3.12(2)

2. V(x)∩A = {V ∩A : V ∈ V(x)} is a filteron A.

Definition 2.10

3. V(x) ∩ A has a cluster point y ∈ A. A is quasi-compact, Definition 8.9(C)

4. y is a cluster point of V(x). y ∈ ⋂V ∈V(x)

cl(V ∩ A) ⊂ ⋂V ∈V(x)

cl(V ).

5. x = y ∈ A. x is a limit and cluster point of V(x), Xis Hausdorff and Proposition 5.8(H5)

6. A is closed. Lemma 3.13(2)

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Homework 8.13Let R be the real line and A ⊂ R. Show that if A is compact, then A is closed and bounded.


Definition 8.14Let (X,TX) be a topological space and Y ⊂ X. We call TY = {U ∩ Y : U ∈ TX} the inducedor subspace topology on Y . We call (Y,TY ) the subspace of (X,TX).

Homework 8.15Let (X,TX) be a topological space and Y ⊂ X. Let TY = {U ∩ Y : U ∈ TX}. Show that TYsatisfies (OI)(OII) as in Definition 2.1.

Example 8.16Let Q be the rational line. Let Z be the ring of integers. Then the subspace Z of Q has thediscrete topology. In fact, {n} is open in Z for all n ∈ Z because {n} = (n− 1

2 , n+ 12) ∩ Z.

Lemma 8.17Let (X,TX) be a topological space. Let (Y,TY ) be its subspace. We have TY ⊂ TX iff Y ∈ TX .

Proof. Suppose TY ⊂ TX . Since Y ∈ TY , Y ∈ TX .Conversely, suppose Y ∈ TX . For all U ∩ Y ∈ TY where U ∈ TX , we have U ∩ Y ∈ TX byDefinition 2.1(OII). Therefore TY ⊂ TX . �

Homework 8.18Let (X,TX) be a topological space. Let (Y,TY ) be its subspace. Show that the following areequivalent: (1) Y is closed in X. (2) For F ⊂ Y , if F is closed in Y , then F is closed in X.

Proposition 8.19If X is quasi-compact and A is closed in X, then A is quasi-compact.Proof.


1. If (Fi)i∈I is a family of closed subsets ofA such that ⋂

i∈IFi = ∅, then (Fi)i∈I is a

family of closed subsets of X.

A is closed in X and Homework 8.18.

2. There exists a finite J ⊂ I such that⋂i∈J

Fi = ∅.X is quasi-compact and Proposi-tion 8.10(C2)

3. A is quasi-compact. Steps 1, 2 and Proposition 8.10(C2)

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72 ZHENGYAO WU

9. November 4, Tychonoff, compact(2)

Lemma 9.1Suppose X is a set with two topologies T1 and T2. If (X,T1) is Hausdorff, (X,T2) is quasi-compactand T1 ⊂ T2, then T1 = T2.Proof.


1. For all closed subset A ⊂ (X,T2), A isquasi-compact in (X,T2).

Proposition 8.19.

2. IdX : (X,T2)→ (X,T1) is continuous. T1 ⊂ T2 and Homework 5.7(2d).

3. IdX(A) = A is quasi-compact in (X,T1). Steps 1, 2 and Proposition 8.11.

4. A is closed in (X,T1). Proposition 8.12.

5. T2 ⊂ T1. So T1 = T2. Steps 1 to 4.

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Homework 9.2Let (X,TX) and (Y,TY ) be topological spaces. Let (X × Y,TX×Y ) be their product space.(1) Suppose Y is quasi-compact and W ∈ TX×Y . Show that if x0 ∈ X and {x0} × Y ⊂ W , thenthere exists U ∈ TX such that x0 ∈ U and U × Y ⊂ W .(2) Show that if X and Y are quasi-compact, then X × Y is quasi-compact. (You must NOT useTheorem 9.6. Hint: use (1) and Proposition 8.10(C3).)

Proposition 9.3Let X be a Hausdorff space. Every subspace A of X is Hausdorff.Proof.


1. For all x 6= y in A, there exist U ∈ V(x)and V ∈ V(y) in X such that U ∩ V = ∅.

Definition 2.21(H).

2. U ∩ A ∈ V(x) and V ∩ A ∈ V(y) in A. Definition 8.14 and Definition 2.6.

(U ∩ A) ∩ (V ∩ A) = ∅. = (U ∩ V ) ∩ A = ∅ ∩ A = ∅.

3. A is Hausdorff. Definition 2.21(H).


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Proposition 9.4Let Xi be a family of nonempty topological spaces. The product space X = ∏

i∈IXi is Hausdorff iff

Xi is Hausdorff for all i ∈ I.

Proof. Suppose Xi is Hausdorff for all i ∈ I.


1. If (xi)i∈I 6= (yi)i∈I in X, then xj 6= yj forsome j ∈ I.

X = ∏i∈IXi

2. There exist U ∈ V(xj) and V ∈ V(yj) inXj such that U ∩ V = ∅.

Xj is Hausdorff and Definition 2.21(H).

3. prj : X → Xj is continuous. Definition 5.14

4. U × ∏i∈I, i6=j

Xi = pr−1j (U) ∈ V((xi)i∈I) and

V × ∏i∈I, i6=j

Xi = pr−1j (V ) ∈ V((yi)i∈I).

Definition 5.1.

(U × ∏

i∈I, i6=jXi

)∩(V × ∏

i∈I, i6=jXi

)=

(U ∩ V )× ∏i∈I, i6=j

Xi = ∅ in X.

Step 2.

5. X is Hausdorff. Step 4 and Definition 2.21(H)

Suppose X is Hausdorff.


1. For all j ∈ I, Xj∼= Xj ×

∏i∈I, i6=j

{xi}. Since Xi 6= ∅, fix xi ∈ Xi.

2. Xj is Hausdorff for all j ∈ I. Xj ×∏

i∈I, i6=j{xi} is a subspace of X, X is

Hausdorff and Proposition 9.3.

�

74 ZHENGYAO WU

Proposition 9.5LetXi be a family of nonempty topological spaces. Let F be a filter on the product spaceX = ∏

i∈IXi

and a = (ai)i∈I ∈ X. The filter F converges to a iff for all i ∈ I, the filter generated by pri(F)converges to ai.

Proof. Suppose V(a) ⊂ F.


1. For all j ∈ I, prj is continuous. X is the product space and Defini-tion 5.14.

2. For all V ∈ V(aj) in Xj, V ×∏

i∈I, i6=jXi =

pr−1j (V ) ∈ V(a) in X.

Step 1, prj(a) = aj and Homework 5.7(1b)

3. V × ∏i∈I, i6=j

Xi ∈ F. Given V(a) ⊂ F.

4. V(aj) ⊂ prj(F). V ∈ prj(F) by step 3.

Conversely, suppose for all i ∈ I, the filter generated by pri(F) converges to ai.


1. There exists a finite J ⊂ I such that a ∈∏j∈J

Uj ×∏

i∈I−JXi ⊂ V where Uj is open in

Xj for all j ∈ J .

By Definition 5.14, every open subset ofXis a union of some sets of the form ∏

j∈JUj×∏

i∈I−JXi (similar to Lemma 5.15).

2. Uj ∈ V(aj) for all j ∈ J . Uj is open in Xj and Proposition 2.7.

Xi ∈ V(ai) for all i ∈ I − J . Xi is open in Xi.

3. pri(∏j∈J

Uj×∏

i∈I−JXi) ∈ pri(F) for all i ∈ I. V(ai) ⊂ pri(F) for all i ∈ I.

∏j∈J

Uj ×∏

i∈I−JXi ∈ F.

4. V ∈ F. So V(a) ⊂ F. Step 1 and Definition 2.10(FI).

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Theorem 9.6 TychonoffLet (Xi)i∈I be a family of nonempty topological spaces with a product space X = ∏

i∈IXi.

(1) X is quasi-compact iff Xi is quasi-compact for all i ∈ I.(2) X is compact iff Xi is compact for all i ∈ I.

Proof. (1) Suppose X is quasi-compact. Since pri is continuous, Xi = pri(X) is quasi-compact byProposition 8.11. Conversely, suppose Xi is quasi-compact for all i ∈ I.


1. Suppose U is an maximal filer on X. Thenpri(U) is a basis of a maximal filter Ui forall i ∈ I.

below.

1.1. If F is a filter on Xi finer than Ui, thenpr−1i (F) is a basis of a filter finer than U.

Lemma 2.15.

1.2. pr−1i (F) is a basis of U. U is maximal.

1.3. F = Ui. pri(pr−1i (F)) is a basis of both Ui and F.

2. Ui is convergent for all i ∈ I. Xi is quasi-compact for all i ∈ I andProposition 8.10(C1)

3. U is convergent. Proposition 9.5

4. X is quasi-compact. Proposition 8.10(C1)

(2) follows from (1), Proposition 9.4 and Definition 8.9. �

Theorem 9.7Let X be a quasi-compact space. Let F be a filter on X. Let A = ⋂

M∈Fcl(M). Then V(A) ⊂ F.

Proof. If not, then there exists V ∈ V(A)− F.


1. For all M ∈ F, M 6⊂ V . Otherwise V ∈ F by Definition 2.10(FI)

2 {M − V : M ∈ F} is a basis of a filter Gon X.

Lemma 2.15

3. G has a cluster point x ∈ X. Definition 8.9(C)

76 ZHENGYAO WU


4. x ∈ ⋂M∈F

cl(M) = A. Step 3 and x ∈ cl(M − V ) ⊂ cl(M) byHomework 3.14(2).

5. x ∈ int(V ). V ∈ V(A) implies that V ∈ V(x).

6. x ∈ cl(X − V ) = X − int(V ), a contradic-tion.

X − V ∈ G and Homework 3.14(1)

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Corollary 9.8Let X be a compact space. A filter F on X is convergent iff F has only one cluster point.

Proof. Suppose F has only one cluster point x. Since X is quasi-compact, V(x) ⊂ F by Theo-rem 9.7. Conversely, suppose F converges to x. Since X is Hausdorff, x is the only cluster pointof F, by Proposition 5.17(H5). �

Lemma 9.9Every compact topological space is regular.

Proof. Let X be a compact space and x ∈ X. Let B be the set of all closed neighborhoods of x.


1. X is Hausdorff. X is compact and Definition 8.9.

2. (OIII) holds, i.e. B is a basis of V(x). below.

2.1. {x} = ⋂B∈B

B. Step 1 and Proposition 5.8(H)(H1)

2.2. x is the only cluster point of B. Step 2.1, B = cl(B) by Lemma 3.13(2)and Definition 3.22.

2.3. B is a basis of a filter F ⊃ V(x). X is compact and Corollary 9.8.

3. (OIII′) holds. Step 2 and Proposition 6.3.

4. X is regular. Steps 1, 3 and Definition 6.1.

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Theorem 9.10Let (X,T) be a compact topological space.(1) V(∆X) in the product topological space X ×X is a uniform structure on X.(2) V(∆X) gives the topology T.(3) If a uniform structure U on X gives the topology T, then U = V(∆X).(4) (X,V(∆X)) is complete.

Proof. (1) This part does not need the assumption “X is compact”.


1. V(∆X) is a filter on X × X, so (FI)(FII)hold.

Example 2.12.

2. ∀V ∈ V(∆X), there exists an open subsetW of X ×X such that ∆X ⊂ W ⊂ V .

Definition 2.6.

3. (UI) holds, i.e. ∀V ∈ V(∆X), ∆X ⊂ V . Step 2.

4. (UII) holds. below.

4.1. ∆X ⊂ W−1 ⊂ V −1. Step 2 and ∆X = ∆−1X .

4.2. W−1 is open in X ×X. Lemma 5.15.

4.3. V −1 ∈ V(∆X). Definition 2.6.

5. (UIII) holds. below.

5.1. If not, then there exists V ∈ V(∆X) suchthat for all W ∈ V(∆X), W ◦W 6⊂ V .

Negation of (UIII).

5.2. {W ◦W − V : W ∈ V(∆X)} is a basis ofa filter F on X ×X.

Lemma 2.15.

5.3. X ×X is quasi-compact. X is quasi-compact and Theorem 9.6(1).(Homework 9.2(2) is enough.)

5.4. F has a cluster point (x, y) ∈ X ×X. Definition 8.9(C)

5.5. x 6= y. (x, y) ∈ cl(X−U) = X−int(U) ⊂ X−∆X

by Homework 3.14(1), steps 1 and 2.

5.6. There exist open neighborhoods U1 ∈V(x) and U2 ∈ V(y) such that U1∩U2 = ∅.

X is Hausdorff and Definition 2.21(H)

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5.7. X is regular. X is compact and Lemma 9.9.

5.8. There exist closed neighborhoods V1 ∈V(x) and V2 ∈ V(y) such that V1 ⊂ U1

and V2 ⊂ U2.

Proposition 6.3(OIII)

5.9. Let U3 = X−V1−V2 andW = (U1×U1)∪(U2×U2)∪ (U3×U3). Then W ∈ V(∆X).

U1 ∪ U2 ∪ U3 = X.

5.10. (V1 × V2) ∩ (W ◦W ) = ∅. below.

5.10.1. If (u, v) ∈ (V1×V2)∩(W ◦W ), then u ∈ V1,v ∈ V2 and there exists w ∈ X such that(u,w) ∈ W and (w, v) ∈ W .

5.10.2. w ∈ U1. u ∈ V1, so (u,w) ∈ U1 × U1.

5.10.3. v ∈ U1 ∪ U3. w ∈ U1, so (w, v) ∈ (U1×U1)∪ (U3×U3).

5.10.4. v ∈ U1 ∪ U3 = X − V2, a contradiction. Steps 5.6, 5.8, 5.9.

5.11. A contradiction to (x, y) ∈ cl(W ◦ W −V ) ⊂ cl(W ◦W ).

Step 10, V1 × V2 ∈ V((x, y)), Defini-tion 3.12(2) and Homework 3.14(2)

6. U is a uniform structure. Definition 1.10, steps 1, 3, 4 and 5.

(2) Suppose V(∆X) gives a topology T′, i.e. for all x ∈ X, U ∈ V(x) in (X,T′) iff ∃V ∈ V(∆X),U = V (x). We need to show that T′ = T. We first show that (X,T′) is Hausdorff.


1. (X ×X,TX×X) is Hausdorff. (X,T) is Hausdorff and Proposition 9.4.(Homework 6.6(2) is enough)

2. For all (x, y) ∈ X ×X, {(x, y)} is closed. Homework 6.6(1).

3. For all x 6= y in X, X × X − {(x, y)} ∈V(∆X).

X×X−{(x, y)} is open, contains ∆X andProposition 2.7.

4. ∆X = ⋂V(∆X)

U . ∆X ⊂⋂

U∈V(∆X)U ⊂ ⋂

x 6=y in X(X × X −

{(x, y)}) = X×X− ⋃x 6=y in X

{(x, y)} = ∆X .



5. (X,U) is Hausdorff, i.e. (X,T′) is Haus-dorff.

Proposition 7.4.

Next, we show that T′ = T.


6. T′ ⊂ T. below

6.1. Suppose U ∈ T′. For all x ∈ U , we haveU ∈ V(x) in (X,T′).

Proposition 2.7.

6.2. There exists V ∈ V(∆X) in (X ×X,TX×X) such that U = V (x).

Def. of T′.

6.3. s : (X,T)→ (X×X,TX×X), s(y) = (x, y)is continuous.

Proposition 5.10.

6.4. U ∈ V(x) in (X,T). U = s−1(V ), steps 6.2, 6.3 and Defini-tion 5.1.

6.5. U ∈ T. Proposition 2.7.

7. T = T′. Steps 5, 6, (X,T) is quasi-compact andLemma 9.1.

(3) We first show that U ⊂ V(∆X).


1. For all U ∈ U, there exists S ∈ S suchthat S ◦∆X ◦ S = S ◦ S ⊂ U .

Lemma 2.16(UI’)(UIII’)

2. S ◦∆X ◦ S ∈ V(∆X) in (X ×X,TX×X). Proposition 3.20(1).

3. U ∈ V(∆X). Steps 1, 2 and Lemma 2.8(VI).

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Next, we show that V(∆X) ⊂ U.


1. If there exists U ∈ V(∆X)− U, then V −U 6= ∅ for all V ∈ U.

Definition 1.10(FI).

2. {V − U : V ∈ U} is a basis of a filter Gon X ×X.

Lemma 2.15.

3. X ×X is quasi-compact. X is quasi-compact and Theorem 9.6(1).(Homework 9.2(2) is enough.)

4. G has a cluster point (a, b) ∈ X ×X. Definition 8.9(C)

5. (a, b) 6∈ ∆X . (a, b) ∈ cl(X−U) = X−int(U) ⊂ X−∆X

by Homework 3.14(1), steps 1 and 2.

6. (a, b) is also a cluster point of U. U ⊂ G by step 2 and Definition 2.10(FI).

7. (a, b) ∈ ⋂V ∈U

V = ∆X , a contradiction tostep 5.

Lemma 6.7; U gives T, (X,U) is Hausdorffand Proposition 7.4.

(4)


1. Every Cauchy filter F on X has a clusterpoint.

X is quasi-compact and Definition 8.9(C).

2. F is convergent. Lemma 4.9.

3. X is complete. Definition 3.6.

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10. November 11, Compact(3), bounded, connected(1)

Theorem 10.1LetX be a compact topological space. Let (X ′,U′) be a uniform space. If f : X → X ′ is continuous,then f : (X,V(∆X))→ (X ′,U′) is uniformly continuous.

Proof. Suppose X ×X and X ′ ×X ′ have the product topologies.


1. f × f : X ×X → X ′ ×X ′ is continuous. f is continuous and Homework 5.7(2d).

2. For all V ′ ∈ U′ ∩ TX′×X′ , (f × f)−1(V ′) ∈TX×X .

Homework 5.7(2d) for f × f .

3. ∆X ⊂ (f × f)−1(V ′). (f × f)(∆X) = ∆f(X) ⊂ ∆X′ ⊂ V ′ byDefinition 1.10(UI).

4. (f × f)−1(V ′) ∈ V(∆X) the unique uni-form structure on X compatible with TX .

Theorem 9.10(1)(2)(3).

5. U′ ∩ TX′×X′ is a basis of U′. Lemma 3.21.

6. f is uniformly continuous. Steps 1 to 5 and Definition 1.19.

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Proposition 10.2Let F be a maximal filter on a set X. For A,B ⊂ X, if A ∪B ∈ F, then A ∈ F or B ∈ F.

Proof. Suppose A 6∈ F. Define S = {M ⊂ X : A ∪M ∈ F}


1. S is a filter on X. Definition 2.10 and below.

1.1. If M ⊂ N and M ∈ S, then N ∈ S. A ∪M ⊂ A ∪N and A ∪M ∈ F.

1.2. If I is finite and Mi ∈ S for all i ∈ I, then⋂i∈IMi ∈ S.

Since A ∪ Mi ∈ F for all i ∈ I, A ∪(⋂i∈IMi) = ⋂

i∈I(A ∪Mi) ∈ F.

1.3. ∅ 6∈ S. A 6∈ F.

2. F ⊂ S. below.

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2.1. For all V ∈ F, A ∪ (V − A) ∈ F. V ⊂ A ∪ (V − A) and Definition 2.10(FI)

2.2. V − A ∈ S. Def. of S.

2.3. V ∈ S. V − A ⊂ V and Definition 2.10(FI)

3. S = F. F is maximal.

4. B ∈ S = F. A ∪B ∈ F.

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Theorem 10.3Let (X,U) be a uniform space. Its completion (X, U) is compact iff for all V ∈ U, there existB1, . . . , Bn ⊂ X such that Bj ×Bj ⊂ V for all 1 ≤ j ≤ n and X = B1 ∪ · · · ∪Bn.

Proof. Let i : X → X, i(x) = V(x) be the canonical map. Suppose X is compact.


1. For all U ∈ U, there exists U ′ ∈ U suchthat U ′ = U ′−1 and U ′ ◦ U ′ ⊂ U .

Example 2.23 and Lemma 2.16(UIII’) forU.

2. (U ′(x))x∈X is a cover of X. Lemma 1.14(VIII).

3. (int(U ′(x)))x∈X is a open cover of X. Definition 2.6.

4. There exist {x1, . . . , xn} ⊂ X suchthat int(U ′(x1)), . . . , int(U ′(xn)) is a opencover of X.

Proposition 8.10(C3).

5. U ′(x1), . . . , U ′(xn) is a cover of X. int(U ′(xj)) ⊂ U ′(xj) for all 1 ≤ j ≤ n.

6. U ′(xj)× U ′(xj) ⊂ U . Step 1.

7. Let V = (i × i)−1(U) and Bj =i−1(U ′(xj)). Then Bj ×Bj ⊂ V .

i−1(U ′(xj)) × i−1(U ′(xj)) = (i ×i)−1(U ′(xj)× U ′(xj)) ⊂ (i× i)−1(U).

8. (Bj)1≤j≤n covers X. Step 5.

9. It is also true for all V ∈ U since U has abasis (i× i)−1(U).

Proposition 4.6.


Conversely, suppose V ∈ U and (Bj)1≤j≤n is a finite cover of X such that Bj ×Bj ⊂ V .


1. Suppose U ∈ U and U is closed in X× X,V = (i × i)−1(U) and Cj = i(Bj). ThenCj × Cj ⊂ U .

i(Bj) × i(Bj) = (i × i)(Bj × Bj) ⊂ (i ×i)(V ).

2. i(X) ⊂ C1 ∪ · · · ∪ Cn. X = B1 ∪ · · · ∪Bn.

3. X = cl(i(X)) = cl(C1 ∪ · · · ∪ Cn) Proposition 4.8.

= cl(C1) ∪ · · · ∪ cl(Cn). Lemma 5.16.

4. cl(Cj)× cl(Cj) = cl(Cj×Cj) ⊂ cl(U) = U

for all 1 ≤ j ≤ n.Lemma 5.16, step 1 and Home-work 3.14(2).

5. If F is a maximal filter on X, then thereexists 1 ≤ k ≤ n such that cl(Ck) ∈ F.

cl(C1) ∪ · · · ∪ cl(Cn) = X ∈ F by step 3;and Proposition 10.2.

6. F is a Cauchy filter on X. Steps 4, 5 and Definition 2.24.

7. F is convergent. X is complete and Definition 3.6.

8. X is quasi-compact. Proposition 8.10(C1).

9. X is Hausdorff. Proposition 4.4.

10. X is compact. Steps 8, 9 and Definition 8.9.

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Corollary 10.4A uniform space (X,U) is compact iff it is Hausdorff, complete, and for all V ∈ U, there existsB1, . . . , Bn ⊂ X such that Bj ×Bj ⊂ V for all 1 ≤ j ≤ n and X = B1 ∪ · · · ∪Bn.

Proof. Suppose (X,U) is compact.


1. X is Hausdorff. Definition 8.9.

2. X is complete. Theorem 9.10.

84 ZHENGYAO WU


3. X ' X is compact. Proposition 7.3 and X is compact.

4. For all V ∈ U, there exists a finite cover(Bj)1≤j≤n of X such that Bj ×Bj ⊂ V .

Theorem 10.3.

Conversely, suppose for all V ∈ U, there exists a finite cover (Bj)1≤j≤n of X such that Bj×Bj ⊂ V .


1. X is compact. Theorem 10.3.

2. X ∼= X is compact. X is complete and Proposition 7.3.

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Proposition 10.5Let (X,U) be a complete uniform space. Let A be a closed subset of X. The subspace A iscomplete.Proof.


1. Let F be a Cauchy filter on A. Then F isa basis of a Cauchy filter G on X.

The inclusion map (A,U ∩ (A × A)) ↪→(X,U) is uniformly continuous and Propo-sition 6.18.

2. G converges to some x ∈ X. X is complete and Definition 3.6.

3. x is a cluster point of F. x is a cluster point of G and F ⊂ G.

4. x ∈ cl(A) = A. A is closed and Lemma 3.13(2).

5. F converges to x. F is Cauchy, steps 3, 4 and Lemma 4.9.

6. A is complete. Definition 3.6.

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Homework 10.6Let R be the real line and A ⊂ R.(1) Show that if A is closed and bounded, then A is compact.(2) Suppose ∅ 6= A ⊂ R. Show that if A is bounded above, then supA exists. (Similarly, if A isbounded below, then inf(A) exists. Hint: use (1))

Example 10.7The real line R is not compact. In fact, {(−n, n) : n > 0, n ∈ Z} is an open cover of R withoutfinite subcover. By Proposition 8.10(C3), R is not quasi-compact.

Example 10.8The open interval (0, 1) in the real line R is not compact. In fact, {( 1

n, 1) : n > 0, n ∈ Z} is an

open cover of (0, 1) without finite subcover. By Proposition 8.10(C3), (0, 1) is not quasi-compact.

Proposition 10.9Let X be a completely regular space. If Y ⊂ X, then Y is completely regular.Proof.


1. Y is Hausdorff. Proposition 9.3(1)

2. For all x0 ∈ Y and for all V ∈ V(x0) inY , there exists U ∈ V(x0) in X such thatU ∩ Y = V .

Lemma 8.17.

3. X is uniformizable. X is completely regular and Definition 8.5.

4. There exists a continuous f : X → [0, 1]such that f(x0) = 0 and f(X −U) = {1}.

X satisfies (OIV) by Theorem 8.4.

5. f |Y : Y → [0, 1] is continuous such thatf(x0) = 0 and f(Y − V ) = {1}.

x0 ∈ Y and Y − V = Y − U ∩ Y = (X −U) ∩ Y ⊂ X − U .

6. Y is uniformizable. Theorem 8.4.

7. Y is completely regular. Steps 1, 6 and Definition 8.5.

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Proposition 10.10Let X be a topological space.

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(1) If X is homeomorphic to a subspace of a compact space, then it is completely regular.(2) If X is completely regular, then it is homeomorphic to a subspace of the compact space [0, 1]Φ,where Φ is the set of all continuous functions X → [0, 1].

Proof. (1) We may assume that Y is compact and X ⊂ Y .


1. Y is Hausdorff. Y is compact and Definition 8.9.

2. Y is uniformizable. Y is compact and Theorem 9.10(1).

3. Y is completely regular. Steps 1, 2 and Definition 8.5.

4. X is completely regular. Proposition 10.9.

(2) Define e : X → [0, 1]Φ, e(x) = (f(x))f∈Φ. For all f ∈ Φ and for all ε > 0, let Uf,ε = {(x, y) ∈X ×X : |f(x)− f(y)| < ε}. Let U be the uniform structure on X with a basis consisting of finiteintersections of Uf,ε.


1. e is continuous. f are all continuous and [0, 1]Φ has theproduct topology.

2. e is injective. below.

2.1. X is Hausdorff. So for all x ∈ X, {x} isclosed.

X is completely regular and Definition 8.5.

2.2. X is uniformizable. X is completely regular and Definition 8.5.

2.3. For all x 6= y in X, there exists f ∈ Φsuch that f(y) = 0 and f(x) = 1.

(OIV) from Theorem 8.4 and X − {x} ∈V(y).

2.4. e(x) 6= e(y). prf (e(x)) = 1 6= 0 = prf (e(y)).

3. e−1 : e(X)→ X is continuous. below.

3.1. Every open subset of X is a union of finiteintersections of g−1(V ), where g ∈ Φ andV is an open subset of [0, 1].

The 2nd part of Theorem 8.4.



3.2. e(g−1(V )) = (V × ∏f∈Φ, f 6=g

[0, 1]) ∩ e(X) is

open in e(X).

V × ∏f∈Φ, f 6=g

[0, 1] is open in [0, 1]Φ and Def-inition 8.14.

3.3. e−1 : e(X)→ X is continuous. Steps 3.1, 3.2, (OI, II) and Home-work 5.7(2d).

4. e : X → e(X) is a homeomorphism. Steps 1, 2, 3 and Definition 6.9.

5. [0, 1]Φ is compact. [0, 1] is compact by Homework 10.6(1);and Theorem 9.6(2).

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Definition 10.11Let (X, d), (X ′, d′) be metric spaces. A map f : X → X ′ is an isometry if it is surjective andd(x, y) = d′(f(x), f(y)) for all x, y ∈ X (this formula and Definition 1.1(ECI) implies that f isinjective).

Example 10.12Let (X, d), (X ′, d′) be metric spaces. Let U,U′ be their uniform structures given by metrics. Iff : X → X ′ is an isometry, then it is an isomorphism of uniform spaces. Conversely, it is possiblethat f is an isomorphism of uniform spaces but not an isometry. If d, d′ are equivalent metrics andd 6= d′ (see Example 1.17), then U = U′. So IdX is an isomorphism of uniform spaces but not anisometry.

Definition 10.13Let (X, d) be a metric space. Let A,B be two nonempty subsets of X. We call d(A,B) =inf(d(a, b) : a ∈ A, b ∈ B) the distance between A and B.

Lemma 10.14Let (X, d) be a metric space and ∅ 6= A ⊂ X. For x ∈ X, d(x,A) = 0 iff x ∈ cl(A).

Proof. d(x,A) = inf(d(x, a) : a ∈ A) = 0 iff ∀ε > 0, ∃a ∈ A, d(x, a) < ε iff ∀ε > 0, the open ballVε(x) ∩ A 6= ∅ iff x ∈ cl(A). �

Homework 10.15Let (X, d) be a metric space and ∅ 6= A ⊂ X. Show that |d(x,A) − d(y, A)| ≤ d(x, y) for allx, y ∈ X.

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Lemma 10.16Let (X, d) be a metric space and ∅ 6= A ⊂ X. The function X → [0,+∞), x 7→ d(x,A) is uniformlycontinuous.Proof.


1. For all ε > 0, take δ = ε, for all (x, y) ∈X × X such that d(x, y) < δ, we have|d(x,A)− d(y, A)| < ε.

|d(x,A) − d(y, A)| ≤ d(x, y) < δ = ε andHomework 10.15.

2. d(•, A) is uniformly continuous. Definition 1.19.

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Definition 10.17Let (X, d) be a metric space and ∅ 6= A ⊂ X. The diameter of A is δ(A) = sup(d(x, y) : x ∈A, y ∈ A). If δ(A) < +∞, then we call A bounded; otherwise we call A unbounded.

Homework 10.18The Euclidean metric on R is d1(x, y) = |x−y|, x, y ∈ R. Define d2(x, y) = |x− y|

1 + |x− y| , x, y ∈ R.Show that d2 is a metric on R and d1 and d2 are equivalent.

Example 10.19

Boundedness is NOT a topological property. Let d1(x, y) = |x− y|, d2(x, y) = |x− y|1 + |x− y| , x, y ∈

R. The real line R is unbounded relative to d1 since δ1(R) = sup(|x− y| : x, y ∈ R) = +∞; R is

bounded relative to d2 since δ2(R) = sup( |x− y|1 + |x− y| : x, y ∈ R) = 1.

Definition 10.20Let X be a topological space. We call X connected if it is not a disjoint union of two nonemptyopen subsets. Otherwise we call X disconnected.

Lemma 10.21Let X be a topological space. The following are equivalent:(1) X is connected, i.e. X is not the disjoint union of two nonempty open subsets.(2) X is not a disjoint union of two nonempty closed subsets.(3) The only open and closed subsets of X are ∅ and X.

Proof. Their negations are equivalent:(1’) X = X1 ∪X2, X1 ∩X2 = ∅, X1 and X2 are open.


(2’) X = X1 ∪X2, X1 ∩X2 = ∅, X1 and X2 are closed.(3’) There exists ∅ ( X1 ( X such that X1 is open and closed. �

Proposition 10.22Let (X,T) be a topological space. Let (Ai)i∈I be a family of connected subsets of X. If ⋂

i∈IAi 6= ∅,

then A = ⋃i∈IAi is connected.

Proof. Suppose A is disconnected.


1. ∃B,C ∈ T, B ∩ A 6= ∅, C ∩ A 6= ∅, A ⊂B ∪ C and A ∩B ∩ C = ∅.

Definition 10.20.

2. ∃x ∈ ⋂i∈IAi ⊂ A ⊂ B∪C. Suppose x ∈ B. ⋂

i∈IAi 6= ∅.

3. Aj is disconnected, a contradiction. Definition 10.20 and below.

3.1. There exists j ∈ I such that Aj ∩ C 6= ∅. A ∩ C 6= ∅.

3.2. Aj ∩B 6= ∅. x ∈ Aj ∩B by steps 2,3.

3.3. Aj ⊂ B ∪ C. Aj ⊂ A ⊂ B ∪ C.

3.4. Aj ∩B ∩ C = ∅. Aj ∩B ∩ C ⊂ A ∩B ∩ C = ∅.

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Proposition 10.23Let (X,T) and (X ′,T′) be topological spaces. Let f : X → X ′ be a continuous map. If A ⊂ X isconnected, then f(A) is connected.

Proof. Suppose f(A) is disconnected.


1. ∃B′, C ′ ∈ T′, f(A)∩B′ 6= ∅, f(A)∩C ′ 6= ∅,f(A) ⊂ B′ ∪ C ′ and f(A) ∩B′ ∩ C ′ = ∅.

Definition 10.20.

2. A is disconnected, a contradiction. Definition 10.20 and below.

2.1. B = f−1(B′)∩A and C = f−1(C ′)∩A areopen in A.

f is continuous, Homework 5.7(2d).

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2.2. B 6= ∅ and C 6= ∅. f(A) ∩B′ 6= ∅ and f(A) ∩ C ′ 6= ∅.

2.3. B ∪ C = A. f(A) ⊂ B′ ∪ C ′

2.4. B ∩ C = ∅. f(A) ∩B′ ∩ C ′ = ∅

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Example 10.24The rational line Q is disconnected since (−∞,

√2)∩Q and (

√2,+∞)∩Q are open, disjoint and

their union is Q.

Homework 10.25Let R be the real line. A subset I ⊂ R is an interval iff for all a, b ∈ I and a ≤ b, the closedinterval [a, b] ⊂ I.


11. November 18, Intervals, extreme / intermediate value theorems, metrizable,first / second countable, basis of a topology

Example 11.1The unit closed interval [0, 1] is connected in R.


1. If [0, 1] is not connected, then [0, 1] = H ∪K, H∩K = ∅ and H,K are open in [0, 1].

Definition 10.20.

2. Suppose 1 ∈ H. Then c = sup(K) < 1. Since H is open in [0, 1], there exists ε > 0such that (1− ε, 1] ⊂ H = [0, 1]−K.

3.1. If c ∈ H, then there exists δ > 0 such that(c− δ, c+ δ) ⊂ H.

H is open in [0, 1].

3.2. c− δ/2 is an upper bound of K, a contra-diction.

c− δ/2 < c = sup(K).

4.1. If c ∈ K, then there exists δ > 0 such that(c− δ, c+ δ) ⊂ K.

K is open in [0, 1].

4.2. c+ δ/2 ∈ K a contradiction. c is an upper bound of K and c+ δ/2 > c.

Example 11.2Let R be the real line. A subset I ⊂ R is connected iff I is an interval.

Proof. Suppose I is connected.


1. If Card(I) = 1, then I = [a, a] for somea ∈ R.

I = {a}.

2. Suppose Card(I) ≥ 2, a, b ∈ I and a < b.If x ∈ (a, b), then x ∈ I. So [a, b] ⊂ I.

Otherwise I ⊂ (−∞, x) ∪ (x +∞), a ∈I ∩ (−∞, x), b ∈ I ∩ (x+∞) and hence Iis disconnected by Definition 10.20.

3. I is an interval. Homework 10.25.

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Conversely, suppose a, b ∈ I, a ≤ b such that [a, b] ⊂ I.


1. For all c, d ∈ R and c < d, [c, d] ∼= [0, 1] isconnected.

A homeomorphism f : [c, d] → [0, 1],f(x) = x− c

d− cand Example 11.1.

2. I is connected. I = ⋃[a,b]⊂[c,d]⊂I

[c, d], ⋂[a,b]⊂[c,d]⊂I

[c, d] =

[a, b] 6= ∅ and Proposition 10.22.

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Homework 11.3Suppose I ⊂ R is an interval and f : I → R is a continuous map. Show that if f is an injection ,then f is strictly monotonic.

Example 11.4(1) Suppose a, b ∈ R and a < b. The open interval (a, b) is homeomorphic to R.(2) Every open interval is homeomorphic to R.

Proof. (1) Define f = f(a,b) : (a, b)→ R, f(x) = 1b− x

− 1x− a

.


1. f is continuous. f ′(x) = 1(b− x)2 + 1

(x− a)2 exists.

2. f is injective. f is strictly increasing by f ′ > 0.

3. f((a, b)) is connected. Proposition 10.23.

4. f((a, b)) is an interval. Example 11.2.

5. f is surjective. limx→a+

f(x) = −∞, limx→b−

f(x) = +∞.

6. f−1 is continuous. Similar to step 3 and 4, for all (c, d) ⊂(a, b), f((c, d)) = (f(c), f(d)).

7. f is a homeomorphism. Definition 6.9.

(2) Suppose J ⊂ R is an interval (possibly unbounded).



1. f(a,b) is a homeomorphism and f−1(a,b)(J) ⊂

I is an open interval.(1) for (a, b).

2. Suppose f−1(a,b)(J) = (c, d). Then f(c,d) is a

homeomorphism.(1) for (c, d).

3. f(c,d) ◦ f−1(a,b) : J → R is a homeomophism. Steps 1 and 2.

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Theorem 11.5 Weierstrass, Extreme Value TheoremLet E 6= ∅ be a quasi-compact topological space. Let f : E → [−∞,+∞] be a continuous function.There exists a, b ∈ E such that f(a) = sup

x∈Ef(x) and f(b) = inf

x∈Ef(x).

Proof.


1. f(E) is quasi-compact in [−∞,+∞]. E is quasi-comapct, f is continuous andProposition 8.11.

2. [−∞,+∞] is Hausdorff. Definition 2.21(H).

3. f(E) is closed in [−∞,+∞]. Steps 1, 2 and Proposition 8.12.

4. infx∈E

f(x), supx∈E

f(x) ∈ cl(f(E)) = f(E). Lemma 3.13(2).

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Corollary 11.6Let (X, d) be a metric space. Suppose A ⊂ X is compact. For all x ∈ X, there exists y ∈ A suchthat d(x,A) = d(x, y).

Proof. Fix x ∈ X, define f : X → [−∞,+∞], f(y) = d(x, y). Then d(x,A) = infa∈A

d(x, a) =infa∈A

f(a). By Theorem 11.5, there exists y ∈ A such that d(x,A) = d(x, y). �

Example 11.7In R, d(0, (0, 1)) = 0 but for all y ∈ (0, 1), d(0, y) = y > 0.

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Theorem 11.8 Bolzano, Intermediate Value Theorem

Let E be a connected topological space. For all a, b ∈ E, let I =

[f(a), f(b)], if f(a) ≤ f(b);[f(b), f(a)], if f(a) > f(b).

Let f : E → [−∞,+∞] be a continuous map. For all c ∈ I, there exists e ∈ E such that f(e) = c.Proof.


1. f(E) is connected. E is connected, f is continuous andProposition 10.23.

2. f(E) is an interval. Similar to Example 11.2.

3. I ⊂ f(E). So ∀c ∈ I, ∃e ∈ E, f(e) = c. f(a), f(b) ∈ I and Homework 10.25.

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Homework 11.9Let (X, d) be a metric space. Suppose A ⊂ X and A is compact; B ⊂ X and B is closed. Showthat if d(A,B) = 0, then A ∩B 6= ∅.

Definition 11.10A topological space (X,T) is metrizable if there exists a metric d compatible with T, i.e. U ∈ Tiff for all x ∈ U , there exists ε > 0 such that {y ∈ X : d(x, y) < ε} ⊂ U .

Proposition 11.11Let X be a metrizable space.(1) Every closed subset of X is the intersection of countably many open sets.(2) Every open subset of X is the union of countably many closed sets.

Proof. Let d be a metric on X compatible with its topology.(1) Suppose A is closed in X. Then

A = cl(A) = {x ∈ X : d(x,A) = 0} =∞⋂n=1{x ∈ X : d(x,A) < 1

n}.

(2) Suppose A is open in X. By (1), X − A =∞⋂n=1{x ∈ X : d(x,X − A) < 1

n}. So

A =∞⋃n=1{x ∈ X : d(x,X − A) ≥ 1

n}

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Definition 11.12A topological space X is first countable if for all x ∈ X, the filter V(x) has a countable basis.

Lemma 11.13Every metrizable space X is first countable.

Proof. Suppose d is a metric on X compatible with the topology of X. For all x ∈ X, {y ∈X : d(x, y) < 1

n}, n ∈ N form a countable basis of V(x). �

Proposition 11.14Suppose X is a first countable space and ∅ 6= A ⊂ X. We have x ∈ cl(A) iff there exists a sequence(xn)n∈N in A such that lim

n→∞xn = x.

Proof. Suppose limn→∞

xn = x and F = {Y ⊂ X : xn ∈ X − Y for finitely many n.}


1. F is a filter and F ⊃ V(x). Example 2.13 and Example 3.2.

2. x is a cluster point of F. Proposition 3.23

3. x ∈ cl(A). For all V ∈ V(x) ⊂ F, there exists k ∈ N,for all n > k, xn ∈ V ∩ A.

Conversely, suppose x ∈ cl(A).


1. V(x) has a countable basis V0 ⊃ V1 ⊃ · · · X is first countable and Definition 11.12

2. Vn ∩ A 6= ∅ for all n ∈ N. x ∈ cl(A) and Definition 3.12

3. For all n ∈ N, take xn ∈ Vn ∩ A. Thenlimn→∞

xn = x.For all V ∈ V(x), there exists k ∈ N suchthat Vk ⊂ V , for all n > k, xn ∈ Vn ⊂Vk ⊂ V .

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Corollary 11.15A metric space X is complete iff every Cauchy sequence is convergent.

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Proof. Suppose X is complete.


1. Every Cauchy sequence (xn) on X corre-sponds a Cauchy filter F.

Example 2.25.

2. F is convergent. X is complete and Definition 3.6.

3. (xn) is convergent. Example 3.2.

Conversely, suppose every Cauchy sequence on X is convergent.


1. If X is not complete, then there exists x ∈X − i(X).

Proposition 7.3.

2. X is a metric space, hence first countable. Theorem 7.7 and Lemma 11.13.

3. x ∈ X = cl(i(X)). Proposition 4.8.

4. There exists a sequence (xn) in X suchthat x = lim

n→∞i(xn).

Steps 2, 3 and Proposition 11.14.

5. (i(xn)) is a divergent Cauchy sequence ini(X), a contradiction.

i is injective and every Cauchy sequenceon X is convergent.

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Definition 11.16Let (X,T) be a topological space and B ⊂ T. We call B a basis of T if every U ∈ T is a union ofelements of B.

Example 11.17A discrete space X has basis {{x} : x ∈ X}.

Example 11.18The real line R has basis {(a, b) : a < b, a, b ∈ R}.


Definition 11.19A topological space (X,T) is second countable if T has a countable basis.

Homework 11.20Let (X,T) be a topological space.(1) Show that B ⊂ T is a basis of the topology T iff for all x ∈ X, the set {B ∈ B : x ∈ B} is abasis of the filter V(x).(2) Show that if X is second countable, then it is first countable.

Example 11.21The real line R is second countable with basis {(a, b) : a < b, a ∈ Q, b ∈ Q}. Furthermore, theEuclidean space Rn is second countable. A differential manifold is second countable (Omit).The discrete space R is first countable but not second countable.

Proposition 11.22Let (X,T) be a metrizable topological space. The following are equivalent:(1) X is second countable.(2) X has a countable dense subset.(3) X is homeomorphic to a subspace of [0, 1]N.

Proof. (1) implies (2). It does not need the “metrizable” assumption.


1. T has a countable basis U1, U2, . . .. (1) and Definition 11.19.

2. Choose xn ∈ Un for each n ∈ N. For allU open in X, xn ∈ Un ⊂ U for some n.

Definition 11.16.

3. The set of x1, x2, . . . is dense in X. Lemma 3.16.

(2) implies (3).


1. Suppose {a1, a2, . . .} is dense in X. (2).

2. X has a metric d′ compatible with T. X is metrizable and Definition 11.10.

3. Let d(x, y) = d′(x, y)1 + d′(x, y) . Then d is a

metric equivalent to d′.

Similar to Homework 10.18.

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4. Define ϕ : X → [0, 1]N, ϕ(x) =(d(x, an))n∈N.

d(x, y) ∈ [0, 1].

5. ϕ is continuous. Each d(•, an) is continuous and [0, 1]N hasthe product topology.

6. For all x ∈ X = cl({a1, a2, . . .}), thereexists a subsequence an1 , an2 , . . . such thatx = lim

k→∞ank

.

X is first countable by Lemma 11.13;Proposition 11.14.

7. ϕ is injective. If ϕ(x) = ϕ(y) and x = limk→∞

ank, then

d(y, x) = d(y, limk→∞

ank) = lim

k→∞d(y, ank

) =limk→∞

d(x, ank) = 0 and hence x = y.

8. ϕ−1 : ϕ(X)→ X is continuous. Definition 5.1 and below.

8.1. For all ε > 0, there exists an such thatd(x0, an) < ε/3.

Step 6.

8.2. LetW = {x ∈ X : |d(x0, an)−d(x, an)| <ε/3.}. ThenW ⊂ {x ∈ X : d(x, x0) < ε}.

By step 8.1 and Definition 1.1(ECIII),d(x, x0) ≤ d(x, an) + d(an, x0) ≤|d(x, an)− d(an, x0)|+ 2d(an, x0) < ε.

8.3. W ∈ V(x0) in X. x0 ∈ W , W is open and Proposition 2.7.

8.4. ϕ(W ) ∈ V(ϕ(x0)) in [0, 1]N. ϕ(W ) = pr−1n ((d(x0, an)−ε/3, d(x0, an)+

ε/3)) is open and Proposition 2.7.

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12. November 25, Quotient space, open map, closed map

(3) implies (1).


1. [0, 1] has a countable basis U1, U2, . . .. Example 11.21.

2. [0, 1]N has a countable basis.∞∏i=1

Vi, Vi ( [0, 1] for finitely many i andVi ∈ {U1, U2, . . .}.

3. X has a countable basis, i.e. second count-able.

(3) and Definition 8.14; Definition 11.19.

Proposition 12.1A metrizable topological space (X,T) is compact iff every sequence in X has a convergent subse-quence.

Proof. Suppose X is compact.


1. For all sequence (xn) in X, its fil-ter F = {A ⊂ X : xn ∈ X −A for finitely many n} has a cluster pointx.

Example 2.13 and Definition 8.9(C)

2. x ∈ cl(x1, x2, . . .). The set of x1, x2, . . . is an element of F.

3. X is first countable, so (xn) has a conver-gent subsequence.

Lemma 11.13 and Proposition 11.14.

Conversely, suppose every sequence in X has a convergent subsequence.


1. X has a metric d compatible with T. X is metrizable and Definition 11.10.

2. Every Cauchy sequence (xn) in X has aconvergent subsequence (xnk

).Every sequence inX has a convergent sub-sequence.

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3. If limk→∞

xnk= x, then lim

n→∞xn = x. ∀ε > 0, ∃N ∈ N, ∀m,n >

N, d(xm, xn) < ε/2; ∃K ∈ N, ∀k >

K, d(xnk, x) < ε/2; ∀n > N, ∃nk >

N, d(xn, x) ≤ d(xn, xnk) + d(xnk

, x) < ε.

4. (X, d) is complete. Steps 1, 3 and Corollary 11.15.

5. If X is not compact, then there exists ε >0 such that there does not exist a finitecover of X of sets of diameters ≤ ε.

Theorem 10.3.

6. There exists (xn) such that d(xm, xn) >ε/2 for all m 6= n.

Induction. Choose x1 randomly and sup-pose x1, . . . , xn are chosen. If d(y, xi) ≤ε/2 for all y ∈ X for at least onexi, (1 ≤ i ≤ n), then open ballsVε/2(x1), . . . , Vε/2(xn) cover X.

7. (xn) does not have a convergent subse-quence, a contradiction.

(xn) does not have a Cauchy subsequence;every convergent sequence is Cauchy.

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Proposition 12.2Let (X,T) be a compact topological space. It is metrizable iff it is second countable.

Proof. Suppose X is metrizable. Let d be a metric on X compatible with T.


1. For all n ∈ N, there exists a finite coverV1/n(aj), a1, . . . , am(n) ∈ X.

Theorem 10.3.

2. Let An = {a1, . . . , am(n)}. Then∞⋃n=1

An iscountable and dense in X.

limn→∞

1n

= 0.

3. X is second countable. Proposition 11.22(1)(2).

Conversely, suppose X is second countable.



1.1. Suppose T has a basis U1, U2, . . .. Definition 11.19.

1.2. The product topology TX×X has basis(Ui × Uj), i, j ∈ Z>0.

Lemma 5.15.

2.1. {Un : x ∈ Un} is a basis of V(x). Step 1.1 and Homework 11.20(1).

2.2. {Ui × Uj : (x, x) ∈ Ui × Uj} is a basis ofV((x, x)).

Step 1.2 and Homework 11.20(1).

3.1. ∀(x, x) ∈ ∆X , ∀V ∈V((x, x)), ∃Ui, Uj, (x, x) ∈ Ui × Uj ⊂ V .

Definition 2.14

3.2. There exists Un, x ∈ Un ⊂ Ui ∩ Uj. Lemma 2.15(BI).

3.3 {Un×Un : x ∈ Un} is a basis of V((x, x)). Steps 3.1, 3.2 and Definition 2.14.

4.1. X ×X is quasi-compact. X is quasi-compact by Definition 8.9;Theorem 9.6(1)

4.2. ∆X is closed in X ×X. X is Hausdorff by Definition 8.9; Proposi-tion 5.17(H2)

4.3. ∆X is quasi-compact. Proposition 8.19.

5.1. For all W ∈ V(∆X) and for all x ∈ X,W ∈ V((x, x)).

(x, x) ∈ ∆X .

5.2. There exists n(x) ∈ Z>0 such that Un(x)×Un(x) ∈ V((x, x)) and Un(x) × Un(x) ⊂ W .

Step 3.

5.3. The cover (Un(x) × Un(x))x∈X of ∆X has afinite subcover (Un(xi) × Un(xi))1≤i≤m.

Proposition 8.10(C3).

5.4. Let finite unions of Un × Un be V1, V2, . . ..Then (Vm)m∈N is a countable basis ofV(∆X) of X compatible with T.

m⋃i=1

(Un(xi) × Un(xi)) ∈ V(∆X) andm⋃i=1

(Un(xi) × Un(xi)) ⊂ W .

6. V(∆X) is the only uniform structure on Xcompatible with T.

Theorem 9.10(1)(2)(3).

7. X is metrizable. X is Hausdorff, step 5, 6 and Proposi-tion 7.10.

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Definition 12.3Let X be a set. We call R ⊂ X ×X a relation on X. When (x, y) ∈ R, we often write x ∼ y.(1) R is an equivalence if ∆X ⊂ R; R−1 ⊂ R; R ◦R ⊂ R. In other words

x ∼ x; x ∼ y =⇒ y ∼ x; x ∼ y, y ∼ z =⇒ x ∼ z.

(2) For x ∈ X, its equivalence class relative to R is ϕ(x) = {y ∈ X : x ∼ y}.(3) We call X/R = {ϕ(x) : x ∈ X} the quotient set of X by R. We call ϕ : X → X/R thecanonical map.

Definition 12.4Let (X,TX) be a topological space. Let R be an equivalence relation on X and ϕ : X → X/R thecanonical map. Suppose a topology TX/R on X/R satisfies(1) ϕ : (X,TX)→ (X/R,TX/R) is continuous.(2) If T′ is a topology on X/R such that ϕ : (X,TX)→ (X/R,T′) is continuous, then T′ ⊂ TX/R.We call TX/R the quotient topology on X/R and (X/R,TX/R) the quotient space of X by R.

Lemma 12.5Let (X,TX) be a topological space. Let R be an equivalence relation on X and ϕ : X → X/R thecanonical map. Let (X/R,TX/R) be the quotient space.(1) TX/R = {U ⊂ X/R : ϕ−1(U) ∈ TX}, i.e. U is open in X/R iff ϕ−1(U) is open in X.(2) F is closed in X/R iff ϕ−1(F ) is closed in X.

Proof. (1) Let T′ = {U ⊂ X/R : ϕ−1(U) ∈ TX}.


1. T′ is a topology on X/R. below and Definition 2.1

(OI) If Ui ∈ T′ for all i ∈ I, then ⋃i∈IUi ∈ T′. ϕ−1(⋃

i∈IUi) = ⋃

i∈Iϕ−1(Ui) ∈ TX .

(OII) If I is finite and Ui ∈ T′ for all i ∈ I, then⋂i∈IUi ∈ T′.

ϕ−1(⋂i∈IUi) = ⋂

i∈Iϕ−1(Ui) ∈ TX .

2. If U ∈ TX/R, then ϕ−1(U) ∈ TX . So U ∈T′. Hence TX/R ⊂ T′.

ϕ : (X,TX) → (X/R,TX/R) is continuousby Definition 12.4(1); Homework 5.7(2d).

3. ϕ : (X,TX)→ (X/R,T′) is continuous. Homework 5.7(2d).

4. T′ ⊂ TX/R. Definition 12.4(2).


(2) X/R− F is open iff ϕ−1(X/R− F ) = X − ϕ−1(F ) is open. �

Definition 12.6Let f : X → Y be a map between topological spaces.(1) We call f open if for all U open in X, f(U) is open in Y .(2) We call f closed if for all F closed in X, f(F ) is closed in Y .

Homework 12.7(1) Let pr1 : R2 → R be the projection pr1(x, y) = x.(1a) Show that pr1 is an open map.(1b) Show that pr1 is not a closed map.(2) Show that pr1 : R × [0, 1]→ R is a closed map.(3) Let f : [0, 2π]→ S1 = {(x, y) ∈ R2 : x2 + y2 = 1}, f(t) = (cos t, sin t).(3a) Show that f is a closed map.(3b) Show that f is not an open map.

Theorem 12.8Let f : X → Y be a continuous surjection between topological spaces. If f is open or closed, thenY is homeomorphic to a quotient space of X.Proof.


1. R = (f × f)−1(∆Y ) = {(x1, x2) ∈ X ×X : f(x1) = f(x2)} is an equivalence re-lation on X.

Definition 12.3.

2. Let ϕ : X → X/R. We have a bijectiong : X/R→ Y , g(ϕ(x)) = f(x).

below.

2.1. g is well-defined. ϕ is surjective; if ϕ(x1) = ϕ(x2), thenf(x1) = f(x2).

2.2. g is injective. If f(x1) = f(x2), then ϕ(x1) = ϕ(x2).

2.3. g is surjective. f is surjective.

3. g : X/R→ Y is continuous. Homework 5.7(2d) and below.

3.1. If U is open in Y , then f−1(U) is open inX.

f is continuous and Homework 5.7(2d).

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3.2. g−1(U) = ϕ(f−1(U)) is open in X/R. ϕ−1(g−1(U)) = ϕ−1(ϕ(f−1(U))) =f−1(U) by step 1.

4. If f is open, then g is open. Definition 12.6(1) and below.

4.1. For all V open in X/R, ϕ−1(V ) is open inX.

ϕ is continuous and Homework 5.7(2d).

4.2. f(ϕ−1(V )) is open in Y . f is open and Definition 12.6(1).

4.3. g(V ) = g(ϕ(ϕ−1(V ))) = f(ϕ−1(V )) isopen in Y .

ϕ is surjective and f = g ◦ ϕ.

5. If f is closed, then g is closed. Definition 12.6(2) and below.

5.1. For all F closed in X/R, ϕ−1(F ) is closedin X.

ϕ is continuous and Homework 5.7(2c).

5.2. f(ϕ−1(F )) is closed in Y . f is closed and Definition 12.6(2).

5.3. g(F ) = g(ϕ(ϕ−1(F ))) = f(ϕ−1(F )) isclosed in Y .

ϕ is surjective and f = g ◦ ϕ.

6. g−1 : Y → X/R is continuous. Step 4 or 5, Homework 5.7(2c) or (2d).

7. g is a homeomorphism. Steps 2, 3, 6 and Definition 6.9.

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Theorem 12.9Let (X,TX) be a topological space. Let R be an equivalence relation on X and ϕ : X → X/R

the canonical map. Let (X/R,TX/R) be the quotient space. A map g : X/R→ Y is continuous iffg ◦ ϕ : X → Y is continuous.

Proof. By Definition 12.4(1), ϕ is continuous. If g is continuous, then g ◦ ϕ is continuous.Conversely, suppose g ◦ ϕ : X → Y is continuous.


1. If U is open in Y , then (g ◦ ϕ)−1(U) =ϕ−1(g−1(U)) is open in X.

g◦ϕ is continuous and Homework 5.7(2d).



2. g−1(U) is open in X/R. Lemma 12.5(1).

3. g : X/R→ Y is continuous. Homework 5.7(2d).

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Definition 12.10Let (X,TX) be a topological space. Let D be a set of some subsets of X. A topological space(D,TD) is called a decomposition space of (X,TX) if(1) X is a disjoint union of elements of D.(2) TD = {E ⊂ D : ⋃

D∈ED ∈ TX}, i.e. E is open in D iff ⋃

D∈ED is open in TX .

Homework 12.11Show that in Definition 12.10, TD is a topology on D.

Lemma 12.12Every decomposition space (D,TD) of (X,TX) is a quotient space.Proof.


1. R = {(x, y) ∈ X × X : ∃D ∈ D, x ∈D, y ∈ D} is an equivalence relation onX.

Definition 12.3, where transitivity followsfrom disjointness of Definition 12.10(1).

2. Define ϕ : X → D, ϕ(x) = D if x ∈ D. ϕ is well-defined by the disjointness of Def-inition 12.10(1).

3. D = X/R. ϕ is surjective by X = ⋃D∈D

D of Defini-tion 12.10(1).

4. TD = {E ⊂ D : ϕ−1(E) ∈ TX}. Step 2 and Definition 12.10(2).

5. TD = TX/R. Lemma 12.5(1).

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106 ZHENGYAO WU

Lemma 12.13Let (X,TX) be a topological space. Let R be an equivalence relation on X and ϕ : X → X/R thecanonical map. Let (X/R,TX/R) be the quotient space. Then (X/R,TX/R) is a decompositionspace.Proof.


1. If ϕ(x) 6= ϕ(y), then ϕ(x) ∩ ϕ(y) = ∅. If z ∈ ϕ(x) ∩ ϕ(y), then (x, z) ∈ R and(y, z) ∈ R. So (x, y) ∈ R and ϕ(x) =ϕ(y).

2. X = ⋃D∈X/R

D. ϕ is surjective.

3. U is open in X/R iff ϕ−1(U) = ⋃D∈U

D isopen in X.

Lemma 12.5(1).

4. (X/R,TX/R) is a decomposition space of(X,TX).

Steps 1, 2, 3 and Definition 12.10.

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Example 12.14A circle is a quotient space of a closed interval. We say that S1 is obtained from [0, 2π] by 0 ∼ 2π.Proof.


1. f : [0, 2π] → S1, f(t) = (cos t, sin t) is aclosed map.

Homework 12.7(3a).

2. f is continuous. cos and sin are continuous.

3. f is a surjection. For (x, y) ∈ S1, if y ≥ 0, then t =arccos(x); if y < 0, then t = π+arccos(x).

4. S1 is homeomorphic to a quotient space of[0, 2π].

Theorem 12.8.

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Example 12.15A cylinder is a quotient space of a rectangle. By Example 12.14, S1 is a quotient space of [0, 2π].A cylinder C ∼= S1 × [a, b] is a quotient space of [0, 2π]× [a, b] by (0, v) ∼ (2π, v), a ≤ v ≤ b.

Example 12.16A two dimensional torus is a quotient space of a rectangle. By Example 12.14, S1 is a quotientspace of [0, 2π]. So S1 × S1 is a quotient space of [0, 2π]× [0, 2π] by

(u, 0) ∼ (u, 2π), 0 ≤ u ≤ 2π; (0, v) ∼ (2π, v), 0 ≤ v ≤ 2π.

We show that S1 × S1 is homeomorphic to a torus

T = {(x, y, z) ∈ R3 : (√x2 + y2 − r2)2 + z2 = r2

1}, (0 < r1 < r2).

Proof. The map f : S1 × S1 → T ,

f((cosu, sin u), (cos v, sin v)) = ((r2 + r1 cosu) cos v, (r2 + r1 cosu) sin v, r1 sin u)

and its inverse f−1 : T → S1 × S1,

f−1(x, y, z) =((√

x2 + y2 − r2

r1,z

r1

),

(x√

x2 + y2 ,y√

x2 + y2

)).

are both elementary and hence continuous. Therefore, f is a homeomorphism. �

Homework 12.17Calculate the surface area and the volume of the torus in Example 12.16. (This is a question inCalculus, not in topology.)

Example 12.18Suppose 0 < r1 < r2. The quotient space of [0, 2πr2]× [−r1, r1] obtained by

(0, v) ∼ (2πr2,−v), −r1 ≤ v ≤ r1

is called the Möbius strip. It has only one side.The Möbius strip is homeomorphic to the cubic surface in R3 (we omit its proof)

{(x, y, z) ∈ R3 : y(y − z)2 + 2x(y − z)(x+ r2)− y(x+ r2)2 = 0}.

108 ZHENGYAO WU

13. December 2nd, Cone, suspension, non-Hausdorff, path connected

Example 13.1The Klein bottle is the quotient space of [0, 2π]× [0, 2π] by

(0, v) ∼ (2π, v), 0 ≤ v ≤ 2π; (u, 0) ∼ (2π − u, 2π), 0 ≤ v ≤ 2π.

The Klein space is a quotient space of a cylinder, it is also a quotient space of a Möbius strip.The Klein bottle has only one side; it is not homeomorphic to a subspace of R3; it is homeomorphicto a subspace of R4. (We omit the proof of the last two facts.)

Example 13.2The following three quotient spaces X1/R1, X2/R2 and X3/R3 are homeomorphic to each other.(1) X1 = R3 − {(0, 0, 0)}; R1 : (a1, a2, a3) ∼ (b1, b2, b3) if there exists λ ∈ R∗ such thatλ(a1, a2, a3) = (b1, b2, b3). We call RP2 = X1/R1 the real projective plane.(2) X2 = S2 = {(x1, x2, x3) ∈ R3 : x2

1 + x22 + x2

3 = 1}; R2 : (a1, a2, a3) ∼ (−a1,−a2,−a3) in S2.(3) X3 = [0, 2π]× [0, 2π]; R3 : (0, v) ∼ (2π, 2π−v), 0 ≤ v ≤ 2π; (u, 0) ∼ (2π−u, 2π), 0 ≤ v ≤ 2π.

Definition 13.3Let X be a topological space. Let X × [0, 1] be the product space. The cone CX over X is thequotient space of X × [0, 1] by (x, 0) ∼ (x′, 0) for all x, x′ ∈ X. We write CX = (X × [0, 1])/(X ×{0}).

Example 13.4The cone over a point is a line segment; The cone over a line interval is a filled-in triangle.The cone over a disc is a solid three dimensional cone.The cone over a circle S1 × {0} = {(x, y, 0) ∈ R3 : x2 + y2 = 1} is {(x, y, z) ∈ R3 : x2 + y2 =(z − 1)2, 0 ≤ z ≤ 1}, which is homeomorphic to a closed disc.The cone over Sn = {(x1, . . . , xn+1) ∈ Rn+1 : x2

1 + · · · + x2n+1 = 1} is homeomorphic to Dn+1 =

{(y1, . . . , yn+1) ∈ Rn+1 : y21 + · · ·+ y2

n+1 ≤ 1}.

Lemma 13.5Let X be a quasi-compact topological space. Let Y be a Hausdorff topological space. If f : X → Y

is a continuous bijection, then it is a homeomorphism.Proof.


1. If F is closed in X, then F is quasi-compact.

X is quasi-compact and Proposition 8.19.

2. f(F ) is quasi-compact. f is continuous and Proposition 8.11.



3. f(F ) is closed in Y . Y is Hausdorff and Proposition 8.12.

5. f−1 is continuous. Homework 5.7(2c).

6. f is a homeomorphism. f is a continuous bijection, step 5 and Def-inition 6.9.

�

Homework 13.6Let X be a compact subspace of the Euclidean space Rn. Suppose y ∈ Rn−X. For each x ∈ X, letLxy be the line segment with end points x and y. Suppose that for all x 6= x′ in X, Lxy∩Lx′y = {y}.Prove that the cone CX is homeomorphic to the subspace ⋃

x∈XLxy of Rn.

Definition 13.7Let X be a topological space. Let X × [0, 1] be the product space. The suspension SX over Xis the quotient space of X × [0, 1] by (x, 0) ∼ (x′, 0) for all x, x′ ∈ X and (y, 1) ∼ (y′, 1) for ally, y′ ∈ X. We write SX = (X × [0, 1])/ ∼.

Example 13.8The suspension of Sn is homeomorphic to Sn+1.

Example 13.9If CardX ≥ 2 has the trivial topology, then X is not Hausdorff. In fact, for all x 6= y in X,V(x) = V(y) = {X}.

Example 13.10The line with two origins is the quotient space of X = R × {0, 1} by S : (x, 0) ∼ (x, 1) for allx ∈ R − {0}. Then X/S is not Hausdorff.Let ϕ : X → X/S be the canonical map. In fact, a basis of V(ϕ((0, 0))) is ϕ((−ε1, ε1)×{0}), (ε1 >

0); a basis of V(ϕ((0, 1))) consists of images of ϕ((−ε2, ε2)× {1}), (ε1 > 0). However,

ϕ((−ε1, ε1)× {0}) ∩ ϕ((−ε2, ε2)× {1}) = ϕ(((−ε3, 0) ∪ (0, ε3))× {0}) 6= ∅, ε3 = min(ε1, ε2).

Homework 13.11Prove that the line of two origins is second countable.

110 ZHENGYAO WU

Proposition 13.12Let (X,TX) be a topological space. Let R be an equivalence relation on X and ϕ : X → X/R thecanonical map. Let (X/R,TX/R) be the quotient space.(1) If X/R is Hausdorff, then R is closed in X ×X.(2) If ϕ is open and R is closed in X ×X, then X/R is Hausdorff.

Proof. (1)


1. ∆X/R is closed in X/R×X/R. X/R is Hausdorff and Proposi-tion 5.17(H2)

2. R = (ϕ× ϕ)−1(∆X/R). (x, y) ∈ R iff ϕ(x) = ϕ(y)

3. R is closed in X ×X. ϕ×ϕ is continuous and Homework 5.7(2c)

(2) Define R × R : (x, y) ∼ (x′, y′) iff x ∼ x′ and y ∼ y′. It is an equivalence relation on X ×X.We have a commutative diagram. Let ψ : X ×X → (X ×X)/(R×R) be the canonical surjection.

X ×Xϕ×ϕ

''

ψ

vv(X ×X)/(R×R)

f// (X/R)× (X/R)


1. There exists a homeomorphism f : (X ×X)/(R×R)→ (X/R)× (X/R).

f(ψ(x, y)) = (ϕ(x), ϕ(y)), Definition 6.9and below.

1.1. f is well-defined. ψ is surjective. If (x, y) ∼ (x′, y′), thenx ∼ x′ and y ∼ y′.

1.2. f is injective. If x ∼ x′ and y ∼ y′, then (x, y) ∼ (x′, y′).

1.3. f is surjective. ϕ× ϕ is surjective.

1.4. f is continuous. ϕ× ϕ is continuous and Theorem 12.9.

1.5. ϕ× ϕ is open. ϕ is open and Definition 5.14.

1.6. f is open. ψ is continuous, Homework 5.7(2d) andstep 1.5.



2. ∆X/R is closed in (X/R)× (X/R). (ϕ× ϕ)−1(∆X/R) = ψ−1(f−1(∆X/R)) = R

is closed in X ×X; Lemma 12.5(2) for ψ.

3. X/R is Hausdorff. Proposition 5.17(H2)

�

Lemma 13.13Let (X,TX) be a topological space. Let R be an equivalence relation on X and ϕ : X → X/R

the canonical map. Let (X/R,TX/R) be the quotient space. If X is quasi-compact, then X/R isquasi-compact.Proof.


1. ϕ is continuous. Definition 12.4(1).

2. ϕ is surjective. Definition 12.3(3).

3. X/R = ϕ(X) is quasi-compact. X is quasi-compact and Proposition 8.11.

�

Example 13.14Circles, cylinders, two dimensional tori, the Möbius strip, the Klein bottle and the real projectiveplane are compact.In fact, they are quasi-compact since they are quotient spaces of a compact space [0, 2π] or[0, 2π] × [0, 2π] and Lemma 13.13; they are Hausdorff since they are subspaces of some Rn andProposition 9.3. So they are compact by Definition 8.9.

Lemma 13.15Let (X,TX) be a topological space. Let R be an equivalence relation on X and ϕ : X → X/R thecanonical map. Let (X/R,TX/R) be the quotient space. If X is connected, then X/R is connected.Proof.

112 ZHENGYAO WU


1. ϕ is continuous. Definition 12.4(1).

2. ϕ is surjective. Definition 12.3(3).

3. X/R = ϕ(X) is connected. X is connected and Proposition 10.23.

�

Definition 13.16Let X be a topological space. A path on X is a continuous map c : [0, 1] → X with initial pointc(0), terminal point c(1). In particular, if c(0) = c(1), then we call c a loop.

Example 13.17(1) Let c, d be paths. If c(1) = d(0), then c ∗ d : [0, 1]→ X

(c ∗ d)(t) =

c(2t), 0 ≤ t ≤ 1

2 .

d(2t− 1), 12 ≤ t ≤ 1.

is the juxtaposed path of c and d.(2) Let c be a path. Its opposite path is c(t) = c(1− t), t ∈ [0, 1].

Definition 13.18A topological space X is path connected if for all x, y ∈ X, there exists a path c : [0, 1] → X

such that c(0) = x and c(1) = y.

Example 13.19Every convex subspace of Rn is path connected. (A subset C of Rn is convex if for all x, y ∈ C,their line segment is a subset of C. For example, a filled-in triangle.)

Proposition 13.20Every path connected topological space X is connected.Proof.


1. Take x0 ∈ X. For all x ∈ X, there existsa path cx such that cx(0) = x0, cx(1) = x.

X is path connected and Definition 13.18

2. cx([0, 1]) is connected. Example 11.1 and Proposition 10.23



3. X = ⋃x∈X

cx([0, 1]) is connected. x0 ∈⋂x∈X

cx([0, 1]) 6= ∅ and Proposi-tion 10.22

�

Lemma 13.21Let X be a topological space. If A ⊂ X is connected and A ⊂ B ⊂ cl(A), then B is connected.Proof.


1. If B is disconnected, then there exists Cand D open in X such that B ∩ C 6= ∅,B ∩D 6= ∅, C ∩D = ∅ and B ⊂ C ∪D.

Definition 10.20.

2. A ∩ C 6= ∅ and A ∩D 6= ∅. B ⊂ cl(A).

3. A is disconnected, a contradiction. Step 2, C ∩D = ∅ and A ⊂ B ⊂ C ∪D.

�

Example 13.22Let G be the graph of y = sin 1

x, x ∈ (0, 1]. We call G ∪ {(0, 0)} topologists’ sine curve. It is

connected and not path connected.

Proof. Since x 7→ (x, sin 1x

) is continuous and (0, 1] is connected, by Proposition 10.23, G is con-

nected. Since limn→∞

( 1nπ

, sin 11/nπ ) = (0, 0), (0, 0) ∈ cl(G) by Proposition 11.14. By Lemma 13.21,

X is connected. Since limx→0+

sin 1x

does not exist, for all x0 ∈ (0, 1], there is no path between (0, 0)

and (x0, sin1x0

). Therefore G ∪ {(0, 0)} is not path connected. �

Homework 13.23In the Eulidean plane R2,(1) Find path connected subspaces A1 and B1 such that A1 ∩B1 is path connected.(2) Find path connected subspaces A2 and B2 such that A2 ∩B2 is not path connected.(3) Find path connected subspaces A3 and B3 such that A3 ∪B3 is path connected.

114 ZHENGYAO WU

(4) Find path connected subspaces A4 and B4 such that A4 ∪B4 is not path connected.

Lemma 13.24Let (X,TX) be a topological space. Let R be an equivalence relation on X and ϕ : X → X/R thecanonical map. Let (X/R,TX/R) be the quotient space. If X is path connected, then X/R is pathconnected.Proof.


1. For all a, b ∈ X/R, there exist x, y ∈ X

such that ϕ(x) = a and ϕ(y) = b.ϕ is surjective by Definition 12.4.

2. There exists a continuous c : [0, 1] → X

such that c(0) = x and c(1) = y.X is path connected and Definition 13.18.

3. ϕ ◦ c : [0, 1]→ X/R is continuous. c is continuous by step 2; ϕ is continuousby Definition 12.4(1).

4. ϕ ◦ c(0) = ϕ(x) = a, ϕ ◦ c(1) = ϕ(y) = b. Steps 1 and 2.

5. X/R is path connected. Definition 13.18.

�

Definition 13.25A topological space X is normal if it is Hausdorff and(OV). If A 6= ∅ and B 6= ∅ are closed in X and A ∩ B = ∅, then there exists a continuous mapf : X → [0, 1] such that f(A) = {0} and f(B) = {1}.

Lemma 13.26Let X be a normal topological space. Let A be a closed subset of X. For all continuous u : A →[−1, 1], there exists a continuous v : X → [−1

3 ,13 ] such that |u(x)− v(x)| ≤ 2

3 for all x ∈ A.Proof.


1. Let B = u−1([−1,−13 ]) and C =

u−1([13 , 1]). Then B,C are closed in A.

u is continuous and Homework 5.7(2c)

2.1. B,C are closed in X. A is closed in X and Homework 8.18



2.2. B ∩ C = ∅. [−1,−13 ] ∩ [1

3 , 1] = ∅

3.1. [−13 ,

13 ] is homeomorphic to [0, 1]. [−1

3 ,13 ]→ [0, 1], x 7→ 3x+1

2 is a homeomor-phism.

3.2. There exists a continuous v : X → [−13 ,

13 ]

such that v(B) = {−13}, v(C) = {1

3}.(OV).

4.1. |u(x)− v(x)| ≤ 23 for all x ∈ B ∪ C. 1− 1

3 = −13 − (−1) = 2

3

4.2. |u(x)− v(x)| ≤ 23 for all x ∈ A− (B ∪C). 1

3 − (−13) = 2

3

�

Theorem 13.27 UrysohnLet X be a topological space. The following are equivalent: (OV).(OV

′) For all A,B closed in X with A ∩ B = ∅, there exist open U ∈ V(A) and open V ∈ V(B)such that U ∩ V = ∅.(OV

′′) For all A closed in X and for all open U ∈ V(A), there exists an open W ∈ V(A) such thatcl(W ) ⊂ U .(U) If A,B are closed in X and A ∩B = ∅, then there exists a family of open subsets (U(t))t∈[0,1]

of X such that (i) A ⊂ U(0); (ii) B ⊂ X − U(1); (iii) For all t < t′, cl(U(t)) ⊂ U(t′).(OV

′′′) Tietze extension. For all A closed in X and for all continuous f : A→ [−∞,+∞], thereexists a continuous g : X → [−∞,+∞] such that g|A = f .Proof. (OV) implies (OV

′). Let U = f−1([0, 12)) and V = f−1((1

2 , 1]).


1. U and V are open. [0, 12) and (1

2 , 1] are open in [0, 1], f is con-tinuous and Homework 5.7(4)

2. U ∈ V(A), V ∈ V(B). 0 ∈ [0, 12), 1 ∈ (1

2 , 1] and Proposition 2.7.

3. U ∩ V = ∅ [0, 12) ∩ (1

2 , 1] = ∅.

(OV′) implies (OV

′′).

116 ZHENGYAO WU


1. B = X − U is closed and A ∩B = ∅. U is open and A ⊂ U .

2. There exists open W ∈ V(A) and openV ∈ V(B) such that W ∩ V = ∅.

(OV′).

3. cl(W ) ⊂ X − V ⊂ X −B = U W ⊂ X − V is closed and B ⊂ V .


14. December 9, Urysohn theorem, Tietze extension, Connected component,Cantor set

(OV′′) implies (U). For n, k ∈ N, we call k

2n a dyadic number. Dyadic numbers in [0, 1] are dense.


(ii). Define U(1) = X −B. It is open in X. B is closed in X.

1. A ⊂ U(1). A ∩B = ∅.

(i) and There exists an open U(0) ∈ V(A). U(1) ∈ V(A) and (OV′′)

(iii) 0 < 1. cl(U(0)) ⊂ U(1).

2. Suppose for all k ∈ [0, 2n] ∩ N, U( k2n ) is

defined and cl(U( k2n )) ⊂ U(k+1

2n ).

3. There exists U(2k+12n+1 ) ∈ V(cl(U( k

2n ))) suchthat cl(U(2k+1

2n+1 )) ⊂ U(k+12n )

U(k+12n ) ∈ V(U( k

2n )) and (OV′′)

(iii) dyadic. For all dyadic r < r′, cl(U(r)) ⊂ U(r′). Induction, steps 2,3 and 2k2n+1 = k

2n

4. For all r ∈ [0, 1], define U(r) =⋃t≤r, t dyadic

U(t). It is increasing.

5. There exists dyadic t, t′ such that r ≤ t <

t′ ≤ r′.Dyadic numbers in [0, 1] are dense.

(iii). For all 0 ≤ r < r′ ≤ 1, cl(U(r)) ⊂ U(r′). cl(U(r)) ⊂ cl(U(t)) ⊂ U(t′) ⊂ U(r′) bythe dyadic case and steps 4 and 5.

(U) implies (OV′). Define

f : X → [0, 1], f(x) =

1, x ∈ X − U(1);

inf{t : x ∈ U(t)}, x ∈ U(1).

Define U(t) = X if t > 1; U(t) = ∅ if t < 0.


1. f(A) = {0}. (U)(i)

2. f(B) = {1}. (U)(ii)

118 ZHENGYAO WU


3. We show that f is continuous at x. below.

3.1. For all ε > 0, cl(U(a − ε)) ⊂ U(a + ε)where f(x) = a.

(U)(iii).

3.2. V = U(a+ε)∩ (X−cl(U(a−ε))) ∈ V(x). Lemma 2.8(VII)

3.3. For all y ∈ V , f(y) ≤ a+ ε. y ∈ U(a+ ε) and

f(y) ≥ a− ε. So |f(y)− a| ≤ ε. y 6∈ cl(U(a− ε)).

(OV′′′) implies (OV). Suppose B,C are nonempty, closed in X and B ∩ C = ∅.


1. Define f : B ∪ C → {0, 1} such thatf(B) = {0}, f(C) = {1}. Then f is con-tinuous.

Homework 5.7(2c).

2. There exists a continuous g : X →[−∞,+∞] such that g|B∪C = f .

(OV′′′)

3.1 Define h(x) = min(max(g(x), 0), 1). Thenh is continuous.

g, 0, 1 are continuous and max,min pre-serves continuity

3.2 h(X) ⊂ [0, 1]. defn of h

3.3 h(B) = {0} and h(C) = {1}. f(B) = {0} and f(C) = {1}

(OV) implies (OV′′′). Since there is a homeomorphism [−1, 1]→ [−∞,+∞] given by x 7→ tan(πx2 ),

we may assume that f : A→ [−1, 1] and need to construct the extension g : X → [−1, 1].Next, we construct a sequence of continuous functions (gn : X → R)n∈N such that for all x ∈ X,limn→∞

gn(x) exists and lies in [−1, 1].


1. For f , ∃g0 : X → [−13 ,

13 ], ∀x ∈ A, |f(x)−

g0(x)| ≤ 23 .

Lemma 13.26

120 ZHENGYAO WU


1. {U ∩V : U ∈ V(A), V ∈ V(B)} is a basisof a filter F on X.

Lemma 2.15.

2. F has a cluster point x ∈ X. X is compact and Definition 8.9(C).

3. X is regular. Compact implies regular byLemma 9.9(2).

4. For all y ∈ X−A, there exists UA ∈ V(A)and Uy ∈ V(y) such that UA ∩ Uy = ∅.

Definition 6.1(OIII′)

5. y is not a cluster point of F. y ∈ int(X−UA) = X−cl(UA) and UA ∈ F

6. y 6= x. Thus x ∈ A. Similary, x ∈ B. Steps 2 and 5.

7. A contradiction to A ∩B = ∅. x ∈ A ∩B.

Finally, X is normal since it is Hausdorff and Theorem 13.27. �

Proposition 14.2Every metrizable space is normal.

Proof. Suppose (X,T) is a metrizable space with a metric d compatible with T. We verify (OV’).Suppose A,B are disjoint and closed in X.


1. X → [0,+∞)2, x 7→ (d(x,A), d(x,B)) iscontinuous.

Lemma 10.16.

2. U = {x ∈ X : d(x,A) < d(x,B)} is openin X.

Step 1, {(x, y) ∈ [0,+∞)2 : x < y} isopen in [0,+∞)2,

V = {x ∈ X : d(x,A) > d(x,B)} is openin X and U ∩ V = ∅.

{(x, y) ∈ [0,+∞)2 : x > y} is open in[0,+∞)2 and Homework 5.7(2d).

3. For all a ∈ A, d(a,B) > 0. B is closed and Lemma 10.14.

For all b ∈ B, d(b, A) > 0. A is closed and Lemma 10.14.

4. U ∈ V(A) and V ∈ V(B). Steps 2 and 3.



5. X is normal. X is Hausdorff by Lemma 2.22; (OV’)holds, and Theorem 13.27.

�

Proposition 14.3Every normal space is completely regular.

Proof. We mainly show that (OV) implies (OIV).


1. X is Hausdorff. Definition 13.25.

2. For all x0 ∈ X, {x0} is closed in X. Homework 6.6(1).

3. For all V ∈ V(x0), there exists open U ∈V(x0) such that x0 ∈ U ⊂ V .

Definition 2.6.

4. There exists a continuous f : X → [0, 1]such that f(x0) = 0 and f(X −U) = {1}.

Steps 2, 3 and Definition 13.25(OV)

5. (OIV) holds since f(X − V ) = {1}. X − V ⊂ X − U .

6. X is uniformizable. Theorem 8.4.

7. X is completely regular. Steps 1, 6 and Definition 8.5.

�

Definition 14.4We define inductively intervals {In,p : n ∈ Z, p ∈ Z, n ≥ 0, 1 ≤ p ≤ 2n}.For n = 0, define I0,1 = (1

3 ,23). Suppose for all 0 ≤ n ≤ m, In,p are defined such that

• [0, 1]−m⋃n=0

2n⋃p=1

In,p =2m+1⋃p=1

Km,p.

• Km,i are disjoint closed intervals of length 13m+1 .

If Km,p = [a, b], define Im+1,p = (2a+b3 , a+2b

3 ). We call K = [0, 1] −∞⋃n=0

2n⋃p=1

In,p Cantor’s triadic

set. Hence K consists of elements of [0, 1] with a triadic expansion∞∑i=1

xi

3i , xi ∈ {0, 2}.

122 ZHENGYAO WU

Homework 14.5The Cantor’s triadic set K is compact as a subspace of the real line R.

Definition 14.6Let X be a topological space and x ∈ X. The connected component of x is the largestconnected subset of X containing x.

Example 14.7If X is connected and x ∈ X, then the connected component of x is X itself.

Lemma 14.8If for all x, y ∈ X, there exists a connected A such that {x, y} ⊂ A ⊂ X, then X is connected.

Proof. Let x ∈ X and C is the connected component of x. If X−C 6= ∅, then take y ∈ X−C andthere exists a connected A such that {x, y} ⊂ A ⊂ X. By Proposition 10.22, C ∪ A is connected.But C ( C ∪ A, a contradiction. Hence X = C. �

Definition 14.9A topological space X is totally disconnected if every x ∈ X has connected component {x}.

Example 14.10Every discrete space is totally disconnected.The rational line Q is totally disconnected but not discrete.

Proposition 14.11Let X be a topological space.(1) For all x ∈ X, the connected component Cx of x is closed.(2) R = {(x, y) ∈ X ×X : x ∈ Cy} is an equivalence relation; Cx is the equivalence class of x.(3) X/R is totally disconnected.

Proof. (1)


1. cl(Cx) is connected. Cx is connected and Lemma 13.21

2. cl(Cx) = Cx. Cx is maximal by Definition 14.6

3. Cx is closed. Lemma 3.13(2)

(2) By Definition 14.6, (x, y) ∈ R iff Cx = Cy.(3) Let F be the connected component of Cx in X/R. Let ϕ : X → X/R be the canonical map.



1. If {Cx} ( F , then ϕ−1(F ) is disconnected. Cx ( ϕ−1(F ) and Definition 14.6.

2. Suppose A,B are nonempty closed in X,ϕ−1(F ) = A ∪B, A ∩B = ∅.

Lemma 10.21

3. A = ϕ−1(ϕ(A)) and B = ϕ−1(ϕ(B)). If a ∈ A, then Ca ⊂ A since A is bothopen and closed in ϕ−1(F ).

4. ϕ(A), ϕ(B) are closed. Lemma 12.5(2)

5. ϕ(A) ∪ ϕ(B) = F . A ∪B = ϕ−1(F )

6. ϕ(A) ∩ ϕ(B) = ∅. A ∩B = ∅ and step 3.

7. a contradiction. F is connected and Definition 10.20.

�

Homework 14.12The Cantor’s triadic set K is totally disconnected as a subspace of the real line R.

124 ZHENGYAO WU

15. December 16, Subbasis, isolated, perfect, Stone-Čech compactification

Lemma 15.1Let X be a compact topological space. For all x ∈ X, the connected component of x is theintersection of all open and closed subsets of X containing x.

Proof. Let Cx be the connected component of x. Let U run through all open and closed subsetsof X containing x. We need to show that Cx = ⋂

U . The proof of Cx ⊂⋂U does not need the

condition “X is compact” and only use that Cx is “connected”.


1. If Cx − U 6= ∅, then it is open in Cx. Cx − U = Cx ∩ (X − U) and U is closed.

2. Cx ∩ U 6= ∅ and is open in Cx. x ∈ Cx ∩ U and Definition 2.1(OII).

3. Cx = (Cx ∩ U) ∪ (Cx − U). X = U ∪ (X − U).

4. (Cx ∩ U) ∩ (Cx − U) = ∅. U ∩ (X − U) = ∅.

5. A contradiction. So Cx ⊂ U for all U . Cx is connected and Definition 10.20.

Conversely, we show that ⋂U ⊂ Cx.


1. If ⋂U is disconnected, then ⋂U ⊂ A∪B,where A,B are closed in X, disjoint, A ∩(⋂U) 6= ∅ and B ∩ (⋂U) 6= ∅.

Lemma 10.21.

2. ⋂U is closed in X. So A∩ (⋂U) and B ∩

(⋂U) are closed in X.U is closed in X.

3. X is normal. Proposition 14.1.

4. There exists open V ∈ V(A ∩ (⋂U)) andopenW ∈ V(B∩(⋂U)) such that V ∩W =∅.

Theorem 13.27(OV’)

5. There exists open and closed subsetsU1, . . . , Un containing x such that U1∩· · ·∩Un ⊂ V ∪W .

X is compact, (⋂U)∩(X−V )∩(X−W ) =∅ and Proposition 8.10(C2).



6. U1∩· · ·∩Un∩V = U1∩· · ·∩Un∩(X−W )is open and closed.

Definition 2.1(OII).

7. If x ∈ A, then B∩ (⋂U) = ∅, a contradic-tion to step 1.

⋂U ⊂ U1 ∩ · · · ∩ Un ∩ V .

8. ⋂U ⊂ Cx.

⋂U is connected and Definition 14.6.

�

Definition 15.2Let (X,T) be a topological space and S ⊂ T. Let B be the set of finite intersections of elements ofS. If B is a basis of T, then we call S a subbasis of the topology T.

Example 15.3Let X and Y be topological spaces. Let C(X, Y ) be the set of continuous maps X → Y . IfK ⊂ X is compact and U ⊂ Y is open, we write V (K,U) = {f ∈ C(X, Y ) : f(K) ⊂ U}. LetS = {V (K,U) : K ⊂ X compact, U ⊂ Y open}. The compact-open topology T on C(X, Y )is the topology with subbasis S.

Theorem 15.4Let (Xi,Ti)i∈I be a family of topological spaces. Let (X = ∏

i∈IXi,TX) be their product space.

(1) TX has a subbasis S = {Ui ×∏

j∈I−{i}Xj : Ui ∈ Ti}.

(2) A map g : Z → X is continuous iff pri ◦g : Z → Xi is continuous for all i ∈ I.

Proof. (1) Let T′ be the topology of unions of finite intersections of S.


1. ∀i ∈ I, pri : (X,TX)→ (Xi,Ti) is contin-uous.

Definition 5.14(1).

2. S ⊂ TX . So T′ ⊂ TX . Homework 5.7(2d), Definition 2.1(OI, II).

3. ∀i ∈ I, pri : (X,T′)→ (Xi,Ti) is continu-ous.

Homework 5.7(2d).

4. T′ ⊂ TX . Definition 5.14(2).

5. T′ = TX . Steps 2 and 4.

126 ZHENGYAO WU

(2) By Definition 5.14(1), pri : (X,TX) → (Xi,Ti) is continuous for all i ∈ I. If g is continuous,then pri ◦g is continuous for all i ∈ I.Conversely, suppose pri ◦g is continuous for all i ∈ I.


1. Suppose U = Ui ×∏

j∈I−{i}Xj.

(pri ◦g)−1(Ui) = g−1(pr−1i (Ui)) = g−1(U)

is open in Z.

pri ◦g is continuous and Home-work 5.7(2d).

2. g is continuous. Homework 5.7(2d) and (1).

�

Homework 15.5The Cantor’s triadic set K as a subspace of the real line R is homeomorphic to the product space{0, 2}N where {0, 2} has the discrete topology. (Hint: Lemma 13.5)

Definition 15.6Let X be a topological space and x ∈ A ⊂ X. We call x an isolated point of A if there existsV ∈ V(x) such that V ∩ A = {x}. A closed set without isolated points is called perfect.

Example 15.7Let X be a topological space. A point x is isolated in X iff {x} is open.

Homework 15.8The Cantor set K is perfect as a subspace of the real line R.

Lemma 15.9Let X be a compact, totally disconnected, perfect topological space. Every open subset of X withat least two elements is disconnected.Proof.


1. Suppose U is open in X and Card(U) > 1.Suppose x 6= y in U .

X is perfect.



2. There exists V open and closed in X suchthat x ∈ V and y 6∈ V .

{x} = ⋂V since X is compact, totally

disconnected and Lemma 15.1.

3. U ∩ V is open. U, V are open and Definition 2.1(OII)

U − V = U ∩ (X − V ) is open. V is closed and Definition 2.1(OII)

4. U is the disjoint union of nonepmty openU ∩ V and U − V .

Steps 2, 3.

�

Theorem 15.10Let K be the Cantor set K as a subspace of the real line R.(1) Every compact, totally disconnected, perfect, metric space is homeomorphic to K.(2) For every compact metric space X, there is a continuous surjection f : K → X.The proof uses the notion “inverse limit of topological spaces”, which we omit.

Proposition 15.11Let X be a topological space. There exists a compact topological space Z and a continuous mapϕ : X → Z such that(1) For all compact topological space Y and for all continuous map g : X → Y , there exists aunique continuous map h : Z → Y such that g = h ◦ ϕ.(2) The pair (Z, ϕ) is unique up to homeomorphism.(3) Let Φ be the set of all continuous maps X → [0, 1]. The map ϕ is injective iff for all a 6= b

in X, there exists f ∈ Φ such that f(a) = 0 and f(b) = 1. (Review that ϕ : X → ϕ(X) is ahomeomorphism iff X is completely regular by Proposition 10.10. )We write βX = Z and call it the Stone-Čech compactification of X.

Proof. Construction: Let ϕ : X → [0, 1]Φ, x 7→ (f(x))f∈Φ. Let Z = cl(ϕ(X)).


1. [0, 1]Φ is compact. [0, 1] is compact and Theorem 9.6.

2. Z is compact. Definition 8.9 and below.

2.1. Z is quasi-compact. [0, 1]Φ is quasi-compact, Z is closed andProposition 8.19.

128 ZHENGYAO WU


2.2. Z is Hausdorff. [0, 1]Φ is Hausdorff and Proposition 9.3.

3. ϕ : X → [0, 1]Φ is continuous. [0, 1]Φ has the product topology and The-orem 15.4.

4. ϕ : X → Z is continuous. Step 3, ϕ(X) ⊂ Z and Z has the subspacetopology.

(1) A special case: Y = [0, 1] and g : X → [0, 1] is continuous.


1. g = prg ◦ϕ. Def. of ϕ.

2. There exists a continuous map h : Z →[0, 1] such that h|ϕ(X) = prg |ϕ(X).

h = prg |Z .

h is unique. Z = cl(ϕ(X)), [0, 1] is Hausdorff andCorollary 5.18(2).

The general case: Y is compact and g : X → Y is continuous. Let Ψ be the set of all continuousmaps Y → [0, 1].

X

g

��

ϕ// Z

h��

hf

##

� � // [0, 1]Φ

YeY // [0, 1]Ψ

prf// [0, 1]


1. Y is homeomorphic to a subspace of[0, 1]Ψ, eY : Y → [0, 1]Ψ, eY (y) =(f(y))f∈Ψ.

Y is compact, Proposition 10.10.

2. For all f ∈ Ψ, prf ◦eY ◦ g : X → [0, 1]extends uniquely to hf : Z → [0, 1].

The special case above.



3. There exists a unique continuous h : Z →[0, 1]Ψ such that prf ◦h = hf for all f ∈ Ψ.

[0, 1]Ψ has the product topology and The-orem 15.4.

4. h ◦ ϕ = eY ◦ g. prf ◦h ◦ ϕ = hf ◦ ϕ = prf ◦eY ◦ g for allf ∈ Ψ.

5. prf (h(Z)) = hf (Z) = hf (cl(ϕ(X)) ⊂cl(hf (ϕ(X))) = cl(prf (eY (g(X)))) ⊂cl(prf (eY (Y ))) for all f ∈ Ψ.

hf is continuous and Homework 5.7(2b).

6. h(Z) = ∏f∈Ψ

prf (h(Z))

⊂ ∏f∈Ψ

cl(prf (eY (Y ))) =

cl( ∏f∈Ψ

prf (eY (Y )))

Step 4 and Proposition 3.20.

= cl(eY (Y )) = eY (Y ). eY (Y ) is closed by Proposition 8.12.

(2) Suppose there is another pair (Z ′, ϕ′) satisfying (1).


1. There exists a unique continuous h : Z →Z ′ such that ϕ′ = h ◦ ϕ.

(1) for (Z, ϕ).

There exists a unique continuous h′ : Z ′ →Z such that ϕ = h′ ◦ ϕ′.

(1) for (Z ′, ϕ′).

2. ϕ = h′ ◦ ϕ′ = h′ ◦ h ◦ ϕ. Step 1.

3. Since (h′ ◦ h) |ϕ(X)= Idϕ(X), h′ ◦ h = IdZ . ϕ(X) is dense in Z, Z is Hausdorff andCorollary 5.18(2).

Similarly, h ◦ h′ = IdZ′ .

4. Z and Z ′ are homeomorphic. Steps 1, 3 and Definition 6.9.

(3)

130 ZHENGYAO WU


1.1. If ϕ is injective, there exists f0 ∈Φ, f0(a) 6= f0(b).

ϕ(a) 6= ϕ(b).

1.2. Define f(x) = f0(x)− f0(a)f0(b)− f0(a) . Then f ∈

Φ, f(a) = 0 and f(b) = 1.

Step 1.

2. Conversely, if a 6= b, then ϕ(a) 6= ϕ(b). There exists f ∈ Φ, prf (ϕ(a)) = f(a) =0 6= 1 = f(1) = prf (ϕ(b)).

�

Remark 15.12The following are equivalent (we omit the proof of the equivalence and assume they are true):The axiom of choice: Let I be a set. For all family of nonempty sets (Si)i∈I , there exists afamily (xi)i∈I such that xi ∈ Si for all i ∈ I.The well-ordering principle: Every set S is well-ordered, i.e. S has a total order ≤ such thatfor all ∅ 6= A ⊂ S, A has a least element.Zorn’s lemma: Let (S,≤) be a partially ordered set. If every totally ordered subset of S has anupper bound, then S has a maximal element.

Remark 15.13Transfinite induction: Let (S,≤) be a well-ordered set. For each x ∈ S, let P (x) be a proposi-tion. Let s0 be the least element of S.Step 1: Prove that P (s0) is true;Step 2: For a ∈ S, suppose P (b) is true for all b < a and b ∈ S;Step 3: Prove that P (a) is true.We conclude that P (x) is true for all x ∈ S.Mathemetical induction is a special case of transfinite induction.

Example 15.14Every vector space V over a field K has a basis.Proof.


1. V has a well-ordering ≤. V is a set and the well-ordering principle.



2. B = {v ∈ V : v 6∈ SpanK{w ∈ V : w <

v}} is linearly independent.below.

2.1. Suppose v1 < . . . < vn ∈ B, k1, . . . , kn ∈K and kn 6= 0 and k1v1 + · · ·+ knvn = 0.

If kn = 0, then we delete kn and vn.

2.2. vn 6∈ B, a contradiction. vn ∈ SpanK{w ∈ V : w < v} by step 3.1.

3. SpanK B = V . Transfinite induction below.

3.1. If v0 is the least element of V , then v0 ∈B ⊂ SpanK B.

6 ∃w < v0.

3.2. Suppose v ∈ V and w ∈ SpanK B for allw ∈ V , w < v.

Inductive hypothesis.

3.3. If v ∈ B, then v ∈ SpanK B. v ∈ B ⊂ SpanK B

If v 6∈ B, then v ∈ SpanK B. v ∈ SpanK{w ∈ V : w < v} ⊂ SpanK Bby step 3.2.

Here dimK V is not necessarily countable. �

132 ZHENGYAO WU

16. December 23, Locally finite, refinement, paracompact, Lindelöf, Sorgenfrey

Definition 16.1Let X be a topological space. A family (Ai)i∈I of subsets of X is locally finite if for all x ∈ X,there exists V ∈ V(x) such that {i ∈ I : V ∩ Ai 6= ∅} is finite.

Example 16.2(1) Every finite family of subsets of X is locally finite.(2) In R, the family {(n,+∞) : n = 1, 2, . . .} is infinite and locally finite.(3) In Q, {[ 1

n, 1] : n = 1, 2, . . .} is not locally finite since lim

n→∞1n

= 0.

Theorem 16.3Let X be a normal topological space. Let Y be a closed subset of X. Let (Ai)i∈I be a locally finiteopen cover of Y . Then there exists an open cover (Bi)i∈I of Y such that cl(Bi) ⊂ Ai.

Proof. By the well-ordering principle, the set I has a well-ordering ≤. We define a family (Bi)i∈Iof open subsets of X such that(a) For all i ∈ I, cl(Bi) ⊂ Ai.(b) For all i ∈ I, (Bj)j≤i and (Aj)j>i is an open cover of Y .Suppose i0 is the least element of I. We first show (i)(ii) for i0.


1. C = (X − Y ) ∪ ⋃j>i0

Aj ∈ V(X − Ai0) isopen.

Y is closed, (Aj)j∈I is an open cover of Y .

2. There exists V ∈ V(X − Ai0) open suchthat cl(V ) ⊂ C.

Theorem 13.27(OV′′)

3. Define Bi0 = X−cl(V ). (a) cl(Bi0) ⊂ Ai0 . cl(Bi0) = cl(X − cl(V )) = X −int(cl(V )) = X − V ⊂ Ai0 .

4. Bi0 ∪ C = X. cl(V ) ⊂ C.

5. (b) Bi0 and (Aj)j>i0 is an open cover of Y . Bi0 ∪ (X − Y ) ∪ ⋃j>i0

Aj = X.

Suppose (a)(b) are true for all i < k. We show that (Bj)j<k and (Aj)j≥k cover Y .




1. For each y ∈ Y , there exists i1 < · · · < in

in I such that y ∈ Ai1 ∩ . . . ∩ Ain .(Ai)i∈I is locally finite.

2. y ∈ ⋃j<k

Bj ∪⋃j≥k

Aj. If in ≥ k, then y ∈ Ain ; If in < k, theny ∈ Bj, (j ≤ in) by (b) for in.

Finally, we define Bk and show (a)(b) for k.


1. C = (X−Y )∪ ⋃j<k

Bj∪⋃j>k

Aj ∈ V(X−Ak)is open.

Y is closed, (Bj)j<k and (Aj)j≥k cover Y .

2. There exists V ∈ V(X − Ak) open suchthat cl(V ) ⊂ C.

Theorem 13.27(OV′′)

3. Define Bk = X − cl(V ). (a) cl(Bk) ⊂ Ak. cl(Bk) = cl(X−cl(V )) = X−int(cl(V )) =X − V ⊂ Ak.

4. Bk ∪ C = X. cl(V ) ⊂ C.

5. (b) (Bj)j≤k and (Aj)j>k is an open coverof Y .

Bk ∪ (X − Y ) ∪ ⋃j<k

Bj ∪⋃j>k

Aj = X.

Since (Ai)i∈I is locally finite, for all y ∈ Y , suppose only y ∈ Ai1 ∩ · · · ∩ Ain . Then for k > in,y ∈ Bj for some j ≤ k. Therefore (Bj)j∈I cover Y . �

Definition 16.4Let X be a topological space. A real function f : X → R has support

Supp(f) = cl({x ∈ X : f(x) 6= 0}).

If (fi : X → R)i∈I is a family of real functions and (Supp(fi))i∈I is locally finite, then the sum∑i∈If(x) is well-defined.

134 ZHENGYAO WU

Definition 16.5Let X be a topological space. A family of real functions (fi : X → R)i∈I is a partition of unityon X if (1) fi are continuous and fi ≥ 0; (2) (Supp(fi))i∈I is locally finite; (3) ∑

i∈Ifi(x) = 1 for all

x ∈ X.

Proposition 16.6Let X be a normal topological space. Let (Ai)i∈I be a locally finite open cover of X. There existsa partition of unity (fi)i∈I on X such that Supp(fi) ⊂ Ai for all i ∈ I.Proof.


1. There exists an open cover (Bi)i∈I of Xsuch that cl(Bi) ⊂ Ai.

Theorem 16.3.

2. (Bi)i∈I is locally finite. Bi ⊂ Ai for all i ∈ I and (Ai)i∈I is locallyfinite.

3. For all i ∈ I, there exists an open Ci ∈V(cl(Bi)) such that cl(Ci) ⊂ Ai.

X is normal and Theorem 13.27(OV′′).

4. For all i ∈ I, there exists a continuousgi : X → [0, 1] such that gi(X −Ci) = {0}and gi(cl(Bi)) = {1}.

X is normal and Definition 13.25(OV)

5. Supp(gi) ⊂ cl(Ci) ⊂ Ai. If g(x) 6= 0 then x ∈ Ci.

6. For all x ∈ X, ∑i∈Igi(x) > 0. Steps 1, 2 and 4.

7. Define fj(x) = gj(x)∑i∈Igi(x) . Then (fj)j∈I is a

partition of unity on X.

Definition 16.5 and below.

7.1. fj are all continuous and fj ≥ 0. gj are all continuous and gj ≥ 0.

7.2. (Supp(fj))j∈I is locally finite. Supp(fj) = Supp(gj) step 5 and Ai is lo-cally finite.

7.3. For all x ∈ X, ∑j∈I

fj(x) = 1. The construction in step 7.

8. Supp(fj) ⊂ Aj for all j ∈ I. Supp(fj) = Supp(gj) and step 5.

�


Definition 16.7Let X be a topological space. Let (Ui)i∈I and (Vj)j∈J be two covers of X. If for all i ∈ I, thereexists j ∈ J such that Ui ⊂ Vj, then we call (Ui)i∈I a refinement of (Vj)j∈J .

Example 16.8Let (Ui)i∈I be an open cover of X and J ⊂ I. The subcover (Uj)j∈J is a refinement of the cover(Ui)i∈I . “Refinement” is a transitive relation.

Definition 16.9A topological space X is paracompact if X is Hausdorff and every open cover of X has a locallyfinite open refinement.

Example 16.10(1) Every compact space is paracompact, because a finite subcover is a locally finite refinement.(2) Every discrete space X is paracompact, because {{x} | x ∈ X} is a locally finite refinement ofevery cover of X.(3) Stone’s theorem: Every metrizable space X is paracompact. We omit its proof.

Lemma 16.11Let X be a paracompact space. Let A and B be disjoint closed subsets of X. If for all x ∈ A,there exists an open Vx ∈ V(x) and an open Wx ∈ V(B) such that Vx ∩Wx = ∅, then there existsan open V ∈ V(A) and an open W ∈ V(B) such that V ∩W = ∅.Proof.


1. The open cover (Vx)x∈A, X − A of X hasa locally finite open refinement (V ′i )i∈I .

X is paracompact and Definition 16.9.

2. If A ∩ V ′i 6= ∅, then there exists f(i) ∈ Asuch that V ′i ⊂ Vf(i).

(V ′i )i∈I is a refinement of (Vx)x∈A, X − Aand V ′i 6⊂ X − A.

3. V = ⋃A∩V ′i 6=∅

V ′i ∈ V(A) is open. (A∩V ′i )i∈I is an open cover of A and Def-inition 2.1(OI).

4. For all y ∈ B, there exists an open Sy ∈V(y) such that J = {i ∈ I : Sy∩A∩V ′i 6=∅} is finite.

(V ′i )i∈I is locally finite and Definition 16.1.

5. W ′y = Sy ∩

⋂i∈J

Wf(i) ∈ V(y) is open. Definition 2.1(OII)

6. If A ∩ V ′i 6= ∅, then V ′i ∩W ′y = ∅. V ′i ∩W ′

y ⊂ Vf(i) ∩Wf(i) = ∅.

136 ZHENGYAO WU


7. V ∩W ′y = ∅ for all y ∈ B. Steps 3 and 6.

8. W = ⋃y∈B

W ′y ∈ V(B) is open. (W ′

y)y∈B is an open cover of B and Defi-nition 2.1(OI).

9. V ∩W = ∅. Steps 7 and 8.

�

Proposition 16.12Every paracompact topological space X is normal.Proof.


1. X is Hausdorff. X is paracompact and Definition 16.9.

2. Let A be a closed subset of X and y ∈X − A. For all x ∈ A, there exists anopen Vxy ∈ V(x) and an open Wxy ∈ V(y)such that Vxy ∩Wxy = ∅.

{y} is closed by step 1 and Home-work 6.6(1).

3. There exists an open Vy ∈ V(A) and anopen Wy ∈ V(y) such that Vy ∩Wy = ∅.

Step 2 and Lemma 16.11 for A, {y}.

4. Let A and B be disjoint closed subsets ofX. There exists an open V ∈ V(A) andan open W ∈ V(B) such that V ∩W = ∅.

Step 3 and Lemma 16.11 for B,A.

5. X is normal. Theorem 13.27(OV′).

�

Corollary 16.13Let X be a paracompact space. Let (Ai)i∈I be an open cover of X. There exists a partition ofunity (fi)i∈I on X such that Supp(fi) ⊂ Ai.Proof.



1. The open cover (Ai)i∈I of X has a locallyfinite open refinement (Uj)j∈J .

X is paracompact.

Suppose f : J → I such that Uj ⊂ Af(j) Definition 16.9.

2. X is normal. X is paracompact and Proposition 16.12.

3. There exists a partition of unity (gj)j∈J onX such that Supp(gj) ⊂ Uj for all j ∈ J .

(Uj)j∈J is locally finite and Proposi-tion 16.6.

4. fi = ∑j∈J, f(j)=i

gj is well-defined. (Supp(gj))j∈J is locally finite by Defini-tion 16.5(2).

5. fi is continuous and ≥ 0 for all i ∈ I. gj is continuous and ≥ 0 by Defini-tion 16.5(1).

6. Bi = ⋃j∈J, f(j)=i

Supp(gj) is closed. below.

6.1. For all x ∈ X −Bi, there exists V ∈ V(x)such that {j ∈ J : f(j) = i, V ∩Supp(gj) 6= ∅} is finite.

(Supp(gj))j∈J is locally finite by Defini-tion 16.5(2).

6.2. For all j ∈ J , f(j) = i, X − Supp(gj) ∈V(x) is open.

Supp(gj) is closed by Definition 16.4 andx 6∈ Bi.

6.3. X − Bi ⊃ V ∩ ⋂j∈J, f(j)=i

(X − Supp(gj)) ∈

V(x).

Step 6.1, 6.2 and Lemma 2.8(VII).

7. Supp(fi) ⊂ Bi ⊂ Ai. For all x ∈ X − Bi, fi(x) =∑j∈J, f(j)=i

gj(x) = 0, step 6 and Supp(gj) ⊂

Uj ⊂ Ai for all j ∈ J, f(j) = i.

8. (Bi)i∈I is locally finite. below.

8.1. For all x ∈ X, there existsW ∈ V(x) suchthat H = {j ∈ J : W ∩ Uj 6= ∅} is finite.

(Uj)j∈J is locally finite by step 1.

8.2. For all i ∈ I − f(H), W ∩Bi = ∅. W ∩Bi ⊂ W ∩Supp(gj) ⊂ W ∩Uj = ∅ forall j ∈ J −H.

9. (Supp(fi))i∈I is locally finite. Steps 7 and 8.

138 ZHENGYAO WU


10. For all x ∈ X, ∑i∈Ifi(x) = 1. ∑

i∈Ifi(x) = ∑

i∈I

∑j∈J, f(j)=i

gj(x) =∑j∈J

gj(x) = 1.

11. (fi)i∈I is a desired partition of unity. Steps 5, 7, 9, 10 and Definition 16.5.

�

Definition 16.14A topological space is X Lindelöf if every open cover of X has a countable subcover.

Proposition 16.15If X is second countable, then X is Lindelöf.

Proof. Suppose X has a countable basis U1, U2, . . .. Let (Vi)i∈I be an open cover of X.


1. For all x ∈ X, there exists Vi 3 x. (Vi)i∈I is a cover of X

2. There exists Un such that x ∈ Un ⊂ Vi. U1, U2, . . . is a basis of X and Defini-tion 11.16.

3. (Un : Un ⊂ Vf(n) for some f(n) ∈ I) is acover of X.

Steps 1 and 2.

4. (Vf(n)) is a countable subcover of (Vi)i∈Iof X.

{n ∈ Z>0 : Un ⊂ Vf(n) for some f(n) ∈I} is countable and step 3.

5. X is Lindelöf. Definition 16.14

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Proposition 16.16Let f : X → Y be a continuous map between topological spaces. If X is Lindelöf, then so is f(X).Proof.



1. Let (Vi)i∈I be an open cover of f(X) in Y .Then (f−1(Vi))i∈I is an open cover of X.

f(X) ⊂ ⋃i∈IVi =⇒ X = ⋃

i∈If−1(Vi) and

Homework 5.7(2d).

2. There exists a countable J ⊂ I such that(f−1(Vi))i∈J covers X.

X is Lindelöf and Definition 16.14.

3. (Vi)i∈J covers f(X). X = ⋃i∈J

f−1(Vi) =⇒ f(X) ⊂ ⋃i∈J

Vi.

4. f(X) is Lindelöf. J is countable, step3 and Definition 16.14.

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Proposition 16.17Every regular, Lindelöf space X is paracompact.

Proof. Let C be an open cover of X.


1. For all x ∈ X, there exists Ux, x ∈ Ux ∈C. Then (Ux)x∈X is an open cover of X.

C is an open cover of X.

2. There exist open Vx ∈ V(x), Wx ∈ V(X−Ux) such that Vx ∩Wx = ∅.

X is regular and Definition 6.1(OIII’).

3. cl(Vx) ⊂ Ux. cl(Vx) ⊂ cl(X−Wx) = X− int(Wx) ⊂ Ux.

4. The open cover (Vx)x∈X has a countablesubcover Vx1 , Vx2 , . . . of X.

X is Lindelöf and Definition 16.14.

5. Let Wn = Uxn −n−1⋃k=1

cl(Vxk). Then

W1,W2, . . . is an open cover of X.Uxn −

n−1⋃k=1

Uxk⊂ Wn ⊂ Uxn by step 3. So

∞⋃n=1

Wn =∞⋃n=1

Uxn ⊃∞⋃n=1

Vxn = X.

6. W1,W2, . . . is a refinement of C. Wn ⊂ Uxn ∈ C.

7. W1,W2, . . . is locally finite. For all x ∈ X, if x ∈ Vxi, then x 6∈ Wn for

all n > i.

8. X is paracompact. Steps 5, 6, 7 and Definition 16.9.

140 ZHENGYAO WU

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Definition 16.18A topological space X is locally connected if for all x ∈ X, V(x) has a basis of connectedneighborhoods.

Example 16.19(1) Recall that the real line R is connected. It is also locally connected.(2) Recall that the topologists’ sine curve {(0, 0)} ∪ {(x, y) ∈ R2 : y = sin 1

x, x ∈ (0, 1]} is

connected, not path connected. But it is not locally connected since every V ∈ V((0, 0)) hasinfinitely many connected components.(3) The discrete space {1, 2} is disconnected, and locally connected.

Definition 16.20A topological space X is locally compact if X is Hausdorff and for all x ∈ X, there exists acompact K ∈ V(x).(iff for all x ∈ X, V(x) has a basis of compact neighborhoods. We omit its proof.)

Example 16.21(1) Every compact space is locally compact.(2) Z with the discrete topology is Hausdorff but not quasi-compact. In fact, {{n} : n ∈ Z} is anopen cover without finite subcover. It is locally compact since {n} is compact.(3) Recall that the real line R is not compact. It is locally compact since every x ∈ R has acompact neighborhood [x− 1, x+ 1].

Homework 16.22This homework will not be graded. Students are encouraged to work on (1)-(13) to prepare forthe final exam. Read the references mentioned in (14) if you like.The Sorgenfrey line is a topological space R` = (R,T) such that T has basis {[a, b) : a < b in R}.(1) Let E be the Euclidean topology on R. Prove that E $ T.(2) For all a < b in R, [a, b) is open and closed in R`. (In (R,E) [a, b) is neither open nor closed.)(2’) For all a ∈ R, (−∞, a) is open and closed in R`. (In (R,E), (−∞, a) is open but not closed.)(2”) For all a ∈ R, [a,+∞) is open and closed in R`. (In (R,E), [a,+∞) is closed but not open.)(3) Prove that R` is totally disconnected. So it is disconnected. ((R,E) is path connected, so it isconnected.)(4) Prove that R` is not locally connected. ((R,E) is locally connected.)(5) Prove that if A ⊂ R` is compact, then A is at most countable. ([0, 1] is compact in (R,E) and[0, 1] is uncountable. )(6) Prove that R` is not locally compact. So it is not compact. ((R,E) is locally compact and notcompact. )(7) Prove that R` is Hausdorff. ((R,E) is also Hausdorff.)


(8) Prove that R` is regular. ((R,E) is also regular.)(9) Prove that R` has a countable dense subset Q. (In (R,E), Q is also a countable dense subset.)(10) Prove that R` is first countable. ((R,E) is first countable since it is second countable andHomework 11.20(2).)(11) Prove that R` is not second countable. ((R,E) is second countable.)(12) Prove that R` is Lindelöf. ((R,E) is Lindelöf since it is second countable and Proposi-tion 16.15.)(13) Prove that R` is paracompact. So it is normal. ((R,E) is also paracompact and normal. )(14) Prove that the product space R` × R` is not normal. So it is not paracompact. SeeR.H.Sorgenfrey, On the topological product of paracompact spaces. Bull. Amer. Math. Soc.53 (1947), 631-632. (The Euclidean space R2 is paracompact and normal. See Bourbaki, Generaltopology, Ch.IX, no.5, Th.4)

notes to point-set topology · 2020-04-30 · notes to point-set topology zhengyaowu abstract....

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