MAS3706: Topology

Dr. Zinaida Lykova

School of Mathematics, Statistics and PhysicsNewcastle University

*Room 3.13 in Herschel Building

These lectures concern metric and topological spaces and continuous maps betweentopological spaces. We shall define these terms and study their properties. This materialis a part of point set topology, also called general topology.

You can consult the following books.

1. W. A. Sutherland, “Introduction to Metric and Topological Spaces”, ClarendonPress, Oxford, 1975.

2. J. R. Munkres, “Topology”, Prentice Hall, New York, 2000.

Topology is used in applications, for example, in many parts of Pure Mathematics,in differential equations, in dynamical systems, in mechanics, in physics and in mathe-matical economics.

The web page:http://www.mas.ncl.ac.uk/∼nzal/teaching.html

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Teaching Information - MAS3706

Venue:

Tuesday 11.00 – 12.00, PERB.G.05Thursday 16.00 – 17.00, HERB.G.LT2Thursday 17.00 – 18.00, HERB.G.LT2

Drop-in Sessions:

Thursday 17.00 – 18.00, HERB.G.LT214.02.2019, 7.03.2019, 21.03.2019, 9.05.2019

Office hours:

Tuesday 12.00 – 13.00, HERB.3 Room 3.13Thursday 13.00 – 14.00, HERB.3 Room 3.13

Assessment:

1. Assessment of course work (10%).

Assignments:

Assignments are to be handed in by 10am on Friday:15.02.2019, 8.03.2019, 22.03.2019, 10.05.2019.

2. 2 hours examination at end of Semester (90%).

Past Exam Papers:Exam Papers Online :https://crypt.ncl.ac.uk/exam.papers/Module code: MAS3706 and MAS3209Title: Topology

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MAS3706 Lectures, Tutorials and Drop-in Sessions with dates:Week 1, 29th January Tuesday 11.00 – 12.00, Lecture, PERB.G.05

31st January Thursday 16.00 – 17.00, Lecture, HERB.G.LT231st January Thursday 17.00 – 18.00, Lecture, HERB.G.LT2

Week 2, 5th February Tuesday 11.00–12.00, Lecture, PERB.G.057th February Thursday 16.00–17.00, Lecture, HERB.G.LT27th February Thursday 17.00–18.00, Tutorial 1, HERB.G.LT2

Week 3, 12th February Tuesday 11.00–12.00, Lecture, PERB.G.0514th February Thursday 16.00–17.00, Lecture, HERB.G.LT214th February Thursday 17.00–18.00, Drop-in, HERB.G.LT2

Week 4, 19th February Tuesday 11.00–12.00, Lecture, PERB.G.0521th February Thursday 16.00–17.00, Lecture, HERB.G.LT222th February Thursday 17.00–18.00, Lecture, HERB.G.LT2

Week 5, 26th February Tuesday 11.00–12.00, Lecture, PERB.G.0528th February Thursday 16.00–17.00, Lecture, HERB.G.LT228th February Thursday 17.00–18.00, Tutorial 2, HERB.G.LT2

Week 6, 5th March Tuesday 11.00–12.00, Lecture, PERB.G.057th March Thursday 16.00–17.00, Lecture, HERB.G.LT27th March Thursday 17.00–18.00, Drop-in, HERB.G.LT2

Week 7, 12th March Tuesday 11.00–12.00, Lecture, PERB.G.0514th March Thursday 16.00–17.00, Lecture, HERB.G.LT214th March Thursday 17.00–18.00, Tutorial 3, HERB.G.LT2

Week 8, 19th March Tuesday 11.00–12.00, Lecture, PERB.G.0521st March Thursday 16.00–17.00, Lecture, HERB.G.LT221st March Thursday 17.00–18.00, Drop-in, HERB.G.LT2

Week 9, 26th March Tuesday 11.00–12.00, Lecture, PERB.G.0528th March Thursday 16.00–17.00, Lecture, HERB.G.LT228th March Thursday 17.00–18.00, Lecture, HERB.G.LT2

Week 10, 30th April Tuesday 11.00–12.00, Lecture, PERB.G.052nd May Thursday 16.00–17.00, Lecture, HERB.G.LT22nd May Thursday 17.00–18.00, Tutorial 4, HERB.G.LT2

Week 11, 7th May Tuesday 11.00–12.00, Lecture, PERB.G.059th May Thursday 16.00–17.00, Revision Lecture, HERB.G.LT29th May Thursday 17.00–18.00, Drop-in, HERB.G.LT2

Week 12 - Revision week, 14th May Tuesday 11.00–12.00, Revision Lecture, PERB.G.0516th May Thursday 16.00–17.00, Revision Lecture, HERB.G.LT216th May Thursday 17.00–18.00, Drop-in, HERB.G.LT2

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Contents1. Open Subsets of R and Continuous Functions Page 62. Metric Spaces Page 133. Topological Spaces Page 274. Hausdorff Spaces and Limits Page 355. Closed Sets Page 396. Separation axioms Page 437. Basic Topological Concepts Page 458. Compactness Page 519. Continuity Page 5610. Metric Spaces Again Page 6011. Completeness in Metric Spaces Page 6212. Connectedness Page 6813. Picard’s Theorem Page 70

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1 Open Subsets of R and Continuous FunctionsDefinition 1.1. A subset U of R is open if around each point p of U there is an openinterval (p− h, p+ h) wholly contained in U :

(p− h, p+ h) ⊆ U.

Remark 1.2. You should think of an open set as containing none of the points at theedge of the set.

Example 1.3. Every open interval (a, b) is an open set in R.

Proof. If p ∈ (a, b), let h be the distance from p to the nearest endpoint:

h = min(b− p, p− a).

Then(p− h, p+ h) ⊆ (a, b).

Example 1.4. R is open (obviously).

Example 1.5. The empty set ∅ is vacuously open.

Example 1.6. The closed interval [a, b] is not open in R.

Proof. If you take p to be a then every open interval (p − h, p + h) contains pointsoutside [a, b].

Example 1.7. A singleton is not open in R.

Proof. Every singleton is a closed interval — since {a} = [a, a] — so no singleton isopen.

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The classic ε− δ definition of continuity is the following.

Definition 1.8. Let S be an open subset of R. A function f : S −→ R is continuouson S ⇐⇒ for each a ∈ S and each ε > 0 there is a δ > 0 such that

|x− a| < δ =⇒ |f(x)− f(a)| < ε.

Remark 1.9. Speaking very loosely, f is continuous at a if f(x) is close to f(a)when x is close to a. Working with this definition is not easy: you spend a lot oftime juggling inequalities. Fortunately, there is an equivalent definition of continuity,couched in the language of sets rather than numbers, which is much easier to work with.More importantly, the use of set-theoretic language allows us to port this definition tomany other “spaces”, including those whose “points” are not numbers. The study ofsuch spaces is called topology.

Definition 1.10. Letf : A −→ B

be a function. If T is a subset of B then the inverse image or pre-image of T under fis the subset of A consisting of all the elements of A that f maps into T :

f−1(T ) = {a ∈ A : f(a) ∈ T}.

In other words,a ∈ f−1(T )⇐⇒ f(a) ∈ T.

Example 1.11. If f : R −→ R is defined by f(x) = x2 then

f−1([0, 9]) = [−3, 3],

f−1([−5,−4]) = ∅ (the empty set)

f−1(R) = R.

Note that the use of the notation f−1 does not imply that the function f possesses aninverse. The inverse image notation can be used with any function whatsoever.

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Theorem 1.12. Let S be an open subset of R. A function f : S −→ R is continuous⇐⇒ the inverse image f−1(U) under f of each open set U is open.

That is, f is continuous⇐⇒ for every open subset U in R, f−1(U) is open.

Proof. [=⇒] Suppose f is continuous and let U be open.It is not necessary that each point of U be the image of a point of S. Indeed, f−1(U)

may very well be empty. Recall that the empty set is open.

If a ∈ f−1(U) then f(a) ∈ U . But U is open, so (f(a)− ε, f(a) + ε) ⊆ U for someε > 0.

By continuity, there exists δ > 0 such that

|x− a| < δ =⇒ |f(x)− f(a)| < ε.

Thus(a− δ, a+ δ) ⊆ f−1((f(a)− ε, f(a) + ε))

⊆ f−1(U).

Therefore f−1(U) is open.

[⇐=] Now suppose that the inverse image of each open set is open.

Let a ∈ S and let ε > 0. We have to prove that there is δ > 0 such that

|x− a| < δ =⇒ |f(x)− f(a)| < ε.

The interval (f(a)− ε, f(a) + ε) is open. Therefore its inverse image

f−1((f(a)− ε, f(a) + ε))

is open, and contains a, so, by the definition of open set, it contains an interval (a −δ, a+ δ) about a for some δ > 0. Thus

|x− a| < δ =⇒ x ∈ (a− δ, a+ δ)

=⇒ x ∈ f−1((f(a)− ε, f(a) + ε))

⇐⇒ f(x) ∈ (f(a)− ε, f(a) + ε)

⇐⇒ |f(x)− f(a)| < ε.

Therefore f is continuous at a.

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Remark 1.13. If f is not continuous one can find an open subset U in the codomainof f such that the inverse image f−1(U) of U under f is not open.

Example 1.14. If f : R −→ R is defined by f(x) = −3 for x ≤ 1 and f(x) = x2 forx > 1, then, for U = (−4, 0), the inverse image of U under f

f−1(U) = (−∞, 1]

is not open.

Thus, instead of using the ε− δ definition, we could define a continuous function tobe one for which the inverse image of each open set is open. We have claimed that thisdefinition is easier to work with. Here is a sample.

The composition of two functions

f : A −→ B and g : B −→ C

is the function (g ◦ f) : A −→ C defined by

(g ◦ f)(a) = g(f(a)), a ∈ A.

Theorem 1.15. Letf : R −→ R

andg : R −→ R

be continuous. Then so isg ◦ f : R −→ R.

Proof.

U ⊆ R open =⇒ g−1(U)

open, by continuity of g,=⇒ f−1(g−1(U))

open, by continuity of f,i.e., (g ◦ f)−1(U) is open.

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Definition 1.16. Let S be an open subset of R. A function f : S −→ R is called openif the image of every open subset V of S is open in R.

Remark 1.17. Note that not every continuous map is open. For example, a constantmapf(x) = 7 is not an open map.

Remark 1.18. If we are to generalise the open set definition of continuity to spacesother than R, we need to isolate the factors that make it work in R. It turns out that,rather than any specific properties of individual open sets, it is the way that open setscan be combined that is important. These are summarised in the following theorem.

We use the notation (Uλ)λ∈Λ to denote a family of sets which is indexed by themembers of a set Λ.

In many cases, Λ will be N, the set of positive integers, and then the sets in the fam-ily can be listed as U1, U2, U3, . . . .

The notation allows Λ to be uncountable, for we shall sometimes need to consideruncountable families.

If (Uλ)λ∈Λ is a family of subsets of a set X , then their union is the subset of X⋃λ∈Λ

Uλ =

{x ∈ X : x ∈ Uλ for at least one λ ∈ Λ}

and their intersection is the subset of X⋂λ∈Λ

Uλ = {x ∈ X : x ∈ Uλ for every λ ∈ Λ}

If Λ = N then we write⋃∞n=1 Un and

⋂∞n=1 Un.

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Theorem 1.19. Let τ be the collection of all open subsets of R. Then

1. τ contains R and the empty set ∅;

2. τ is closed under arbitrary unions:

Uλ ∈ τ for all λ ∈ Λ =⇒⋃λ∈Λ

Uλ ∈ τ ;

3. τ is closed under finite intersections:

U1, U2, . . . , Un ∈ τ =⇒ U1 ∩ U2 ∩ . . . ∩ Un ∈ τ.

Proof. 1. R is open, since if p ∈ R then

(p− h, p+ h) ⊆ R

for every h > 0. The empty set ∅ is open.

2. Let Uλ be open for all λ ∈ Λ and let p ∈⋃λ∈Λ Uλ. Then p ∈ Uλ0 for some

λ0 ∈ Λ. By assumption, Uλ0 is open and, for some h > 0, we have

(p− h, p+ h) ⊆ Uλ0 ⊆⋃λ∈Λ

Uλ.

Therefore⋃λ∈Λ Uλ is open.

3. Let U1, U2, . . . , Un be open and let

p ∈ U1 ∩ U2 ∩ . . . ∩ Un.

Then, for each i: 1 ≤ i ≤ n, there existshi > 0 such that

(p− hi, p+ hi) ⊆ Ui.

Leth = min(h1, h2, . . . , hn) > 0.

Then(p− h, p+ h) ⊆ (p− hi, p+ hi) ⊆ Ui

for all 1 ≤ i ≤ n, so

(p− h, p+ h) ⊆ U1 ∩ U2 ∩ . . . ∩ Un.

Thus U1 ∩ U2 ∩ . . . ∩ Un is open.

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Note the restriction in 3 to finite intersections of open sets. This is because theintersection of an infinite family of open sets need not be open.

Example 1.20. The intersection of the infinite family of open intervals

(− 1

n,

1

n) (n = 1, 2, 3, . . .)

is the singleton {0}, and no singleton is open.

Example 1.21. If a, b ∈ R, then

(−∞, a) =∞⋃n=1

(a− n, a)

and

(b,∞) =∞⋃n=1

(b, b+ n)

are unions of open sets, and so are open.

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2 Metric SpacesDefinition 2.1. A metric on a set X is a function

d : X ×X −→ R,

(x, y) 7→ d(x, y),

which assigns a distance d(x, y) between each pair of points x, y ∈ X , satisfying thefollowing three axioms:

[M1] For all x, y ∈ X , d(x, y) > 0 if x 6= y and d(x, x) = 0;

[M2] d(x, y) = d(y, x) for all x, y ∈ X .

[M3] d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X .The latter is called the triangle inequality.

Remark 2.2. [M1] says that each point is distant 0 from itself, but distinct points areat positive distance apart.

[M2] says that the distance from x to y is the same as the distance from y to x.[M3] gets its name from the fact that the length of any side of a triangle is less than

the sum of the lengths of the other two sides. It says that a journey from x to z doesn’tget any shorter if you have to take a detour via y, but may possible get longer.

The axioms [M1-[M3] have been chosen so as to give metrics just enough propertiesto define a topology.

Definition 2.3. A metric space is a pair (X, d), where X is a set and d is a metric onX .

In practice, we usually refer to “the metric space X”.

Examples of Metric Spaces

In most cases, verification of [M1] and [M2] is trivial, and we need only check [M3].

Example 2.4. The real line R with the usual metric

d(x, y) = |x− y|

is a metric space.

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Proof. Let us verify [M1]-[M3].[M1] For all x, y ∈ R,d(x, y) = |x− y| > 0 if x 6= y andd(x, x) = |x− x| = 0.

[M2] For all x, y ∈ R,

d(x, y) = |x− y| = |y − x| = d(y, x).

[M3] For all x, y, z ∈ R,

d(x, z) = |x− z|= |(x− y) + (y − z)|≤ |x− y|+ |y − z|= d(x, y) + d(y, z).

Therefore d is a metric and (R, d) is a metric space.

Example 2.5. The complex plane C with the usual metric

d(z1, z2) = |z1 − z2|

is a metric space. The verification of [M1]-[M3] is the same as in the previous example.

Example 2.6. The plane

R2 = {a = (a1, a2) : a1, a2 ∈ R}

with the taxicab metric

d1((a1, a2), (b1, b2)) = |a1 − b1|+ |a2 − b2|

is a metric space.

Proof. Let us verify [M1]-[M3].[M1] For all a = (a1, a2), b = (b1, b2) ∈ R2,d1(a, b) = |a1 − b1|+ |a2 − b2| > 0 if a 6= b and

d1(a, a) = d1((a1, a2), (a1, a2))

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= |a1 − a1|+ |a2 − a2| = 0.

[M2] For all a = (a1, a2), b = (b1, b2) ∈ R2,

d1(a, b) = d1((a1, a2), (b1, b2))

= |a1 − b1|+ |a2 − b2| = |b1 − a1|+ |b2 − a2|

= d1((b1, b2), (a1, a2)) = d1(b, a).

[M3] For all a = (a1, a2), b = (b1, b2), c = (c1, c2) ∈ R2, observe that

d1((a1, a2), (b1, b2)) = |a1 − b1|+ |a2 − b2|

= |(a1 − c1) + (c1 − b1)|+ |(a2 − c2) + (c2 − b2|

≤ |a1 − c1|+ |c1 − b1|+ |a2 − c2|+ |c2 − b1|

= d1((a1, a2), (c1, c2)) + d1((c1, c2), (b1, b2)).

Therefore d1 is a metric and (R2, d1) is a metric space.

Example 2.7. Euclidean n-space

Rn = {a = (a1, a2, . . . , an) : a1, . . . , an ∈ R}

with the euclidean metric

d2((a1, a2, . . . , an), (b1, b2, . . . , bn))

=√

(a1 − b1)2 + · · ·+ (an − bn)2

is a metric space.

Example 2.8. Let X be the set of points on the surface of a sphere, and let the distancebetween two points be the length of the shorter of the two great circle arcs joining thetwo points. Then (X, d) is a metric space. This is the metric used in measuring distanceson the Earth’s surface, e.g., when we are told that Sydney is 10,840 miles from London1.

Example 2.9. The complex plane C with the French Railway metric2:1The distance between Sydney and London is less than 8,000 miles if we use the metric in the previous

example.2Folklore suggests that the shortest rail journey between any two French towns is via Paris.

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df (z1, z2) =

{0 z1 = z2

|z1|+ |z2| z1 6= z2

is a metric space.

Example 2.10. We can define at least one metric, the discrete metric, on any set X:

d(x, x) = 0 and d(x, y) = 1 if x 6= y,

that is, distinct points are distance 1 apart.

Proof. Verification of [M1] and [M2] is trivial, and we need only check [M3]: for allx, y, z ∈ X ,

d(x, z) ≤ d(x, y) + d(y, z).

If the LHS is 0, then there is nothing to prove.Otherwise LHS = 1 so x 6= z. But then y 6= x or y 6= z; in either case the RHS is

at least 1, and so LHS ≤ RHS.

Example 2.11. Let C[0, 1] be the set of all continuous complex-valued functions de-fined on the closed interval [0, 1], and let

d∞(f, g) = sup{|f(x)− g(x)| : 0 ≤ x ≤ 1},

and

d1(f, g) =

∫ 1

0

|f(x)− g(x)|dx.

Then d∞ and d1 are metrics on C[0, 1] and(C[0, 1], d∞) and (C[0, 1], d1) are metric spaces.

Proof. We shall prove that d1 is a metric. Let us verify [M1]-[M3].[M1] For all f, g ∈ C[0, 1], by properties of integrals,d1(f, g) =

∫ 1

0|f(x)− g(x)|dx > 0 if f 6= g

andd1(f, f) =

∫ 1

0|f(x)− f(x)|dx =

∫ 1

00dx = 0.

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[M2] For all f, g ∈ C[0, 1],

d1(f, g) =

∫ 1

0

|f(x)− g(x)|dx

=

∫ 1

0

|g(x)− f(x)|dx = d1(g, f).

[M3] For all f, g, h ∈ C[0, 1] and all x ∈ [0, 1], we have

|f(x)− g(x)| ≤ |f(x)− h(x)|+ |h(x)− g(x)|.

Thus, for all f, g, h ∈ C[0, 1],

d1(f, g) =

∫ 1

0

|f(x)− g(x)|dx

≤∫ 1

0

|f(x)− h(x)|+ |h(x)− g(x)|dx

(by properties of integrals)

=

∫ 1

0

|f(x)− h(x)|dx+

∫ 1

0

|h(x)− g(x)|dx

(by the linearity of integrals)

= d1(f, h) + d1(h, g).

Therefore d1 is a metric and (C[0, 1], d1) is a metric space.

Open Balls in Metric Spaces

Our next task is to show how a metric gives rise to a topology.

Definition 2.12. Let (X, d) be a metric space, let a ∈ X , and let r ∈ R, r > 0. Theopen ball with centre a and radius r is the set

B(a, r) = {x ∈ X : d(a, x) < r}

consisting of all points of X whose distance from a is less than r.

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The motivating example here comes from Euclidean 3-space, where an open ball isthe interior of a sphere. The following examples show that open balls in other metricspaces can look quite different.

Example 2.13. Consider the metric space (R, d) from Example 2.4. An open ballB(a, r) in the real line R

B(a, r) = {x ∈ R : d(a, x) < r}

= {x ∈ R : |x− a| < r}

is an open interval (a− r, a+ r).

Example 2.14. Consider the metric space (C, d) from Example 2.5. For a ∈ C andr > 0, the open ball B(a, r) in the complex plane C

B(a, r) = {z ∈ C : d(a, z) < r}

= {z ∈ C : |z − a| < r}

is the disc consisting of all points interior to the circle with centre a and radius r.

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Example 2.15. Consider the metric space (R3, d2) from Example 2.7. For a =(a1, a2, a3) ∈ R3 and r > 0, the open ball in Euclidean 3-space

B(a, r) = {x ∈ R3 : d2(a, x) < r}

= {x = (x1, x2, x3) ∈ R3 :√(x1 − a1)2 + (x2 − a2)2 + (x3 − a3)2 < r}

really does look like a ball.

Example 2.16. Consider the metric space (R2, d1) with the taxicab metric d1 fromExample 2.6. For a = (a1, a2) ∈ R2 and r > 0, the open ball

B(a, r) = {b ∈ R2 : d1(a, b) < r}

= {b = (b1, b2) ∈ R2 : |b1 − a1|+ |b2 − a2| < r}.

For 0 = (0, 0) ∈ R2 and r > 0, the open ball

B(0, r) = {(x, y) ∈ R2 : d1((0, 0), (x, y)) < r}

= {(x, y) ∈ R2 : |x− 0|+ |y − 0| < r}

consists of all points interior to the square enclosed by the lines

x+ y = r, x− y = r, y − x = r, −y − x = r.

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Example 2.17. Consider the complex plane C with the French Railway metric dffrom Example 2.9. For a ∈ C and r > 0, the open ball

B(a, r) = {z ∈ C : df (a, z) < r}.

Recall that

df (a, z) =

{0 if z = a

|a|+ |z| if z 6= a.

Thus the open ballB(a, r) = {a} if r < |a|,

andB(a, r) = {a} ∪ {z ∈ C : |z| < r − |a|} if r ≥ |a|.

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Example 2.18. Consider the metric space (X, d) with the discrete metric d from Ex-ample 2.10. For a ∈ X and r > 0, the open ball

B(a, r) = {z ∈ X : d(a, z) < r}.

Recall thatd(x, x) = 0 and d(x, y) = 1 if x 6= y.

Therefore

B(a, r) =

{{a} if r ≤ 1

X if r > 1.

Remark 2.19. Open balls need not be spherical!

Open Sets in Metric Spaces

We shall now use open balls to define a topology on any metric space. The methodis the one we used to define the usual topology on the real line.

Definition 2.20. A subset U of a metric space (X, d) is open if and only if for eachu ∈ U there is a positive real number r such that B(u, r) ⊆ U .

Thus an open set in a metric space is one which contains an open ball about each ofits points.

In the case of R with the usual metric, this is just the definition of open set given in1.1. In fact, the following theorem, and its proof, are modelled on Theorem 1.19.

Theorem 2.21. Let (X, d) be a metric space and let τ be the collection of all opensubsets of X . Then

1. τ contains X and the empty set ∅;

2. τ is closed under arbitrary unions:

Uλ ∈ τ for all λ ∈ Λ =⇒⋃λ∈Λ

Uλ ∈ τ ;

3. τ is closed under finite intersections:

U1, U2, . . . , Un ∈ τ =⇒ U1 ∩ U2 ∩ . . . ∩ Un ∈ τ.

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The open sets in a metric space (X, d) form a topology which is called the metrictopology.

Proof. 1. X is obviously open, and ∅ is open by the vacuous truth principle.

2. Let Uλ be open for all λ ∈ Λ and let

x ∈⋃λ∈Λ

Uλ.

Then x ∈ Uλ0 for some λ0 ∈ Λ.By assumption, Uλ0 is open and, for some r > 0, we have B(x, r) ⊆ Uλ0 . Thus

B(x, r) ⊆ Uλ0 ⊆⋃λ∈Λ

Uλ.

Therefore⋃λ∈Λ Uλ is open.

3. Let U1, U2, . . . , Un be open and let

x ∈ U1 ∩ U2 ∩ . . . ∩ Un.

By assumption, for each i: 1 ≤ i ≤ n, Ui is open. Therefore, there exists ri > 0 suchthat

B(x, ri) ⊆ Ui.

Letr = min(r1, r2, . . . , rn) > 0.

ThenB(x, r) ⊆ B(x, r1) ⊆ U1,

B(x, r) ⊆ B(x, r2) ⊆ U2,

. . .

B(x, r) ⊆ B(x, rn) ⊆ Un.

HenceB(x, r) ⊆ U1 ∩ U2 ∩ . . . ∩ Un.

Therefore U1 ∩ U2 ∩ . . . ∩ Un is open.

22

In the case of R with the usual metric, B(u, r) = (u− r, u+ r), so the usual metricinduces the usual topology.

Open Balls Are Open Sets

We can now justify the name “open ball”.

Theorem 2.22. Every open ball B(a, r) in a metric space (X, d) is an open set.

Proof. Let u ∈ B(a, r) and let

r′ = r − d(a, u) > 0.

For all x ∈ B(u, r′),

d(x, u) < r′

=⇒d(x, a) ≤ d(x, u) + d(u, a) < r′ + d(u, a) = r.

Then B(u, r′) ⊆ B(a, r), so B(a, r) is open.

Corollary 2.23. A subset of a metric space is open⇐⇒ it is a union of open balls.

Proof. =⇒ If U is open then for each u ∈ U there is an open ball B(u, ru) ⊆ U . Itfollows that

U =⋃u∈U

B(u, ru),

a union of open balls.⇐= Any union of open balls is open by Theorem 2.21(2).

Example 2.24. Consider the metric space (Cn, d1), where

d1(~v, ~w) =n∑j=1

|vj − wj|,

23

for ~v = (v1, . . . , vn), ~w = (w1, . . . , wn) ∈ Cn.Give the definition of an open ball in the metric space (Cn, d1).

For ~a ∈ Cn and r > 0, the open ball

B(~a, r) = {~v ∈ Cn : d1(~a,~v) < r}

= {~v = (v1, . . . , vn) ∈ Cn :n∑j=1

|vj − aj| < r}.

Are the following subsets of the metric space (Cn, d1) open?

(i)

S1 = {~v ∈ Cn :n∑j=1

|vj − (1 + ji)| < 5},

where i2 = −1;

(ii)

S2 =25⋃p=1

{~v ∈ Cn :n∑j=1

|vj − 1| < 2p}.

Justify your answer.

Proof. (i) The set

S1 = {~v ∈ Cn :n∑j=1

|vj − (1 + ji)| < 5}

is the open ball B(~a, 5) with centre

~a = (1 + i, 1 + 2i, . . . , 1 + ni)

and radius r = 5 in the metric space (Cn, d1) and therefore, by Theorem 2.22, it is open.

(ii) The set

S2 =25⋃p=1

{~v ∈ Cn :n∑j=1

|vj − 1| < 2p}

is the union of open balls B(~b, 2p), p = 1, 2, . . . , 25, with centre

~b = (1, 1, . . . , 1)

and radius r = 2p in (Cn, d1) and therefore, by Theorem 2.21, it is open.

24

Example 2.25. Consider the metric space(C[0, 1], d∞), where C[0, 1] is the set of all continuous complex-valued functions de-fined on the closed interval [0, 1], and

d∞(f, g) = sup{|f(x)− g(x)| : 0 ≤ x ≤ 1}.

Are the following subsets of the metric space (C[0, 1], d∞) open?

(i) S1 = {f ∈ C[0, 1] : sup0≤t≤1

|f(t)− exp(t)| < 10}

and

(ii) S2 =∞⋂p=1

{f ∈ C[0, 1] : sup0≤t≤1

|f(t)− t2| < 1

p}.

Justify your answer.

Proof. (i) The set

S1 = {f ∈ C[0, 1] : sup0≤t≤1

|f(t)− exp(t)| < 10}

is the open ball B(g, 10) with centre g(t) = exp(t), t ∈ [0, 1], and radius r = 10 in themetric space (C[0, 1], d∞) and therefore, by Theorem 2.22, it is open.

(ii) Note that

g ∈∞⋂p=1

{f ∈ C[0, 1] : sup0≤t≤1

|f(t)− t2| < 1

p}

⇐⇒g ∈ {f ∈ C[0, 1] : sup

0≤t≤1|f(t)− t2| < 1

p}

for all p ∈ N⇐⇒

sup0≤t≤1

|g(t)− t2| < 1

pfor all p ∈ N

⇐⇒sup

0≤t≤1|g(t)− t2| = 0

25

⇐⇒g(t) = t2, t ∈ [0, 1].

Therefore S2 = {g} is a singleton and is not open in the metric space (C[0, 1], d∞),because every open ball B(g, r) in (C[0, 1], d∞) contains also all continuous functionsh such that h(t) = g(t) + p, where p ∈ C : |p| < r.

26

3 Topological Spaces

Definition 3.1. Let X be a set. Then a topology on X is a family τ of subsets of Xsuch that

[T1] τ includes X and ∅;

[T2] τ is closed under arbitrary unions:

Uλ ∈ τ for all λ ∈ Λ =⇒⋃λ∈Λ

Uλ ∈ τ ;

[T3] τ is closed under finite intersections:

U1, U2, . . . , Un ∈ τ =⇒ U1 ∩ U2 ∩ . . . ∩ Un ∈ τ.

Remark 3.2. A topology on X is a collection of subsets of X which contains X and ∅and is closed under arbitrary unions and finite intersections.

Here the phrase “arbitrary unions” allows the index set Λ in [T2] to be any setwhatsoever, no matter how large. By contrast [T3] allows only intersections of finitelymany sets U1, U2, . . . , Un.

Definition 3.3. A topological space is a pair (X, τ), whereX is a set and τ is a topologyon X .

It is customary to refer to “the topological space X” when it is obvious which topol-ogy τ is involved. The elements of X are usually called points. The sets which makeup the topology τ are called the τ -open sets, or, where no confusion arises, simply theopen sets3.

Note that [T3] has the alternative formulation:

[T3∗] : U1, U2 ∈ τ =⇒ U1 ∩ U2 ∈ τ.since closure under finite intersections follows by repetition. This alternative version iseasier to use when we wish to verify that a collection of sets is a topology.

3In the early days of topology, there were several competing sets of topology axioms (all more or lessequivalent). The eventual triumph of [T1]-[T3] and the adoption of open set as the fundamental conceptwere largely due to the influence of the mythical French mathematician Nicolas Bourbaki.

27

Examples of Topological Spaces

Example 3.4. Let X = R. It is proved in Theorem 1.19 that the open subsets of R, asdefined in 1.1, form a topology τd on R. This topology is called the usual topology onR.

From now on, unless stated otherwise, R will always be assumed to have the usualtopology.

Example 3.5. Consider a metric space (X, d). It is proved in Theorem 2.21 that theopen subsets of X , as defined in 2.20, form a topology τd on X , called the metrictopology on (X, d).

Example 3.6. Let X be any set, and let τ be the collection of all subsets of X:

τ = {S ⊆ X}.

Clearly τ satisfies [T1]-[T3]:[T1] τ includes X and ∅.

[T2] τ is closed under arbitrary unions, since any union of subsets of X is again is asubset of X .

[T3] τ is closed under finite intersections, since any intersection of subsets of X isagain is a subset of X .

Therefore τ is a topology. We call this the discrete topology on X .

In the discrete topology, every set is an open set.

Example 3.7. Let X be any set, and let

τ = {X, ∅}.

Then it is easy to see that τ is a topology on X .[T1] τ includes X and ∅.

28

[T2] τ is closed under arbitrary unions, since

X ∪ ∅ = X ∈ τ.

[T3] τ is closed under finite intersections, since

X ∩ ∅ = ∅ ∈ τ.

It is called the indiscrete topology.

Example 3.8. Let X be any nonempty set, so X has at least one member p. Let τpconsist of the empty set plus all subsets of X which have p as a member:

τp = {∅} ∪ {S ⊆ X : p ∈ S}.

Then τp is a topology on X . It is called a particular point topology.

Example 3.9. Another topology on R. Let τs.i. consist of R, ∅, and all the semi-infiniteopen intervals (a,∞), a ∈ R. Then τs.i. is a topology on R.

Proof. [T1]: τs.i. includes R and ∅.

[T2]: Let (aλ,∞), λ ∈ Λ, be a family of members of τs.i.. Then(i) if the aλ, λ ∈ Λ are bounded below, their union is (inf aλ,∞) ∈ τs.i.;

(ii) if they are not bounded below, their union is R ∈ τs.i..[T3]

n⋂i=1

(ai,∞) = (max(a1, a2, . . . , an),∞) ∈ τs.i..

Remark 3.10. Note that every semi-infinite open interval (a,∞) is an open subset ofthe topological space R with the usual topology, so

τs.i. ⊂ τd.

On the other hand, the finite open interval (a, b) is not in τs.i.. Thus two topologicalspaces (R, τd) and (R, τs.i.) are different.

29

Equivalent Metrics

Definition 3.11. Two metrics d1 and d2 on a set X are topologically equivalent if theygenerate the same topology on X , that is, a subset U of X is d1-open if and only if it isd2-open.

Remark 3.12. This implies that open sets are the same in both cases, as are the closedsets, the compact sets, the connected sets and the continuous functions which we willconsider later.

Theorem 3.13. Suppose that there are strictly positive real numbers K1 and K2 suchthat, for every x, y ∈ X ,

d1(x, y) ≤ K2d2(x, y)

andd2(x, y) ≤ K1d1(x, y).

Then the metrics d1 and d2 are topologically equivalent.

Proof. Suppose that U ⊆ X is open in the topology induced by d1. We claim that U isopen in the topology induced by d2.

Let u ∈ U . By assumption, there exists ε > 0 such that

B1(u, ε) = {x ∈ X : d1(u, x) < ε} ⊆ U.

Then, for all x ∈ X such that d2(x, u) <ε

K2

, we have

d1(x, u) ≤ K2d2(x, u) < K2 ×ε

K2

= ε.

Thus, for r =ε

K2

,

B2(u, r) = {x ∈ X : d2(u, x) < r}

⊆ B1(u, ε) ⊆ U.

30

Therefore U is open with respect to d2.The same proof works for the following statement: if U ⊆ X is open in the topology

induced by d2, then U is open in the topology induced by d1.

Therefore d1 and d2 generate the same topology on X .

Example 3.14. Consider the following metrics on Rn:

d1(x,y) = Σni=1|xi − yi|,

d2(x,y) =√

Σni=1|xi − yi|2

andd∞(x,y) = sup{|xi − yi| : 1 ≤ i ≤ n}

for x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) from Rn. We shall show that d1, d2

and d∞ are topologically equivalent.

Proof. 1. For all x = (x1, x2, . . . , xn) andy = (y1, y2, . . . , yn) ∈ Rn, we have

(i) d∞(x,y) = max{|xi − yi| : 1 ≤ i ≤ n}

≤ Σni=1|xi − yi| = d1(x,y)

and(ii) d1(x,y) = Σn

i=1|xi − yi|

≤ Σni=1 max{|xi − yi| : 1 ≤ i ≤ n}

= Σni=1d∞(x,y) = n · d∞(x,y).

Thus, by Theorem 3.13, the metrics d1 and d∞ are topologically equivalent.

2. For all x = (x1, x2, . . . , xn) andy = (y1, y2, . . . , yn) ∈ Rn, we have

(i) d∞(x,y) = max{|xi − yi| : 1 ≤ i ≤ n}

≤√

Σni=1|xi − yi|2 = d2(x,y)

and(ii) d2(x,y) =

√Σni=1|xi − yi|2

31

≤√

Σni=1 max{|xi − yi|2 : 1 ≤ i ≤ n}

=√n · d∞(x,y).

Thus, by Theorem 3.13, the metrics d2 and d∞ are topologically equivalent.

3. By Part 1, for all x,y ∈ Rn,

d1(x,y) ≤ n · d∞(x,y)

and, by Part 2,d∞(x,y) ≤ d2(x,y).

Thus, for all x,y ∈ Rn,d1(x,y) ≤ n · d2(x,y).

By Part 2, for all x,y ∈ Rn,

d2(x,y) ≤√n · d∞(x,y)

and, by Part 1,d∞(x,y) ≤ d1(x,y).

Thus, for all x,y ∈ Rn,d2(x,y) ≤

√n · d1(x,y).

Therefore, by Theorem 3.13, the metrics d2 and d1 are topologically equivalent.

It follows that all three metrics generate the same topology.

Example 3.15. Let C[0, 1] be the set of all continuous complex-valued functions de-fined on the closed interval [0, 1], and let

d∞(f, g) = sup{|f(x)− g(x)| : 0 ≤ x ≤ 1},

and

d1(f, g) =

∫ 1

0

|f(x)− g(x)|dx

be metrics on C[0, 1]. Then d∞ and d1 are NOT topologically equivalent.

32

Proof. 1. We shall show that, for every f, g ∈ C[0, 1],

d1(f, g) ≤ d∞(f, g).

By definition of d∞, for all x ∈ [0, 1],

|f(x)− g(x)| ≤ supy∈[0,1]

|f(y)− g(y)| = d∞(f, g).

Hence by integration theory,

d1(f, g) =

∫ 1

0

|f(x)− g(x)|dx

≤∫ 1

0

d∞(f, g)dx = d∞(f, g).

This implies that, for every f ∈ C[0, 1] and r > 0,

B∞(f, r) = {g ∈ C[0, 1] : d∞(f, g) < r}

⊆ {g ∈ C[0, 1] : d1(f, g) < r} = B1(f, r).

Thus if U ⊆ C[0, 1] is open in the topology induced by d1, then U is open in thetopology induced by d∞.

2. Let 0̌ denote the constant function with value 0 for all t ∈ [0, 1]. We shall showthat d∞-open ballB∞(0̌, 1) is not open in the topology induced by d1 (i.e., not d1-open).

For any ε > 0, there exists a continuous function

gε(t) =

{−nt+ 1 if t ∈ [0, 1

n]

0 if t ∈ [ 1n, 1],

where 12n< ε. Then it is easy to see that

d1(0̌, gε) =

∫ 1

0

|0− gε(x)|dx

=

∫ 1n

0

(−nt+ 1)dt =1

2n< ε

and

33

d∞(0̌, gε) = supt∈[0,1]

|0− gε(t)|

supt∈[0,1]

| − nt+ 1| = 1.

Thus, for every ε > 0, there is gε ∈ B1(0̌, ε), but gε does not belong to B∞(0̌, 1).Therefore B∞(0̌, 1) is not d1-open. It means that d∞ and d1 are not topologically equiv-alent.

34

4 Hausdorff Spaces and Limits

Definition 4.1. A topological space (X, τ) is Hausdorff if for each pair of distinctpoints x1 and x2 in X there are disjoint open sets U1 and U2 such that x1 ∈ U1 andx2 ∈ U2.

Example 4.2. The topological space R with the usual topology τd is Hausdorff.

Proof. Let x1 and x2 be distinct points in R, let r = |x1 − x2|/4 and let U1 = (x1 −r, x1 + r) and U2 = (x2 − r, x2 + r).

We know that U1 and U2 are open in the usual topology τd. It is clear that x1 ∈ U1

and x2 ∈ U2. For u ∈ U1 ∩ U2, we have

4r = |x1 − x2| ≤ |x1 − u|+ |u− x2| < r + r = 2r,

a contradiction. Hence U1 ∩ U2 = ∅.

Example 4.3. The topological space X with the discrete topology τ is Hausdorff.

Proof. Let x1 and x2 be distinct points in X . Recall that the discrete topology τ is thecollection of all subsets of X; see Example 3.6.

Let U1 = {x1} and U2 = {x2}. By definition of the discrete topology, U1 and U2

are in τ , so open. It is clear that x1 ∈ U1, x2 ∈ U2 and U1 ∩ U2 = ∅.

Example 4.4. The topological spaceX with the indiscrete topology τ is not Hausdorffprovided X contains more than one point.

Proof. Let x1 and x2 be distinct points in X . Recall that the indiscrete topology τ ={X, ∅}; see Example 3.7.

Every open set U1 ∈ τ such that x1 ∈ U1 has to be X and therefore x2 ∈ U1. Thusthere are no disjoint open sets U1 and U2 in the indiscrete topology such that x1 ∈ U1

and x2 ∈ U2.

35

Theorem 4.5. Every metric space (X, d) is Hausdorff.

For a proof see Solutions 2, Qu.11.

Limits in Metric and Topological Spaces

Recall the definition of limit from the basic analysis course.

Definition 4.6. We say a sequence (tn)∞n=1 of real numbers converges to t ∈ R asn→∞ if for every ε > 0 there is N ∈ N such that for all n ≥ N ,

|tn − t| < ε.

Definition 4.7. Let (X, d) be a metric space. We say a sequence (xn)∞n=1, xn ∈ X ,converges to x ∈ X as n → ∞ if for every ε > 0 there is N ∈ N such that, for alln ≥ N ,

d(xn, x) < ε,

that is, xn ∈ B(x, ε).

We writelimn→∞

xn = x

with respect to the metric d.

Remark 4.8. It is easy to see that a sequence (xn)∞n=1, xn ∈ X , converges to x ∈ X asn→∞ if and only if the sequence of non-negative real numbers tn = d(xn, x)→ 0 asn→∞.

Example 4.9. Consider the metric space (C, d), where d(z1, z2) = |z1 − z2|. Showthat the sequence (zn)∞n=1, where zn = (1− 1

n) + i

2n∈ C, converges to 1 as n→∞.

36

Proof. Note thatd(zn, 1) = |zn − 1|

= |(1− 1

n) +

i

2n− 1| = | − 1

n+

i

2n|

=

{(− 1

n

)2

+

(1

2n

)2}1/2

=

{1

n2+

1

22n

}1/2

→ 0

as n→∞. Thus limn→∞ zn = z with respect to the metric d.

Theorem 4.10. In a metric space (X, d), every convergent sequence has a uniquelimit.

Proof. Suppose that (xn)∞n=1, xn ∈ X , converges to x ∈ X and also to y ∈ X asn→∞.

If x 6= y, we may choose ε > 0, say

ε =1

2· d(x, y),

so that the open ballsB(x, ε) andB(y, ε) are disjoint. Since xn is supposed to belong toboth of those open balls for sufficiently large n, we get a contradiction. Thus x = y.

Definition 4.11. Let (X, τ) be a topological space. We say a sequence (xn)∞n=1, xn ∈ X ,converges to x ∈ X as n → ∞ if for every open set U containing x, there is N ∈ Nsuch that xn ∈ U for all n ≥ N .

We write limn→∞ xn = x with respect to the topology τ .

Example 4.12. Let (X, τ) be the topological space with the indiscrete topology τ .Then any sequence (xn)∞n=1, xn ∈ X , converges to every point x ∈ X as n→∞.

37

Proof. Recall that the indiscrete topologyτ = {X, ∅}; see Example 3.7.

Every open set U ∈ τ such that x ∈ U has to be X and therefore xn ∈ U for alln ≥ 1. Thus limn→∞ xn = x with respect to the indiscrete topology.

Recall that the topological space (X, τ) with the indiscrete topology is not Hausdorffif X has more than one point.

Theorem 4.13. In a Hausdorff space (X, τ), every convergent sequence has a uniquelimit.

Proof. Suppose that (xn)∞n=1, xn ∈ X , converges to x ∈ X and also to y ∈ X asn→∞ with respect to the topology τ ,

If x 6= y, the Hausdorff condition implies that there are disjoint open sets U1 and U2

such that x ∈ U1 and y ∈ U2. Since xn is supposed to belong to both of those open setsfor sufficiently large n, we get a contradiction. Thus x = y.

38

5 Closed Sets

Definition 5.1. A subset F of a topological space (X, τ) is called closed if its comple-ment F c is open.

Recall that, for S ⊆ X , the complement of S in X is

X \ S = {x ∈ X : x 6∈ S}.

In practice, we often work with subsets of a fixed set X and then we write Sc instead ofX \ S. We write Scc instead of (Sc)c; it is obvious that Scc = S.

Complementation obeys de Morgan’s Laws:

(A ∪B)c = Ac ∩Bc, (A ∩B)c = Ac ∪Bc,

and more generally (⋃λ∈Λ

Uλ

)c

=⋂λ∈Λ

U cλ,

(⋂λ∈Λ

Uλ

)c

=⋃λ∈Λ

U cλ.

Example 5.2. Every closed interval [a, b] of R is a closed set in the usual topology. Itscomplement [a, b]c is the union (−∞, a) ∪ (b,∞) of two open sets and so is open.

Example 5.3. Every open interval (a, b) of R is not closed in the usual topology. Itscomplement (a, b)c is the union (−∞, a] ∪ [b,∞) which is not an open subset of R.

Proposition 5.4. Every singleton {y} of a metric space (X, d) is closed in the metrictopology.

Proof. It is clear that the complement of {y} is the set

S = X \ {y} = {x ∈ X : d(x, y) > 0}.

We have to show that, for each u ∈ S, there is an open ball B(u, r′) wholly containedin S. Let r′ = d(u, y) > 0. It is obviuos that y does not belong B(u, r′). ThereforeB(u, r′) ⊂ X \ {y}, so S is open. Hence {y} is closed.

39

Definition 5.5. Let (X, d) be a metric space, let a ∈ X , and let r ∈ R, r > 0. Theclosed ball with centre a and radius r is the set

B(a, r) = {x ∈ X : d(a, x) ≤ r}

consisting of all points of X whose distance from a is less than or equal to r.

Proposition 5.6. The closed ballB(a, r) of a metric space (X, d) is closed in the metrictopology.

Proof. It is clear that the complement of B(a, r) is the set

{x ∈ X : d(x, a) > r},

which is open; see Solution 2, Qu.4.

Example 5.7. Consider the metric space(C[0, 1], d∞), where C[0, 1] is the set of all continuous complex-valued functions de-fined on the closed interval [0, 1], and

d∞(f, g) = sup{|f(x)− g(x)| : 0 ≤ x ≤ 1}.

Are the following subsets of the metric space (C[0, 1], d∞) closed?

(i) S1 = {f ∈ C[0, 1] : sup0≤t≤1

|f(t)− exp(t)| ≤ 10}

and

(ii) S2 =∞⋂p=1

{f ∈ C[0, 1] : sup0≤t≤1

|f(t)− t2| < 1

p}.

Justify your answer.

40

Proof. (i) The set

S1 = {f ∈ C[0, 1] : sup0≤t≤1

|f(t)− exp(t)| ≤ 10}

is the closed ball B(g, 10) with centre g(t) = exp(t), t ∈ [0, 1], and radius r = 10 inthe metric space (C[0, 1], d∞) and therefore, by Proposition 5.6, it is closed.

(ii) We proved in Example 2.25 that∞⋂p=1

{f ∈ C[0, 1] : sup0≤t≤1

|f(t)− t2| < 1

p} = {g},

where g(t) = t2, t ∈ [0, 1]. By Proposition 5.4, the singleton {g} in the metric space(C[0, 1], d∞) is closed.

Every statement involving open sets can be translated into a corresponding state-ment involving closed sets, simply by taking complements.

Warning: In topology, a set can be(i) open and closed, e.g., every point in X with the discrete topology;(ii) open and not closed, e.g., an open inverval (a, b) in R with the usual topology;(iii) closed and not open, e.g., a closed inverval [a, b] in R with the usual topology;

or(iv) neither open nor closed, e.g., an inverval (a, b] in R with the usual topology.In “most” spaces, “most” sets are neither open nor closed.

Theorem 5.8. Let κ be the family of all closed subsets in a topological space (X, τ).Then

[T1′] κ includes ∅ and X;

[T2′] κ is closed under arbitrary intersections:

if each Fλ, λ ∈ Λ, is closed,

then⋂λ∈Λ

Fλ is closed;

[T3′] κ is closed under finite unions:

if all F1, F2, . . . , Fn are closed,

then F1 ∪ F2 ∪ . . . ∪ Fn is closed.

41

Proof. [T1′] ∅ is closed since its complement X is open. X is closed since its comple-ment ∅ is open.

[T2′] Each F cλ is open, so, by [T2], the union

⋃λ∈Λ F

cλ is open, and hence its com-

plement(⋃λ∈Λ

F cλ)c =

⋂λ∈Λ

F ccλ =

⋂λ∈Λ

Fλ

is closed.

[T3′] F c1 , F

c2 . . . , F

cn are open, so, by [T3], the finite intersection

F c1 ∩ F c

2 ∩ . . . ∩ F cn

is open, and hence its complement

(F c1 ∩ F c

2 ∩ . . . ∩ F cn)c = F cc

1 ∪ F cc2 ∪ . . . ∪ F cc

n

= F1 ∪ F2 ∪ . . . ∪ Fnis closed.

42

6 Separation axiomsIn metric spaces (X, d), the metric dmeasures the separation of two distinct points, and,in general, it is a very useful tool. Since most topological spaces are not metrisable, thistool is not available to us. However we can generalise useful properties of metrics, inthe form of separation axioms.

Definition 6.1. Let (X, τ) be a topological space. Then (X, τ) is

(0) T0 if for each pair of points x 6= y ∈ X , there is an open set containing one of thepoints and not the other.

(1) T1 if for each pair of points x 6= y ∈ X , there are open sets about each of thepoints not containing the other point.

(2) T2 (also called Hausdorff ) if for each pair of points x 6= y ∈ X , there are opensets U 3 x and V 3 y such that U ∩ V = ∅.

(Reg.) Regular if for each given a closed set A and a point x /∈ A, there are open setsU ⊇ A and V 3 x such that U ∩ V = ∅.

(3) A space which is regular and T1 is called T3.

Remark 6.2. In this context we make an abuse of terminology. The word “axiom”is used here in the meaning of “requirement” contained in a definition (which can befulfilled or not, depending on the cases).

Proposition 6.3. The following holds

T3 =⇒ T2 =⇒ T1 =⇒ T0.

Proof. The proof of the implications is left as an exercise.

Proposition 6.4. A topological space X is T1 if and only if the points are closed.

Proof. (⇒) Suppose that X is T1. For any y 6= x, there is an open set Uy containing yand not x. ThenX \{x} =

⋃y 6=x Uy, which being the union of open sets is open. Hence

its complement {x} is closed.(⇐) If points are closed and x, y are distinct points in X , then X \ {x} is an open

set containing y and not x, while X \ {y} is an open set containing x and not y. ThusX is T1.

Lemma 6.5. There exists a topological space that satisfies none of the separation ax-ioms.

43

Proof. Let X be a set with more than two elements. Then (X, τind) with the indiscretetopology τind = {X, ∅} is such a space. Prove that (X, τind) is not T0.

Lemma 6.6. There exists a topological space that is T0 but it is not T1.

Proof. Let X be a set that contains at least two points. Fix p ∈ X and let the particularpoint topology on p. That is

τp = {∅} ∪ {S ⊆ X : p ∈ S}.

Then (X, τp) is a T0 space. But it is not T1 since p cannot be separated by any otherpoint x ∈ X (every open set containing x ∈ X must contain p as well).

Lemma 6.7. There exists a space that is T1 but it is not T2.

Proof. Let X be an infinite space. Let the co-fnite topology on X defined by

τ = {∅} ∪ {S ⊆ X : X \ S is finite}.

Then (X, τ) is a T1 space. Indeed for x, y ∈ X let U = X \ {y} and V = X \ {x}.Then U and V are open, with x ∈ U and y ∈ V , and y /∈ U and x /∈ V . However (X, τ)is not T2, since it does not contain disjoint open sets. Indeed let A,B ⊆ X be open andsuppose that A ∩B = ∅. Then, by de Morgan’s Laws,

X = X \ (A ∩B) = (X \ A) ∪ (X \B).

However X \ A and X \ B are finite, hence we would have that X is finite which is acontradiction.

Example 6.8. Consider the space X = {a, b, c}. Define the topology

τ = {∅, {a}, {b, c}, X}.

Then (X, τ) is regular.Hint: Consider all possible combinations for the closed subsets and the points which

are not contained there. These closed subsets are open as well.

Lemma 6.9. There exists a regular space that is not T2.

Proof. Consider the space X = {a, b, c}. Define the topology

τ = {∅, {a}, {b, c}, X}.

We have shown above that (X, τ) is regular. By Proposition 6.4, (X, τ) is not T1 since{b} is not a closed set; hence it is not even T2.

44

7 Basic Topological Concepts

Interior of a Set

Definition 7.1. Let A be a subset of a topological space (X, τ). Then a point p is aninterior point of A if there is an open set U such that p ∈ U and U ⊆ A.

Remark 7.2. Let A be an open set. Then all points p of A are interior points of A.

Definition 7.3. The subset A◦ of A consisting of all the interior points of A is calledinterior of A.

Example 7.4. Consider R with the usual topology and the closed interval A = [a, b] ⊆R. Then every point r of the open interval (a, b) is an interior point of [a, b]. One cansee that r ∈ (a, b) and the open subset (a, b) ⊆ A, but a and b are not interior points.Therefore [a, b]◦ = (a, b).

Example 7.5. Consider R with the usual topology and the subset Z ⊆ R. Every pointof Z is NOT an interior point of Z, since every nonempty open interval contains pointswhich are not in Z. Therefore Z◦ = ∅.

Proposition 7.6. Let (X, τ) be a topological space and let A ⊆ X . Then A◦ is theunion of all open subsets of A, and, hence, it is the largest open set.

Proof. By definition, p is an interior point of A⇐⇒ it lies in an open subset of A.The =⇒ part tells us thatA◦ is contained in the union of the open subsets Uλ, λ ∈ Λ,

of A:A◦ ⊆

⋃λ∈Λ

Uλ.

45

The ⇐= part tells us that the union of all the open subsets Uλ, λ ∈ Λ, of A iscontained in A◦: ⋃

λ∈Λ

Uλ ⊆ A◦.

By the definition of a topology T2, any union of open subsets is open. Thus A◦ isthe largest open subset of A.

Definition 7.7. A neighbourhood of x in a topological space (X, τ) means an open setcontaining x.

Closure of a Set

Definition 7.8. Let A be a subset of a topological space (X, τ). Then a point p is aclosure point of A if every open set containing p meets A.

Definition 7.9. Let A be a subset of a topological space (X, τ). The set of all closurepoints of A is called the closure of A, and is denoted by A.

Example 7.10. Consider R with the usual topology and A = (0, 1).The point 0 is a closure point of (0, 1) since every open set containing 0 contains an

open interval (−ε, ε) and hence a point of (0, 1).The point 1 is also a closure point of (0, 1) since every open set containing 1 contains

an open interval (1− ε, 1 + ε) and hence a point of (0, 1).The closure of A = (0, 1) is [0, 1].

Example 7.11. Consider R with the usual topology and A = Q ⊆ R the set of rationalnumbers. Then, every point of R is a closure point for Q, since every nonempty openinterval contains points which are in Q. Thus the closure of Q is R.

Theorem 7.12. Let A be a subset of a topological space (X, τ). Then

46

(i) (A◦)c = Ac;(ii) (A)c = (Ac)◦,

that is, the complement of the interior is the closure of the complement, and the comple-ment of the closure is the interior of the complement.

Proof. (i) x ∈ (A◦)c⇐⇒ x 6∈ A◦⇐⇒ no open set containing x is wholly contained in A⇐⇒ every open set containing x meets Ac

⇐⇒ x ∈ Ac.

(ii) x ∈ (A)c ⇐⇒ x 6∈ A⇐⇒ some open set containing x does not meet A⇐⇒ some open set containing x is wholly contained in Ac

⇐⇒ x ∈ (Ac)◦.

Theorem 7.13. Let A be a subset of a topological space (X, τ). Then(i) A is the intersection of all the closed supersets of A;(ii) A is a closed set;(iii) A ⊆ A;(iv) A is the smallest closed superset of A.

Proof. (i) By the first part of the previous theorem

(A)c = (Ac)◦ =⋃{U : U open and U ⊆ Ac}.

Take complements:

A = (A)cc = (⋃{U : U open and U ⊆ Ac})c

(by de Morgan’s Laws)

=⋂{U c : U c closed and A ⊆ U c}

=⋂{F : F closed and A ⊆ F}.

(ii) follows from (i) and [T2′].

(iii) follows from (i).

(iv) follows from (i) and (ii).

47

Isolated points of a Set

Definition 7.14. Let A be a subset of a topological space (X, τ). An element a of A isan isolated point of A if there is an open set containing a which contains no point of Aother than a itself.

Example 7.15. Consider R with the usual topology and

A = { 1,1

2,1

3,1

4, . . .}.

The point 1 is a closure point and an isolated point of A, since the open set (3

4,5

4)

contains no point of A other than 1. In fact all points of this set A are isolated.

Boundary of a Set

Definition 7.16. A point p is a boundary point of a set A if it lies in the closure of Aand the closure of its complement Ac.

Definition 7.17. The set ∂A of all boundary points of A is called the boundary of A.Thus

∂A = A ∩ Ac.

Example 7.18. Consider the metric space (C, d), the complex plane C with d(z1, z2) =|z1 − z2|. Let A be an open ball B(a, r), for some a ∈ C and r > 0, in the C. Then theboundary of A is the circle with centre a and radius r

∂A = {z ∈ C : |z − a| = r}.

48

Example 7.19. Consider R with the usual topology and A = Q ⊆ R. Then, since everyopen interval of R contains both rational and irrational numbers, every point of R liesin the closure of Q and the closure of its complement R \Q. Thus ∂Q = R.

Topologies on Subspaces

Definition 7.20. Let (X, τ) be a topological space and let A be a subset of X . The opensubsets for the induced topology on A are the intersections of A with the open subsetsof X , that is,

τA = {U ∩ A : U open in X}.

The topological space (A, τA) with the induced topology is called a subspace of (X, τ).

Theorem 7.21. The family τA forms a topology on A.

Proof. [T1] Note that A = X ∩ A ∈ τA and ∅ = ∅ ∩ A ∈ τA.

[T2] Let Uλ ∩ A ∈ τA, λ ∈ Λ. Then its union is⋃λ∈Λ

(Uλ ∩ A) = (⋃λ∈Λ

Uλ) ∩ A ∈ τA,

which belongs to τA, since⋃λ∈Λ Uλ is open in (X, τ).

[T3] Let Ui ∩ A ∈ τA, 1 ≤ i ≤ n. Then its intersection is

n⋂i=1

(Ui ∩ A) = (n⋂i=1

Ui) ∩ A ∈ τA,

which belongs to τA, since⋂ni=1 Ui is open in (X, τ).

Thus τA forms a topology on A.

Theorem 7.22. The closed subsets of (A, τA) are the intersections of A with the closedsubsets of X .

49

Proof. It is easy to see that if B ⊆ X then complement of B ∩ A in A is Bc ∩ A:

A \ (B ∩ A) = Bc ∩ A.

By definition, S is a closed subset of (A, τA) if its complement A \ S is an opensubset U ∩ A of A, where U is open in (X, τ). Thus

S = A \ (A \ S) = A \ (U ∩ A) = U c ∩ A,

which is the intersection of A with a closed subset of (X, τ).Conversely, if F is closed in (X, τ), then F c is open in (X, τ) and

F c ∩ A = A \ (F ∩ A)

is open in (A, τA). Therefore F ∩ A is closed in (A, τA).

50

8 Compactness

Definition 8.1. Let S be a subset of a topological space (X, τ). An open cover of S isa family Uλ, λ ∈ Λ, of open subsets of X such that

S ⊆⋃λ∈Λ

Uλ.

Example 8.2. Thus the family of open intervals

(−n, n) (n = 1, 2, 3, . . .)

is an open cover of R.

Definition 8.3. A subfamily of an open cover which is itself an open cover is called asubcover.

Example 8.4. For example, the family

(−2n, 2n) (n = 1, 2, 3, . . .)

is a subcover of the open cover above.

Definition 8.5. A subsetK of a topological space (X, τ) is compact if every open coverof K has a finite subcover.

Example 8.6. Consider C with the usual topology. The subset

B(1, 1) = {z ∈ C : |z − 1| < 1}

is not compact, since

Un = {z ∈ C : |z − 1| < 1− 1

n},

n = 1, 2, . . . , is an open cover of B(1, 1) in C with the usual topology, but it does nothave a finite subcover.

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Proposition 8.7. Every finite subset of a topological space is compact.

Proof. Let F = {x1, x2, . . . , xn} be a finite set, and let Uλ, λ ∈ Λ, be an open cover ofF . Then for 1 ≤ i ≤ n, there is a λi ∈ Λ such that xi ∈ Uλi . Then

{Uλ1 , Uλ2 , . . . , Uλn}

is a finite subcover. Thus every finite subset of a topological space is compact.

Theorem 8.8. A closed subset of a compact set is compact.

Proof. Let C be a closed subset of a compact set K in a topological space (X, τ), andlet Uλ, λ ∈ Λ, be an open cover of C.

Then, by Theorem 7.22, C = K ∩ F for some closed subset F of (X, τ). Thus F c

is open in (X, τ).Therefore Uλ, λ ∈ Λ, together with the open set F c, is an open cover of K. But

K is compact, so we can extract a finite subcover of K. If F c occurs in this subcover,discard it. We are left with a finite subcover of C, extracted from Uλ, λ ∈ Λ. ThereforeC is compact.

Example 8.9. Consider R with the usual topology.(i) The set R is not compact, since

{(−n, n) : n = 1, 2, 3, . . .}

is an open cover with no finite subcover.(ii) The half-open interval [0, 1) is not compact, since

{(−1/n, 1− 1/n) : n = 2, 3, . . .}

is an open cover with no finite subcover.

Proposition 8.10. A finite union of compact sets is compact.

52

Proof. Exercise.

Theorem 8.11. Let K be a compact subset of a Hausdorff space (X, τ) and let x 6∈ K.Then there exist disjoint open sets U and V such that x ∈ U and K ⊆ V .

Proof. Since (X, τ) is a Hausdorff topological space, for each k ∈ K, there exist dis-joint open sets Uk and Vk with x ∈ Uk and k ∈ Vk.

Then (Vk)k∈K is an open cover ofK, so it contains a finite subcover Vk1 , Vk2 , . . . Vkn .

Let U = ∩nj=1Ukj and V = ∪ni=1Vki . Then U and V are open subsets x ∈ U , K ⊆ Vand

U ∩ V = (∩nj=1Ukj) ∩ (∪ni=1Vki)

= ∪ni=1

(Vki ∩ (∩nj=1Ukj)

)= ∅,

since, for every k, Vk ∩ Uk = ∅.

Theorem 8.12. A compact subset K of a Hausdorff space X is closed.

Proof. By the previous theorem, for eachx ∈ Kc there is an open set Ux such that

x ∈ Ux ⊆ Kc.

Then Kc is the union of all such Ux. The union of open sets is open. Therefore K isclosed.

Example 8.13. Compact subsets of non-Hausdorff topological spaces need not be closed:in {a, b} with the indiscrete topology both of the singletons are compact but not closed.

Theorem 8.14. (Heine-Borel) Every bounded closed interval [a, b] of R with the usualtopology is compact.

53

Proof. Let (Uλ)λ∈Λ be an open cover of [a, b], and let

S = {s : s ≥ a and ∃ a finite subcover of [a, s]}.

It is clear that S is nonempty, since a ∈ S.

Claim: There is an s ∈ S with s > b.

Suppose it is not true, s ≤ b for all s ∈ S. Then S is nonempty and bounded aboveby b.

Let σ = supS. Then σ ∈ [a, b], so σ ∈ Uλ0 for some λ0 ∈ Λ. Since Uλ0 is open, forsome ε > 0,

(σ − ε, σ + ε) ⊆ Uλ0 .

Recall that σ = supS, thus there is x ∈ S with

σ − ε < x ≤ σ,

so there is a finite subcover of [a, x], and if we add Uλ0 we get a finite subcover of[a, σ + ε/2]. This means that σ + ε/2 ∈ S, contradicting the fact that σ = supS. Thisproves the claim.

Note that any finite subcover of [a, s], withs > b is a finite subcover of [a, b]. This proves the theorem.

Theorem 8.15. A subset K of R with the usual topology is compact ⇐⇒ it is closedand bounded.

Proof. [=⇒] Let K be compact in R.By Theorem 8.12, the compact subset K of the Hausdorff space R is closed.Note that

{(−n, n) : n = 1, 2, 3, . . .}is an open cover of K. Thus it has a finite subcover. The union of the subcover is of theform (−N,N) for some integer N . Thus, for allx ∈ K, so |x| ≤ N , that is, K is bounded.

[⇐=] If K is bounded and closed then it is contained in some closed interval [a, b].By the Heine-Borel theorem, [a, b] is compact. Therefore K is a closed subset of thecompact set [a, b]. By Theorem 8.8, K is compact.

54

Example 8.16. Consider the topological space R with usual topology. The sets R,(−∞, 1] and [−5,∞) are not bounded and therefore, by Theorem 8.15, are not compact.

Corollary 8.17. A nonempty compact subset K of R with the usual topology containsits supremum and infimum.

Proof. Let σ = supK. Then for every ε > 0 there exists k ∈ K with

σ − ε < k ≤ σ.

Thus every neighbourhood of σ meets K, that is, σ ∈ K. By Theorem 8.12, the com-pact subset K of the Hausdorff space R is closed, so σ ∈ K.

Similarly inf K ∈ K.

55

9 Continuity

Definition 9.1. Let X, Y be topological spaces. Then f : X −→ Y is continuous atx ∈ X if, for each open set V ⊆ Y containing f(x), there is an open set U ⊆ Xcontaining x such that f(U) ⊆ V .

Definition 9.2. A function f : X −→ Y is continuous on X ⇐⇒ it is continuous ateach point of X .

Theorem 9.3. f : X −→ Y is continuous on X ⇐⇒ for every open subset V in Y ,f−1(V ) is open in X .

Proof. [=⇒] Suppose f is continuous on X , and let V be open in Y . Then, for eachx ∈ f−1(V ), V ⊆ Y is an open set containing f(x). By Definition 9.1, there is an openset U ⊆ X containing x such that f(U) ⊆ V , that is, x ∈ U ⊆ f−1(V ). Thus f−1(V )is open in X .

[⇐=] Suppose, for every open subset W in Y , f−1(W ) is open in X . Claim: f iscontinuous on X .

Let x ∈ X and let V ⊆ Y be an open set containing f(x). Then, by assumption,f−1(V ) is open in X . Thus there is an open set U = f−1(V ) containing x and f(U) ⊆V . Hence f is continuous at each point of X , that is, f is continuous on X .

Remark 9.4. In Theorem 1.12 we proved that a function f : R −→ R is ε−δ continuouson R if and only if it is continuous in the sense of Theorem 9.3.

Example 9.5. If X has the discrete topology and Y is any topological space then everyfunction f : X −→ Y is continuous. If V is open in Y then f−1(V ) is open in X , sinceevery subset of X is open in the discrete topology.

Example 9.6. If Y has the indiscrete topology and X is any topological space thenevery function f : X −→ Y is continuous. The only open sets in Y are Y and ∅, andf−1(Y ) = X , and f−1(∅) = ∅, both open in X .

56

Theorem 9.7. Let (X, dX) and (Y, dY ) be metric spaces. Then f : X −→ Y is contin-uous at x⇐⇒ for each ε > 0 there is δ > 0 such that f(B(x, δ)) ⊆ B(f(x), ε).

Proof. It is similar to Theorem 1.12.

Theorem 9.8. A continuous image of a compact space is compact, that is, if X is com-pact and

f : X −→ Y

is continuous then f(X) is compact.

Proof. Let (Uλ)λ∈Λ be an open cover of f(X), that is,

f(X) ⊆⋃λ∈Λ

Uλ.

ThusX ⊆

⋃λ∈Λ

f−1(Uλ).

By Theorem 9.3, since f is continuous on X , (f−1(Uλ))λ∈Λ is an open cover of X . Byassumption, X is compact and so has a finite subcover (f−1(Uλi))

ni=1, that is,

X ⊆n⋃i=1

f−1(Uλi).

Therefore

f(X) ⊆n⋃i=1

Uλi .

Hence (Uλi)ni=1 is a finite subcover of f(X).

Theorem 9.9. Let K be compact and letf : K → R be continuous. Then f is bounded on K and attains its bounds.

Proof. By Theorem 9.8, f(K) is compact in R and, by Theorem 8.15, it is bounded.By Corollary 8.17, f(K) attains its supremum and infimum.

57

Homeomorphisms

Definition 9.10. Let (X, τX) and (Y, τY ) be topological spaces. Then a function

f : X −→ Y

is a homeomorphism if(i) f is a bijection (so f−1 is also a bijection);(ii) both f and f−1 are continuous.

Definition 9.11. When such a homeomorphism exists, we say thatX is homeomorphicto Y .

Remark 9.12. A homeomorphism sets up(i) a one-to-one correspondence between the points of X and the points of Y and(ii) a one-to-one correspondence between the open sets of the two spaces.

Proposition 9.13. Let (X, τX) and (Y, τY ) be topological spaces, and let f : X −→ Ybe a homeomorphism. Then U is open in X ⇐⇒ f(U) is open in Y .

Proof. [=⇒]: By Theorem 9.3, since

f−1 : Y → X

is continuous on Y , U open in X =⇒ f(U) = (f−1)−1(U) is open in Y .

[⇐=]: By Theorem 9.3, since f is continuous on X , f(U) open in Y =⇒ U =f−1(f(U)) is open in X .

Example 9.14. The real line R is homeomorphic to the open interval (0,∞), since thefunction x 7→ ex is a homeomorphism of R onto (0,∞). One can check that bothx 7→ ex and its inverse x 7→ log x are continuous bijections.

58

Example 9.15. The open interval (0,∞) is homeomorphic to the open interval (1,∞),since the function x 7→ x+ 1 is a homeomorphism. One can check that both x 7→ x+ 1and its inverse x 7→ x− 1 are continuous bijections.

Example 9.16. The open interval (0, 1) is homeomorphic to the open interval (0,∞)

via the homeomorphism x 7→ 1

x−1. One can check that both x 7→ 1

x−1 and its inverse

y 7→ 1

y + 1are continuous bijections.

59

10 Metric Spaces AgainIn the case of metric spaces some topological concepts can be translated into the lan-guage of sequences.

Theorem 10.1. If A is a subset of a metric space (X, d) then x ∈ A ⇐⇒ there is asequence (an)∞n=1 in A which converges to x.

Proof. [=⇒] If x ∈ A then every open ball B(x, 1n) about x contains a point an of A.

Thend(x, an) <

1

n,

so d(x, an)→ 0 as n −→∞. Therefore

limn→∞

an = x

with respect to the metric d.[⇐=] Suppose that there is a sequence (an)∞n=1 in A which converges to x. Let U be

any open set containing x. By the definition of open set in metric spaces, ∃ε > 0 suchthat B(x, ε) ⊆ U . Since limn→∞ an = x, for ε > 0, ∃N ≥ 1 such that

n > N =⇒ d(an, x) < ε

=⇒ an ∈ B(x, ε) ⊆ U.

Thus every open set U containing x meets A, and so x ∈ A.

Theorem 10.2. Let (X, dX) and (Y, dY ) be metric spaces. Then a function f : X → Yis continuous at x ∈ X if and only if

limn→∞

xn = x =⇒ limn→∞

f(xn) = f(x).

Proof. [=⇒] Suppose that f is continuous at x and that

limn→∞

xn = x.

Claim: for each ε > 0 there is N ≥ 1 such that dY (f(xn), f(x)) < ε for all n ≥ N .

Since f is continuous, for ε > 0, there is δ > 0 such that f(B(x, δ)) ⊆ B(f(x), ε).

60

Since limn→∞ xn = x, for δ > 0, ∃N ≥ 1 such that

n ≥ N =⇒ dX(xn, x) < δ

=⇒ xn ∈ B(x, δ) =⇒ f(xn) ∈ B(f(x), ε)

=⇒ dY (f(xn), f(x)) < ε.

Thus limn→∞ f(xn) = f(x).

[⇐=] By assumption,

limn→∞

xn = x =⇒ limn→∞

f(xn) = f(x).

Suppose that f is NOT continuous at x. We shall show that there is a sequence(xn)∞n=1 such that limn→∞ xn = x, but f(xn) 6→ f(x) as n→∞.

Since f is not continuous there exists ε > 0 such that, for every δ > 0, there is apoint a such that dX(a, x) < δ but dY (f(a), f(x)) ≥ ε.

Choose δ successively equal to 1, 12, 1

3, . . . and get a sequence of points x1, x2, x3, . . .

such thatd(x, xn) <

1

nand d(f(x), f(xn)) ≥ ε.

Clearly limn→∞ xn = x but f(xn) 6→ f(x) as n −→ ∞. This is a contradiction to theassumption.

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11 Completeness in Metric Spaces

Definition 11.1. A sequence (xn)∞n=1 in a metric space (X, d) is a Cauchy sequence iffor every ε > 0 there exists N ≥ 1 such that, for all m,n ≥ N , d(xm, xn) < ε.

Informally, (xn)∞n=1 is Cauchy if d(xm, xn)→ 0 as min(m,n)→∞.

Theorem 11.2. Every convergent sequence in a metric space is a Cauchy sequence.

Proof. Let (xn)∞n=1 be a convergent sequence in (X, d), and let limn→∞ xn = x. Thus,for every ε > 0, ∃N ≥ 1 such that n ≥ N =⇒d(xn, x) < ε/2. Hence, for all m,n ≥ N ,

d(xm, xn) ≤ d(xm, x) + d(x, xn)

< ε/2 + ε/2 = ε.

Therefore (xn)∞n=1 is a Cauchy sequence.

Proposition 11.3. Every Cauchy sequence in C with the usual metric is bounded.

Proof. Let (xn)∞n=1 be a Cauchy sequence. Take ε = 1, ∃N ≥ 1 such that m,n ≥ Nwe have |xm − xn| < 1. Note that, for all n ≥ N ,

|xn| ≤ |xn − xN |+ |xN | < 1 + |xN |.

Thus (xn)∞n=1 is bounded by

M = max1≤i≤N

{|xi|, (1 + |xN |)}.

Example 11.4. The sequence (xn)∞n=1,xn = 3n− 4ni, where i2 = −1, is not bounded in C, since |xn| = 5n→∞ as n→∞.Thus, by Proposition 11.3, the sequence is not a Cauchy sequence.

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Definition 11.5. A metric space (X, d) is complete if every Cauchy sequence in (X, d)converges in (X, d).

Example 11.6. 1. Note that limn→∞(2 − 1n) = 2 in R with the usual metric. Thus, by

Theorem 11.2, (xn)∞n=1, xn = 2− 1n, is Cauchy sequence in R and so in (0, 2).

2. (0, 2) with the usual metric is not complete, because (xn)∞n=1, xn = 2 − 1n, is

Cauchy sequence in (0, 2) which does not converge in (0, 2).

Example 11.7. Metric spaces R and C with the usual metric are complete.

Proposition 11.8. A closed subspace (Y, d) of a complete metric space (X, d) is com-plete.

Proof. Let (yn)∞n=1 be a Cauchy sequence in (Y, d), and so in (X, d). Since (X, d) iscomplete, the sequence (yn)∞n=1 coverges to a limit y ∈ X . By assumption, Y is closed.Therefore the limit of the convergent sequence (yn)∞n=1 from Y belongs to Y , that is,

limn→∞

yn = y ∈ Y.

Thus (yn)∞n=1 is convergent in (Y, d), and so (Y, d) is complete.

Completeness of C(K)

Let K be a compact Hausdorff topological space and let C(K) be the set of contin-uous real-valued functions on K. For f ∈ C(K), let

||f ||∞ = sup{|f(x)| : x ∈ K}.

By Theorem 9.9, f is bounded on K, since K is compact. One can check that

d∞(f, g) = ||f − g||∞

defines a metric d∞ on C(K).

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Theorem 11.9. (C(K), d∞) is a complete metric space.

Proof. 1. Let (fn)∞n=1 be a Cauchy sequence in (C(K), d∞). Then, for every ε > 0,there is an integer N ≥ 1 such that

m,n ≥ N =⇒ ||fm − fn||∞ < ε

=⇒ for all x ∈ K, |fm(x)− fn(x)| < ε.

Hence, for every x ∈ K, (fn(x))∞n=1 is a Cauchy sequence in R with the usual met-ric. Since R is complete, for each x ∈ K, the sequence (fn(x))∞n=1 converges in R.Denote its limit limn→∞ fn(x) by f(x).

2. By Part 1, for every ε > 0, there is an integer N ≥ 1 such that

m,n ≥ N =⇒ for all x ∈ K, |fm(x)− fn(x)| < ε.

Let m→∞, then, for all x ∈ K and for all n ≥ N ,

|f(x)− fn(x)| ≤ ε,

since limm→∞ fm(x) = f(x).Hence, for all n ≥ N ,

||fn − f ||∞ ≤ ε.

Therefore (fn)∞n=1 converges to the function f with respect to the metric d∞.

3. Next we show that f ∈ C(K), that is f is continuous on K. Let x0 ∈ K, and letε > 0. By Part 2, there is N such that, for all n ≥ N ,

||fn − f ||∞ < ε/3.

For the continuous function fN there is a neighbourhood U of x0 such that for allx ∈ U ,

|fN(x0)− fN(x)| < ε/3.

Then, for all x ∈ U , we have

|f(x)− f(x0)| =

|f(x)− fN(x) + fN(x)− fN(x0) + fN(x0)− f(x0)| ≤

|f(x)− fN(x)|+ |fN(x)− fN(x0)|+ |fN(x0)− f(x0)|

< ||f − fN ||∞ + ε/3 + ||fN − f ||∞

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< ε/3 + ε/3 + ε/3 = ε.

Thus f ∈ C(K).

Therefore (C(K), d∞) is complete, since every Cauchy sequence in (C(K), d∞)converges to a continuous function on K.

Example 11.10. By Theorem 11.9, (C(K), d∞) is complete. By Proposition 5.6, theclosed ball

B∞(g, r) = {f ∈ C(K) : d∞(f, g) ≤ r},

where g ∈ C(K) and r ∈ R, r > 0, of a metric space (C(K), d∞) is closed. Therefore,by Proposition 11.8, the metric space (B∞(g, r), d∞), is complete.

Contractions and the Contraction Mapping Principle

Definition 11.11. Let (X, d) be a metric space and T : X −→ X be a mapping. T is acontraction or contraction mapping if there exists k ∈ R, 0 ≤ k < 1, such that

d(Tx, Ty) ≤ kd(x, y) for all x, y ∈ X.

Remark 11.12. Note that contractions T are automatically continuous, since, for everyε > 0, there is δ = ε such that d(x, y) < δ implies that

d(Tx, Ty) ≤ kd(x, y) < ε.

Remark 11.13. We shall write Tx for T (x),T 2(x) for T (T (x)), T 3(x) for T (T (T (x))), and so on. . . . Let T be a contraction.Repeated applications of the following

d(Tx, Ty) ≤ kd(x, y) for all x, y ∈ X,

shows that if r is a positive integer then

d(T rx, T ry) ≤ krd(x, y) for all x, y ∈ X.

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Definition 11.14. A fixed point of a mapping

T : X −→ X

is a point x ∈ X such that Tx = x.

Theorem 11.15. The Contraction Mapping Principle or Banach Fixed Point Theo-rem.Every contraction mapping T on a complete metric space (X, d) has a unique fixedpoint.

Proof. Let x0 be any point of X . Define a sequence (xn)∞n=1 by x1 = Tx0, x2 = T 2x0,and so on.

Step 1: Claim: (xn)∞n=1 is a Cauchy sequence.For n > m, we have

d(xm, xn) = d(Tmx0, Tnx0) =

d(Tmx0, Tm ◦ T n−mx0) ≤ kmd(x0, T

n−mx0)

≤ km[d(x0, Tx0) + d(Tx0, T2x0)+

d(T 2x0, T3x0) + . . .+ d(T n−m−1x0, T

n−mx0)] ≤

km[d(x0, Tx0) + kd(x0, Tx0) + k2d(x0, Tx0)+

. . .+ kn−m−1d(x0, Tx0)] =

km[1 + k + k2 + . . .+ kn−m−1] · d(x0, Tx0) =

km(

1− kn−m

1− k

)· d(x0, Tx0) <

(km

1− k

)· d(x0, Tx0).

We can make the last expression as small as we please by taking m (and so also n)sufficiently large. Therefore (xn)∞n=1 is a Cauchy sequence.

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By assumption, (X, d) is complete, so (xn)∞n=1 converges to some x ∈ X withrespect to the metric d.

Step 2: Claim: x is a fixed point of T . We have limn→∞ xn = x. Since T iscontinuous, we have also limn→∞ Txn = Tx.

By the definition, Txn = xn+1 and so

limn→∞

Txn = limn→∞

xn+1 = x.

By uniqueness of limits for the sequence (Txn)∞n=1, we must have Tx = x, that is, x isa fixed point of T .

Step 3: Claim: x is the only fixed point of T .If x′ is a fixed point distinct from x, then

0 < d(x, x′) = d(Tx, Tx′) ≤ k · d(x, x′) < d(x, x′),

a contradiction.

Example 11.16. Let f : [a, b] → [a, b] be a function such that |f ′(x)| ≤ K < 1 for allx : a ≤ x ≤ b. Then f has a unique fixed point on [a, b].

Proof. Note that [a, b] is a closed subspace of the complete metric space R with theusual metric. Hence, by Proposition 11.8, [a, b] is complete.

By the Mean Value Theorem, for all x1, x2 ∈ [a, b], we have

|f(x1)− f(x2)| = |f ′(ξ)| · |x1 − x2| ≤ K · |x1 − x2|

where ξ lies between x1 and x2. Since K < 1, f is a contraction mapping. Therefore,by the Contraction Mapping Theorem, f has a unique fixed point.

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12 ConnectednessDefinition 12.1. A topological space (X, τ) is disconnected if there are disjoint nonemptyopen sets U and V in X such that X = U ∪ V .

In this case, we say that (X, τ) is disconnected by U and V .

Definition 12.2. A topological space (X, τ) is connected if it cannot be expressed asthe union of two disjoint nonempty open sets U and V .

Remark 12.3. Since the complement of an open set is closed, it is clear that if U and Vdisconnect X , then U and V are closed as well as open. Such sets are said to be clopen.

Consequently, we could equally use closed sets in the definition of a disconnectedspace.

In a connected space, the only clopen sets are therefore the empty set and the wholespace.

Example 12.4. If (X, τdisc) has more than one point and the discrete topology, then itis disconnected.

Example 12.5. Let (X, τ) be a T1 space with more than one point. If (X, τ) has anisolated point p, then (X, τ) is disconnected by U = p and V = X \ p. By Proposition6.4, p is clopen.

Proposition 12.6. A subspace of R with the usual topology is connected if and only if itis an interval or a point.

Proof. See Solutions 4.

Lemma 12.7. Suppose (X, τ) is disconnected with separation U and V . If E ⊆ X isconnected, then either E ⊆ U or E ⊆ V .

Proof. Since E = (E ∩ U) ∪ (E ∩ V ), and the sets E ∩ U and E ∩ V are open in therelative topology on E, they separate E unless one of them is empty and the other is allof E.

We next consider how connectedness behaves under continuous maps and subspacesof connected spaces.

Theorem 12.8. Let (X, τX) be a connected topological space, and let be

f : (X, τX)→ (Y, τY )

be a continuous map. Then the continuous image of f is connected.

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Proof. Suppose τY -open subsets U and V disconnect f(X). Then, by continuity of f ,f−1(U) and f−1(V ) are τX-open, disjoint, and f−1(U) ∪ f−1(V ) = f−1(f(X)) = X .This is a contradiction to the connectedness of (X, τX).

Corollary 12.9. A topological space (X, τX) is connected if and only if there is nocontinuous surjection from (X, τX) onto {0, 1} (with the discrete topology).

Proof. The space {0, 1} (with the discrete topology) is disconnected by {0} and {1}. If(X, τX) is connected, then, by Theorem 12.8, the image of a continuous map on X to{0, 1} is connected, so it is not onto.

On the other hand, if there is a continuous surjection f from X onto {0, 1}, thenf−1({0}) and f−1({1}) give a separation of X , so (X, τX) is not connected.

Corollary 12.10. Let A be a connected subspace of a topological space (X, τX), andsuppose A ⊆ B ⊆ A. Then B is connected. In particular, the closure of a connectedspace is connected.

Proof. Suppose that B is not connected. By Corollary 12.9, there exists a continuoussurjection f : B → {0, 1}. Since A is connected, we may assume that f |A takes A to 0.By assumption, B is not connected, and so there is a point x ∈ B such that f(x) = 1.It is clear that x /∈ A and there is an open set U ⊆ f−1({1}) containing x. Note thatU ∩A 6= ∅. This is a contradiction to the fact that x is in A. Hence B is connected.

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13 Picard’s Theorem

Theorem 13.1. Picard’s Theorem. Let

f : S → R

and∂f

∂ybe continuous on the closed rectangle S = [a1, a2]× [b1, b2] in R2 and let

a1 < x0 < a2, b1 < y0 < b2.

Then the ordinary differential equation (ODE)

dy

dx= f(x, y), y(x0) = y0

has a unique solution y = g(x) on an interval about x0.

Proof. We shall find a function g(x) which satisfies the integral equation

g(x) = y0 +

∫ x

x0

f(t, g(t))dt.

If we find such a function then(i) set x = x0: this shows that g(x0) = y0;(ii) differentiate: this shows that

dg(x)

dx= f(x, g(x)),

so g(x) satisfies the ODE.

By assumption, f : S → R and∂f

∂yare continuous on the closed rectangle

S = [a1, a2]× [b1, b2]

in R2. Thus they are bounded on S, say, by K and by M , respectively. One can find apositive real number a such that

(i) a <1

Mand(ii) |x− x0| ≤ a, |y − y0| ≤ Ka

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implies that (x, y) ∈ S.

Consider the metric space (C[x0 − a, x0 + a], d∞), of continuous real-valued func-tions on

[x0 − a, x0 + a]

with the metricd∞(f, g) = sup

x0−a≤t≤x0+a|f(t)− g(t)|.

LetX = {g ∈ C[x0 − a, x0 + a] : d∞(g, y0) ≤ Ka}

and let a map T : X → X be defined by

(Tg)(x) = y0 +

∫ x

x0

f(t, g(t))dt.

Let us check that T is well defined.

1. Let g ∈ X . We claim Tg ∈ C[x0 − a, x0 + a].Let x ∈ [x0 − a, x0 + a] and let ε > 0. Take δ = ε/K. Then, for all x′ such that|x− x′| < δ,

|(Tg)(x)− (Tg)(x′)| =∣∣∣∣∣[y0 +

∫ x

x0

f(t, g(t))dt

]−

[y0 +

∫ x′

x0

f(t, g(t))dt

]∣∣∣∣∣=

∣∣∣∣∫ x

x′f(t, g(t))dt

∣∣∣∣≤ |x− x′| ·K <

ε

K·K < ε.

Therefore Tg is continuous on [x0 − a, x0 + a], and so

Tg ∈ C[x0 − a, x0 + a].

2. Claim: Tg ∈ X . By properties of integrals,

d∞(Tg, y0) =

supx0−a≤x≤x0+a

∣∣∣∣[y0 +

∫ x

x0

f(t, g(t))dt

]− y0

∣∣∣∣= sup

x0−a≤x≤x0+a

∣∣∣∣∫ x

x0

f(t, g(t))dt

∣∣∣∣ ≤ aK,

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so Tg ∈ X .

3. Claim: T : X → X is a contraction.

For t ∈ [x0 − a, x0 + a] and g1, g2 ∈ X , by the Mean Value Theorem,

|f(t, g1(t))− f(t, g2(t))| ≤M · |g1(t)− g2(t)|,

since∣∣∣∣∂f∂y (t, s)

∣∣∣∣ ≤M on S.

Thus, for |x− x0| ≤ a,|(Tg1)(x)− (Tg2)(x)| =∣∣∣∣y0 +

∫ x

x0

f(t, g1(t))dt− y0 −∫ x

x0

f(t, g2(t))dt

∣∣∣∣ =∣∣∣∣∫ x

x0

[f(t, g1(t))− f(t, g2(t))]dt

∣∣∣∣≤M ·

∫ x

x0

|g1(t)− g2(t)|dt

≤ d∞(g1, g2) ·M · a.

Thusd∞(Tg1, T g2) =

supx0−a≤x≤x0+a

|(Tg1)(x)− (Tg2)(x)|

≤ (M · a) · d∞(g1, g2),

where M · a < 1. Thus T : X −→ X is a contraction mapping.

4. Claim: there is a function g ∈ X which satisfies the integral equation

g(x) = y0 +

∫ x

x0

f(t, g(t))dt.

By Theorem 11.9, the metric space(C[x0 − a, x0 + a], d∞) is complete.

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By Proposition 5.6,X = B∞(y0, a ·K) is a closed subset. We know that a closed subsetof a complete metric space is complete. Therefore X is complete. Thus T : X −→ Xis a contraction mapping on the complete metric space X . By the Contraction MappingPrinciple, T has a unique fixed point g ∈ X . Therefore

g(x) = Tg(x) = y0 +

∫ x

x0

f(t, g(t))dt.

We have found the unique function g ∈ X which satisfies the integral equation

g(x) = y0 +

∫ x

x0

f(t, g(t))dt.

Thus, for x = x0, we have g(x0) = y0. Let us differentiate the integral equation: itshows that

dg(x)

dx= f(x, g(x)),

and so g(x) satisfies the ordinary differential equation

dy

dx= f(x, y), y(x0) = y0

and the solution y = g(x) is unique on an interval about x0. The theorem is proved.

Remark 13.2. The Contraction Mapping Theorem also tells us that if we start with anyfunction g0 in X , then the sequence of functions

g0, T g0, T2g0, T

3g0, . . .

delivers better and better approximations to the unique solution g, and that these ap-proximations converge to g in the metric

d∞(f, h) = ||f − h||∞.

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