where we are in your textbook - cpb-us-w2.wpmucdn.com · p goes around each of a, b and c with...
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where we are in your textbook
I you should’ve read chapter 1, sections 1,3,6
I chapter 1, sections 4,5 are good background reading for realanalysis; rest of chapter 1 is optional
I continue with chapter 2, sections 1,2,3
I will focus on chapter 2, section 6 (compactness)
Tibor Beke metric spaces
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answer to problem 1
Let a, b, c be three distinct points in the plane. Can you drawcurves P and Q in the plane, avoiding the points a, b, c , such that
• P goes around each of a, b and c with winding number +2• Q goes around each of a, b and c with winding number +2• P cannot be continuously deformed into Q (without passingthrough one of the points a, b, c).
Yes. Can be done in infinitely many different ways, in fact.(“Different” here means “non-deformable into each other”.We’ll make this precise!)
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answer to problem 2(a)(b)
(a) (U ∪ V ) rW = (U rW ) ∪ (V rW ) is equivalent to
(U ∪ V ) ∩W = (U ∩W ) ∪ (V ∩W )
true by virtue of the distributive law for ∩ with respect to ∪.
(b) (U ∩ V ) rW = (U rW ) ∩ (V rW ) is equivalent to
(U ∩ V ) ∩W = (U ∩W ) ∩ (V ∩W )
which follows from properties of ∩.
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answer to problem 2(c)
(c) the next two could differ, however . . .
U r (V rW ) = U ∩ V rW = U ∩ V ∩W = U ∩ (V ∪W ) =
U ∩{
(V ∩W ) t (V ∩W ) t (V ∩W )}
=
(U ∩ V ∩W ) t (U ∩ V ∩W ) t (U ∩ V ∩W )
while(U r V ) rW = (U ∩ V ) ∩W = U ∩ V ∩W
Above, t stands for ‘disjoint union’.
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problem 4
The ordinary notion of euclidean distance in the plane correspondsto the formula
d(〈x1, x2〉, 〈y1, y2〉) =
√(x1 − y1)2 + (x2 − y2)2
Prove using algebra, not geometry that this distance functionsatisfies the triangle inequality. Concretely, prove the algebraicinequality√
(x1 − z1)2 + (x2 − z2)2 6√(x1 − y1)2 + (x2 − y2)2 +
√(y1 − z1)2 + (y2 − z2)2
When is there an equality?
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answer to problem 4 using “on the nose” algebra
Set ui = xi − yi , vi = yi − zi , then ui + vi = xi − zi (i = 1, 2).Need to prove√
(u1 + v1)2 + (u2 + v2)2 6√
u21 + u22 +√
v21 + v22
Square, simplify, divide by 2 to obtain
u1v1 + u2v2 6√
u21 + u22
√v21 + v22
Square, expand, simplify to obtain
2u1v1u2v2 6 u21v22 + u22v
21
0 6 u21v22 − 2u1v1u2v2 + u22v
21
0 6 (u1v2 − u2v1)2
Steps are reversible(!), equality iff the vectors 〈u1, u2〉 and 〈v1, v2〉are parallel, iff the points (x1, x2), (y1, y2), (z1, z2) are collinear.
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The above proof generalizes to Rn, though the algebra is trickier.The second stage becomes
n∑i=1
uivi 6
√√√√ n∑i=1
u2i
√√√√ n∑i=1
v2i
( n∑i=1
uivi
)26( n∑
i=1
u2i
)( n∑i=1
v2i
)the beautiful Cauchy-Schwarz inequality (over R). A one-line proofof this inequality is the algebraic identity( n∑
i=1
uivi
)2+
∑16i<j6n
(uivj − ujvi )2 =
( n∑i=1
u2i
)( n∑i=1
v2i
)the n = 2 case of which you discovered on the previous slide.
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“textbook” solution to problem 4
We work in Rn. For vectors u = 〈u1, u2, . . . , un〉 andv = 〈v1, v2, . . . , vn〉, inner (a.k.a. ‘scalar’ or ‘dot’) product
u · v :=n∑
i=1
uivi
and norm ‖u‖ :=√
u · u defined as usual.
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Cauchy-Schwarz inequality
Let α, β be arbitrary scalars. Using non-negativity of squares andbilinearity of inner product
0 6 (αu− βv) · (αu− βv) = α2 u · u− 2αβ u · v + β2 v · v
Choose the particular scalars α = ‖v‖ and β = ‖u‖ and substitute:
0 6 ‖v‖2‖u‖2 − 2‖v‖‖u‖u · v + ‖u‖2‖v‖2
‖v‖‖u‖u · v 6 ‖u‖2‖v‖2
u · v 6 ‖u‖‖v‖
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Cauchy-Schwarz inequality
u · v 6 ‖u‖‖v‖Using this on −u and v gives
−u · v 6 ‖u‖‖v‖
The two cases can be compressed into the single inequality
|u · v| 6 ‖u‖‖v‖
equivalently
(u · v)2 6 ‖u‖2‖v‖2 = (u · u)(v · v)
‘long-hand’ form
( n∑i=1
uivi)2
6( n∑i=1
u2i)( n∑
i=1
v2i)
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triangle inequality
In Rn, we wish to prove
‖x− z‖ 6 ‖x− y‖+ ‖y − z‖
Write u := x− y and v := y − z so u + v = x− z. Want to prove
‖u + v‖ 6 ‖u‖+ ‖v‖
‖u + v‖2 6 (‖u‖+ ‖v‖)2
(u + v) · (u + v) 6 ‖u‖2 + 2‖u‖‖v‖+ ‖v‖2
‖u‖2 + 2 u · v + ‖v‖2 6 ‖u‖2 + 2‖u‖‖v‖+ ‖v‖2
u · v 6 ‖u‖‖v‖
Steps reversible; equality iff equality in Cauchy-Schwarz iff u, vlinearly dependent (with non-negative inner product), iff x, y, zcollinear (in this order).
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different metrics, same opens
Recall our three examples of metrics on R2:
d1(〈x1, x2〉, 〈y1, y2〉) =√
(x1 − y1)2 + (x2 − y2)2
d2(〈x1, x2〉, 〈y1, y2〉) = |x1 − y1|+ |x2 − y2|
d3(〈x1, x2〉, 〈y1, y2〉) = max{|x1 − y1|, |x2 − y2|}
Proposition With respect to any one of these metrics, exactly thesame subsets of R2 are open.
Let Bi (x , r) be the ‘open ball’ with center x ∈ R2 and radius r > 0in the metric di (i = 1, 2, 3). Enough to prove that Bi (x , r) is anopen set when considered with respect to the metric dj , i 6= j .Use elementary geometry!
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strongly equivalent metrics
Definition Let d1 and d2 be two metrics on the set X .They are said to be strongly equivalent (sometimes, coarselyequivalent) if there exist constants 0 < k < K <∞ (dependingonly on the metrics) such that for all distinct points x , y ∈ X ,
k <d1(x , y)
d2(x , y)< K
This implies thatB1(x , kr) ⊆ B2(x , r)
B2(x ,r
K) ⊆ B1(x , r)
for all x ∈ X and r > 0, where Bi (−,−) means the open ball withrespect to the metric di .
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topologically equivalent metrics
Proposition Suppose d1 and d2 are strongly equivalent. ThenU ⊆ X is an open subset in (X , d1) if and only if it is open in(X , d2).
Two metrics d1 and d2 on a set X with the property that theydefine the same open sets are said to be topologically equivalent.
We’ve just seen that strong equivalence is a sufficient (but notnecessary) condition for topological equivalence.
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why abstract away from metric spaces?
I the continuity of a map f : (X , dX )→ (Y , dY ) between metricspaces only depends on what the open subsets of X and Y are
I lots of metrics give rise to the same open sets; there is no‘simplest’ or ‘best’ metric in general
I there are spaces on which continuity makes sense that do notarise from a metric (e.g. Frechet spaces; non-Hausdorff spacesin combinatorial and algebraic geometry; infinite products ofmetric spaces; many function spaces)
I there is no natural metric on quotients by equivalencerelations and ‘gluings’ of metric spaces
I we will not abandon metric spaces, really; they will continueto be the main source of examples and motivation for ourdefinitions.
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topology on a set of points
Seeing that the notion of continuity can be encapsulated in thealgebra of open sets, we introduce
Definition Let X be a set (to be called ‘the set of points’).A topology on X is a collection of subsets of X (to be called ‘thecollection of open subsets of X ’) satisfying the axioms
I The empty set is open; X itself is open (in X ).
I If each Ui is open, i ∈ I (arbitrary collection!), so is theirunion
⋃i∈I Ui .
I If each Ui is open, i = 1, 2, . . . , n (finite collection!), so istheir intersection
⋂ni=1 Ui .
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topological spaces
A topological space is a set equipped with a topology.
Any metric space is automatically a topological space: given themetric, define the notion of open set with the help of ε-balls asbefore. The open subsets of any metric space satisfy the axiomsfor a topology.
Example The following is a topological space:
I X = {?, •}I the open subsets are the empty set ∅, X itself and {•}.
This topology does not come from any metric on the set X .
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interior
Let X be a topological space and W an arbitrary subset of X .
The point x ∈W is an interior point of W if there exists an openset U such that x ∈ U ⊆W . The set of interior points of W is theinterior of W , denoted int(W ).
I int(W ) is always an open set
I int(W ) is the largest open set contained in W (the union ofall open sets contained in W )
I a set is open if and only if it equals its interior.
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closed sets; closure
Let X be a topological space. A subset of X is closed (bydefinition) if its complement is open. ∅ and X are closed; arbitraryintersection of closed sets is closed; finite union of closed sets isclosed.
Let W be an arbitrary subset of X . The point x ∈ X is a point ofclosure of W if every open set containing x intersects Wnon-trivially. The set of points of closure of W is the closure ofW , denoted W .
I W is always a closed set
I W is the smallest closed set containing W (the intersection ofall closed sets containing W )
I a set is closed if and only if it equals its closure.
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For any topological space X and any subset S ⊆ X ,
int(S) t X r S = X
S t int(X r S) = X
(Recall that ‘t’ stands for ‘disjoint union’.)
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some terminology
Let X be a topological space, x ∈ X a point. An openneighborhood U of x is any open subset of X containing x .A neighborhood S of x is any subset of X that contains an openneighborhood of x . (So, x ∈ U ⊆ S ⊆ X and U is open in X .)
Remark The meaning of ‘neighborhood’ varies slightly fromtextbook to textbook, but ‘open neighborhood’ means the same toall authors.
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continuity; homeomorphism
Let f : X → Y be a function from one topological space toanother. f is continuous (over its entire domain) if for every opensubset U of Y , its preimage f −1(U) is an open subset of X .
The composite of two continuous maps is continuous.
f : X → Y is a homeomorphism if it is a continuous bijection witha continuous inverse.
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compactness of the closed unit interval (version one)
Theorem Let (ai , bi ), i ∈ I , be a family of open intervals suchthat
[0, 1] ⊂⋃i∈I
(ai , bi ) .
Then there exist finitely many of these open intervals, say(ai1 , bi2), . . . , (ain , bin), so that already
[0, 1] ⊂n⋃
k=1
(aik , bik ) .
Slogan: “ Every cover of [0, 1] by open intervals has a finitesubcover ”
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Homework Problem 12
(a) List all possible topologies on the 3-element set {a, b, c}.(b) How many different homeomorphism types of 3-point spacesare there?
Due Wed, Feb 12, in class.
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Homework Problem 13
Your textbook, Chapter 2, Section 1, Exercise 2 on page 63.
Due Wed, Feb 12, in class.
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Homework Problem 14∗ (optional)
Your textbook, Chapter 2, Section 1, Exercise 11 on page 63.
Due (optionally) Wed, Feb 19, in class.
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Homework Problem 15
Your textbook, Chapter 2, Section 3, Exercise 2 on page 68.
Due Wed, Feb 12, in class.
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Homework Problem 16
Let (X , dX ) and (Y , dY ) be metric spaces. On the cartesianproduct X × Y , define
dX×Y(〈x1, y1〉, 〈x2, y2〉
)= dX (x1, x2) + dY (y1, y2)
(a) Show that dX×Y is a metric on the set X × Y .(b) Show that the projection maps X × Y → X and X × Y → Yare continuous in this metric.(c)∗ (optional) Let (Z , dZ ) be an arbitrary metric space andf : Z → X and g : Z → Y two functions. Show that the maph : Z → X × Y that takes z ∈ Z to 〈f (z), g(z)〉 is continuous ifand only if both of the coordinate functions f and g arecontinuous.
Due Wed, Feb 12, in class.
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