Solutions to Assignment-3 - UCB Mathematics

Solutions to Assignment-3

September 19, 2017

1. Let (X, d) be a metric space, and let Y X be a metric subspace with the induced metric dY . Let E Y.

(a) Show that a set U Y is open in Y if and only if there is a subset V X open in X such that U = V Y . As an example, consider X = R, Y = [0, 1]. Then U = [0, 1/2) is an open subset of Y with the induced metric. In this case we can take V = (-1, 1/2). Then V is open in R and U =Y V.

Solution: We will denote the balls of radius r around p Y with respect to metrics dY and d by BrY (p) and BrX (p) respectively. That is

BrY (p) = {y Y | d(p, y) < r}, BrX (p) = {x X | d(p, x) < r}.

The key observation is that

BrY (p) = BrX (p) Y.

As an illustration take X = R and Y = Q. Then B1Q(0) consists of all the rationals in the interval (-1, 1) while B1R(0) is the whole interval (-1, 1) and clearly B1Q(0) = B1R(0) Q.

Coming back to the problem, suppose U Y is open (with respect to the subspace metric dY ). Then for each p U , there exists an rp > 0 such that BrYp (p) U. So we can write

U = pU BrYp (p).

Let V = pU BrXp (p). Then clearly V is open in X since arbitrary union of open sets is open, and by the above property of balls, it is clear that U = V Y .

For the converse, suppose U = V Y where V is open in X. Let p U . Then p V , and since V is open in X, there exists rp > 0 such that BrXp (p) V . But then BrYp (p) = BrXp (p) Y U . This shows that for any p U we have a ball BrYp (p) U , and so U is open in Y .

(b) Show that E is compact subset of Y (with respect to the metric dY ) if and only if it is a compact subset of X (with respect to the metric d).

Solution: Suppose E Y X is a compact subset of Y (with respect to dY ). Let {V} be a cover of E by open subsets of X. By the above part, U = V Y will be a cover of E by open subsets of (Y, dY ). Since E is compact with the subspace metric, there exists 1, ? ? ? N such that

E Nk=1Uk Nk=1Vk .

So we have managed to extract a finite sub-cover of {V}. This shows that E is compact in (X, d). The converse is also similar.

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(c) Show that E is a connected subset of Y (with respect to the metric dY ) if and only if it is a connected subset of X (with respect to the metric d).

Solution: Again we show one direction leaving the converse as an excerise. Suppose E Y X is a connected subset of (Y, dY ) but not (X, d). Then there exists non-empty subsets A and B of X such that E = A B but AX B = BX A = . Here we are taking closures with resect to the metric d. Claim. AY B = BY A = . Clearly A B = . Suppose p AY B. Then p is a limit point of A (in the metric dY ). So for any r > 0, BrY (p) A = . Since BrY (p) BrX (p), this shows that BrX (p) A = A for any r > 0. This shows that p is a limit of A even in the metric d. This is a contradiction since AX B = . This completes the proof of the claim. But then we have written E = A B where A and B are non-empty separated sets with respect to the metric dY , contradicting the fact that E is connected with respect to dY . Hence E must be connected with respect to the metric d too.

2. A subset E Rn is called convex if for any two points p, q E, the straight line

l(t) = (1 - t)p + tq, t [0, 1]

joining the two points is contained completely in E. Show that any convex set is connected. Hint. Argue by contradiction.

Solution: If not, then we can write E = A B, where A and B are non-empty and separated, that is, A, B = , and A B = A B = . Recall that this also means that A and B are both open and closed in E.

Choose p A and q B. Since E is convex the straight line l joining them is contained in E. Intuitively, there will be a first point where the line exits A and enters B. This point will lie in A B, a contradiction. More rigorously, let

T = sup{t | l(t) A for all s t}.

Then T is the maximum time such that the line is always in A for any time smaller than T . Claim-1. l(T ) / A. Proof. Now suppose l(T ) A. Since A is open in E, there is an > 0 such that

{x E | |x - l(T )| < } A.

Now note that

|l(s) - l(T )| = |(T - s)p + (s - T )q| |s - T |||p| + |s - T ||q| = |s - T |(|p| + |q|).

So

if

s

[T, T

+

|p|+|q|

),

then

|s

-

T|

<

/(|p|

+

|q|),

and

so

|l(s) - l(T )| < .

By

our

choice

of

and

T,

this

shows

that

l(s)

A

for

all

s

[0, T

+

|p|+|q|

),

contradicting

the

maximality of T . This proves the claim.

Now, since E = A B, the claim implies that l(T ) B. In particular T > 0, since l(0) = p A.

Claim-2. l(T ) is a limit point of A.

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Proof. Let > 0. We need to show that there is some x B(l(T )) A. Now let

s=T -

.

2(|p| + |q|)

Then by the estimate above,

|l(s) - l(T )| < ,

2

and so l(s) B(l(T )). But by definition of T , since s < T , automatically l(s) A, hence proving the claim.

So we have now shown that l(T ) A B, a contradiction.

3. A metric space (X, d) is called separable is it has a countable dense subset. A collection of open sets {U} is called a basis for X if for any p X and any open set G containing p, p U G for some I. The basis is said to be countable if the indexing set I is countable. (a) Show that Rn is countable. Hint. Q is dense in R.

Solution: Let Qn Rn be the set of all points with all rational coordinates. Then Qn is dense in Rn, and is countable. So Rn is separable.

(b) Prove that a metric is separable if and only if it has a countable basis of open sets. Hint. One direction is not hard (which one?). For the other direction, think of smaller and smaller balls of rational radii.

Solution: Proof of = . Suppose (X, d) has a countable basis {Uk}. We can assume without loss of generality that none of the basis elements are identical open sets. Pick a point pk Uk. Let P be the collection of points {pk}. Note that unless X is finite (in which case the proof is trivial anyway), we can always pick points which are not repeated. We claim that X = P . To see this, let x X \ P . We have to show that x is a limit point of P . Let G be any open set containing x. Then by definition of basis, there is a k such that x Uk G. But then pk G P . So every open set around x intersects P at a point different from x itself (since x / P ). This shows that x is a limit point of P .

Proof of = . Let S = {x1, x2, ? ? ? } be the countable dense subset, and let Un,j = B1/n(xj). We claim that {Un,j} forms a basis, and since there are clearly countable number of sets, this completes the proof. To see that it forms a basis, let x X and G be an open set containing x. Since G is open, there is an integer N such that the ball B4/N (x) G. Since S is dense, xj B1/N (x) for some j. Choose an integer N/2 < M < N Then we claim that

x B1/M (xj ) B4/N (x).

The first inclusion follows since M < N and d(x, xj) < 1/N < 1/M . For the second inclusion, let y B1/M (xj). Then

In particular, completing the proof.

11 3

d(x, y) d(x, xj) + d(y, xj) <

N

+

M

<

. N

x UM,j G,

(c) Prove that every compact metric space is separable. Hint. First, cover the metric space by balls of

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radius 1. By compactness only a finitely many such balls are needed (see problem 1). Continue to do this for balls of radius 1/n for larger and larger integers. Does this give you a countable dense subset?

Solution: Let X be a compact metric space. By problem 1, for each integer n there is a number K(n) and points xn,1, ? ? ? , xn,K(n) such that

X Kk=(n1)B1/n(xn,k).

We let Un,k = B1/n(xn,k). Claim. S := {xn,k} is a dense subset. Since it is clearly countable, this will prove that X is separable. Proof. We need to show that for any x X and any open set G, G S is non-empty. Since G is open, there exists an integer N such that B1/N (x) G. But since the collection of balls {B1/N (xN,k)}kK=(N1 ) is a cover for X, there is some k such that x B1/N (xN,k). But then trivially xN,k B1/N (x) G, and so G S is non-empty. This completes the proof of the claim.

4. The aim of this exercise is to complete the proof that compactness and limit point compactness are equivalent. Let (X, d) be a limit point compact metric space. (a) Show for every > 0, X can be covered by finitely many balls of radius . (Note that this is easy for a set already known to be compact; see problem 4 from the previous assignment).

Solution: Pick any point x1. Then pick x2 such that d(x2, x1) . If there is no such point then already X = B(x1) and the claim is proved with N = 1. Now inductively, having picked x1, x2, ? ? ? , xn-1 pick an xn such that d(xn, xj) > for all j = 1, ? ? ? n - 1. We claim that the process terminates in a finitely many steps, thus proving the claim. If not, then we have a sequence {xn}. Since X is limit point compact, there is a limit point p X. So there is an infintely many terms of the sequence in the ball B/2(p). If xn and xm are two such points, then d(xn, xm) d(xn, p) + d(xm, p) /2 + /2 = contradicting the fact that d(xn, xm) > . So the claim is proved.

(b) If Fn is a collection of non-empty closed subsets of X such that Fn+1 Fn for all n, then show that n=1Fn is non-empty.

Solution: Choose points xn Fn. If the range of the sequence {xn} is finite, then clearly there is a point, say xN , which is in infinitely many of the sets Fn. Then, since the sets are decreasing, clearly xN will lie in the common intersection. So suppose the range {xn} is infinite. Then since X is limit point compact, there is a limit point p. Claim. p n=1Fn. Proof. If not, then there exists and N such that p / FN . Since FN is closed, there is a ball Br(p) such that Br(p) FN = . Since Fn FN for all n > N , clearly Br(p) Fn = for all n > N . On the other hand, since p is a limit point, Br(p) {xn} has infinitely many points. In particular there is an n > N such that xn Br(p). But xn Fn which is a contradiction since Br(p) Fn = .

(c) Prove that limit point compactness implies compactness.

Solution: The proof is similar to the proof that closed and bounded sets in Rn are compact. We proceed by contradiction. Suppose there is an open cover {G} such that no finite sub

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collection covers X. By the proof of part (b) there exists a finite collection of points x1, ? ? ? xN such that X Nj=1B1(xj). At least one of the balls cannot be covered by a finite sub collection of {G}. Label this ball B1. Clearly B1 is also limit point compact (Why?), and so we can cover B1 with finitely many balls of radius 1/2 and pick a ball which cannot be covered by a finite sub-collection from {G}. We continue and obtain a sequence of balls {Bj} such that

? Bj+1 Bj for all j. ? The radius of Bj is 1/j. ? No finite sub collection of {G} covers Bj. In particular, no Bj is contained in any of the

Gs. Now, pick a point xj Bj. If the range {xj} is finite, then one of the points, say x1, belongs to infinitely many of the balls Bj. If x1 G, then since G is open, there is a j big enough so that x1 Bj G. But this contradicts property 3 above. If the range {xj} is infinite, then by limit point compactness, there exists a limit point p X. Let such that p G. Since G is open there is an r > 0 such that Br(p) G. Moreover, since p is a limit point of {xj}, there is subsequence {xjk } such that xjk Br/2(p). Choose jk big enough such that 1/jk < r/2. Then by triangle inequality, Bjk Br(p) G contradicting property 3.

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