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Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 5


Exercise 5.1

Solution:

(a) Closed. As it has no limit points, Z is closed. Clearly, it is not open. Z is not compact since Z is not bounded (Heine-Borel Theorem).

(b) Closed and compact. It is closed as zero is only limit point. It is also bounded and hence compact by Heine-Borel Theorem.

(c) Open and Closed. Not compact because R is not bounded.

(d) None of them. 0 is a limit point but not in it. Hence it is not closed. 3 is a limit point of its complement but not in the complement. Hence its complement is not closed. Therefore it is not open. Because it is not closed so it is not compact.

(e) None of them.

(f) Closed and compact (because it is bounded).


Exercise 5.2

Solution:

(a) Let αUα be an arbitrary open cover of AB. Then αUα is an open cover of A. Since A is compact, we have a finite open cover αF1Uα of A, where F1 is a finite set.

Similarly, because αUα is an open cover of B and B is compact, we have a finite open cover αF2Uα of A, where F2 is a finite set.

Let F=F1F2, then F is finite and αF1UααF2Uα=αFUα.Hence ABαF1UααF2Uα=αFUα.Therefore αFUα is a finite open cover of AB. Since αUα is arbitrary, we conclude that AB is compact.

(b) See Exercise 5.10.


Exercise 5.3

Solution:

(a) Let Un=(11n,1+1n), then n=1Un={1} is not open.

(b) Let Un=[1n,11n], then n=1Un=(0,1) is not closed.


Exercise 5.4

Solution:

(a) If Ui is closed, then Uic is open. By Proposition 5.3, i=1nUic is also open. Hence (i=1nUic)c is closed. By De Morgan’s law (Fact 5.11), we have(i=1nUic)c=i=1n(Uic)c=i=1nUi.Hence i=1nUi is closed.

(b) If Uα is closed, then Uαc is open. By Proposition 5.3, αUαc is also open. Hence (αUαc)c is closed. By De Morgan’s law (Fact 5.11), we have(αUαc)c=α(Uαc)c=αUα.Hence αUα is closed.


Exercise 5.5

Solution: Suppose x is a limit of A. Then there exists a sequence (an) converging to x such that anA{x} for all n. For any ϵ>0, since (an)x, we can find an n such that |anx|<ϵ. Since anA{x} for n, we conclude that an is a point in the intersection of A and ϵ-neighborhood of x intersecting other than x.

Conversely, suppose for any ϵ>0, there is a point in the intersection of A and ϵ-neighborhood of x other than x. For each nN, there exists a point an in the intersection of A and 1n-neighborhood of x other than x. Hence |anx|<1n and anA{x}. It is clear that the sequence (an) converges to x and anA{x} for all n. Thus x is a limit point of A.


Exercise 5.6

Solution: LetUk=(1k,41k),then we have U1U2U3.Taking any finitely many sets Un1,,Unm. Let =max{n1,,nm}. Then ni for all 1im. Then UniU. Hencei=1mUniU.Note that U is a proper subset of (0,4). Therefore Un1,,Unm do not cover (0,4).

It says that for this special open cover there are no finite open covers for (0,4). Hence (0,4) is not compact.


Exercise 5.7

Solution: If A is closed and bounded, then every sequence from A is bounded. Hence by Bolzano-Weierstrass Theorem, it has a convergent subsequence. Let x be the limit of this convergent subsequence. If xA, then we are done. If xA, then A=A{x}. Hence x is a limit point of A. But A is closed, we must have xA. We get a contradiction. Hence we always have xA.

If  every sequence from  A has a subsequence that converges to a point in A, we would like to show that A is closed and bounded. Let x be a limit point, then there exists a sequence from A converging to x. By assumption, this sequence has a subsequence converging to a point in A. But by Proposition 3.32, this subsequence must converge to x as well. Hence xA. This shows that A contains its limit points. Hence by Theorem 5.10, A is closed.

Now we show that A is bounded. Suppose A is not bounded, e.g. A is not bounded above. Then for any nN, there exists anA such that an>n. Hence the sequence (an) from A diverges to , so do its any subsequences by Exercise 3.23. Hence we get a contradiction. This implies that A must be bounded, completing the proof.


Exercise 5.8

Solution: Since A is compact, by Heine-Borel Theorem, A is bounded and closed. Because UA, U is bounded as well. Since U is also closed, we conclude that U is both closed and bounded. It follows from Heine-Borel Theorem that U is compact.

This statement also holds for general topological space. A closed subspace of a compact space is compact.

Proof: Let T be a compact space. Let C be a closed subspace of T. Let U be an open cover of C. Since C is closed, it follows by definition of closed that TC is open in T. So if we add TC to U, we see that U{TC} is also an open cover of T. As T is compact, there is a finite subcover of U{TC}, say V={U1,U2,,Ur}. This covers C by the fact that it covers T. If TC is an element of V, then it can be removed from V and the rest of V still covers C. Thus we have a finite subcover of U which covers C, and hence C is compact.


Exercise 5.9

Solution: Let A=[0,1], then it is clear that A is closed, see Example 5.7. Since A is also bounded, by Heine-Borel Theorem, A is compact. Let B=(0,1), then B is bounded. Note that B is not closed. This is because 0 is a limit point of U as limn1n=0. We have AB=B which is not closed. Hence by Heine-Borel Theorem B is not compact. This shows that AB is not compact.


Exercise 5.10

Solution: Since Uα is compact for each α, by Heine-Borel Theorem, each Uα is closed and bounded. Recall from Exercise 5.4 (b), the intersection αUα is closed as well. Note that αUα is a subset of Uα0 for some α0. As Uα0 is bounded, so is αUα. Therefore αUα is both closed and bounded, by Heine-Borel Theorem, αUα is compact.


Exercise 5.11

Solution: Suppose XR is nonempty, open and closed. Let x0X. Finally suppose that XR; then there is some yX; WLOG we can assume that y>x0.

Then the set Z={xR:x>x0,xX} is bounded below (by x0) and nonempty (yZ). Therefore infZ=z exists.

  • Suppose zX. Then since X is open, it contains an open neighborhood (zϵ,z+ϵ). Recall the definition of z=infZ, then there would be a sequence zn>z, |zzn|<1n, znZznX. This is not possible, because [z,z+ϵ)X.
  • Suppose zX. Then since X is closed, its complement is open, therefore there is an open neighborhood (zϵ,z+ϵ) contained in RX. We claim that zϵ2>x0 otherwise zϵ2x0<z which contradicts that (zϵ,z+ϵ) is contained in RX. Therefore zϵ2>x0 and this contradicts the inf definition of z.

It follows that X=R.


Exercise 5.12

Solution: Let U=(0,1]. We show that U is neither open nor closed.

U is not closed. This is because 0 is a limit point of U as limn1n=0, but 0U. It follows from Theorem 5.10 that U is not closed.

U is not open. Consider Uc=(,0](1,]. Since limn(1+1n)=1, we conclude that 1 is a limit point of Uc. But 1Uc, it follows from Theorem 5.10 that Uc is not closed. Hence U is not open by definition 5.6.

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