4.6.E: Problems on Compact Sets
- Page ID
- 23723
Complete the missing details in the proof of Theorem 5.
Verify that any infinite set in a discrete space is closed and bounded but not compact.
[Hint: In such a space no sequence of distinct terms clusters.]
Show that \(E^{n}\) is not compact, in three ways:
\(\left.\text { (i) from definitions (as in Example }\left(\mathrm{a}^{\prime}\right)\right)\) ;
(ii) from Theorem \(4 ;\) and
(iii) from Theorem 5, by finding in \(E^{n}\) a contracting sequence of infinite
closed sets with a void intersection. For example, in \(E^{1}\) take the
closed sets \(F_{m}=[m,+\infty), m=1,2, \ldots(\text { Are they closed?) }\)
Show that \(E^{*}\) is compact under the metric \(\rho^{\prime}\) defined in Problems 5 and 6 in Chapter 3, §11. Is \(E^{1}\) a compact set under that metric?
[Hint: For the first part, use Theorem 2 of Chapter 2, §13, noting that \(G_{q}\) is also a \(\left.\text { globe under } \rho^{\prime} . \text { For the second, consider the sequence } x_{n}=n .\right]\)
Show that a set \(A \subseteq(S, \rho)\) is compact iff every infinite subset \(B \subseteq A\) has a cluster point \(p \in A .\)
[Hint: Select from \(B\) a sequence \(\left\{x_{m}\right\}\) of distinct terms. Then the cluster points of \(\left\{x_{m}\right\}\) are also those of \(B .\) (Why?)]
Prove the following.
(i) If \(A\) and \(B\) are compact, so is \(A \cup B,\) and similarly for unions of \(n\) sets.
(ii) If the sets \(A_{i}(i \in I)\) are compact, so is \(\bigcap_{i \in I} A_{i},\) even if \(I\) is infinite.
Disprove (i) for unions of infinitely many sets by a counterexample.
\( \text { [ Hint: For (ii), verify first that } \bigcap_{i \in I} A_{i} \text { is sequentially closed. Then use Theorem } 1 . ]\)
Prove that if \(x_{m} \rightarrow p\) in \((S, \rho),\) then the set
\[
B=\left\{p, x_{1}, x_{2}, \ldots, x_{m}, \ldots\right\}
\]
is compact.
[Hint: If \(B\) is finite, see Example (b). If not, use Problem \(5,\) noting that any infinite \(\left.\text { subset of } B \text { defines a subsequence } x_{m_{k}} \rightarrow p, \text { so it clusters at } p .\right]\)
Prove, independently, the principle of nested intervals in \(E^{n},\) i.e., Theorem 5 with
\[
F_{m}=\left[\overline{a}_{m}, \overline{b}_{m}\right] \subseteq E^{n} ,
\]
where
\[
\overline{a}_{m}=\left(a_{m 1}, \ldots, a_{m n}\right) \text { and } \overline{b}_{m}=\left(b_{m 1}, \ldots, b_{m n}\right) .
\]
Fixing \(k,\) let \(A_{k}\) be the set of all \(a_{m k}, m=1,2, \ldots .\) Show that \(A_{k}\) is bounded above by each \(b_{m k},\) so let \(p_{k}=\sup A_{k}\) in \(E^{1} .\) Then
\[
(\forall m) \quad a_{m k} \leq p_{k} \leq b_{m k} . \text { (Why?) }
\]
Unfixing \(k,\) obtain such inequalities for \(k=1,2, \ldots, n .\) Let \(\overline{p}=\left(p_{1}, \ldots, p_{k}\right) .\) Then
\[
(\forall m) \quad \overline{p} \in\left[\overline{a}_{m}, \overline{b}_{m}\right], \text { i.e., } \overline{p} \in \bigcap F_{m}, \text { as required. }
\]
Note that the theorem fails for nonclosed intervals, even in \(E^{1} ;\) e.g., take \(F_{m}=\) \((0,1 / m]\) and show that \(\bigcap_{m} F_{m}=\emptyset .\)]
From Problem \(8,\) obtain a new proof of the Bolzano-Weierstrass theorem.
\(\left[\text { Hint: Let }\left\{\overline{x}_{m}\right\} \in[\overline{a}, \overline{b}] \subseteq E^{n} ; \text { put } F_{0}=[\overline{a}, \overline{b}] \text { and set }\right.\)
\[
d F_{0}=\rho(\overline{a}, \overline{b})=d \quad\left(\text { diagonal of } F_{0}\right) .
\]
Bisecting the edges of \(F_{0},\) subdivide \(F_{0}\) into \(2^{n}\) intervals of diagonal \(d / 2 ;\) one of them must contain infinitely many \(x_{m} .\) (Why?) Let \(F_{1}\) be one such inter val; make it closed and subdivide it into \(2^{n}\) subintervals of diagonal \(d / 2^{2} .\) One of them, \(F_{2},\) contains infinitely many \(x_{m} ;\) make it closed, etc.
Thus obtain a contracting sequence of closed intervals \(F_{m}\) with
\[
d F_{m}=\frac{d}{2^{m}}, \quad m=1,2, \ldots
\]
From Problem \(8,\) obtain
\[
\overline{p} \in \bigcap_{m=1}^{\infty} F_{m} .
\]
\(\left.\text { Show that }\left\{\overline{x}_{m}\right\} \text { clusters at } \overline{p} .\right]\)
\(\Rightarrow 10 .\) Prove the Heine-Borel theorem: If a closed interval \(F_{0} \subset E^{n}\) is covered by a family of open sets \(G_{i}(i \in I),\) i.e.,
\[
F_{0} \subseteq \bigcup_{i \in I} G_{i} ,
\]
then it can always be covered by a finite number of these \(G_{i}\).
[Outline of proof: Let \(d F_{0}=d .\) Seeking a contradiction, suppose \(F_{0}\) cannot be covered by any finite number of the \(G_{i}\).
As in Problem \(9,\) subdivide \(F_{0}\) into \(2^{n}\) intervals of diagonal \(d / 2 . A t\) least one of them cannot be covered by finitely many \(G_{i} .\) (Why?) Choose one such interval, make it closed, call it \(F_{1},\) and subdivide it into \(2^{n}\) subintervals of diagonal \(d / 2^{2}\). One of these, \(F_{2},\) cannot be covered by finitely many \(G_{i} ;\) make it closed and repeat the process indefinitely.
Thus obtain a contracting sequence of closed intervals \(F_{m}\) with
\[
d F_{m}=\frac{d}{2^{m}}, \quad m=1,2, \ldots
\]
\(\text { From Problem } 8 \text { (or Theorem } 5),\) get \(\overline{p} \in \bigcap F_{m}\).
As \(\overline{p} \in F_{0}, \overline{p}\) is in one of the \(G_{i} ;\) call it \(G .\) As \(G\) is open, \(\overline{p}\) is its interior point, so let \(G \supseteq G_{\overline{p}}(\varepsilon) .\) Now take \(m\) so large that \(d / 2^{m}=d F_{m}<\varepsilon\). Show that then
\[
F_{m} \subseteq G_{\overline{p}}(\varepsilon) \subseteq G .
\]
\(\left.\text { Thus (contrary to our choice of the } F_{m}\right) F_{m}\) is covered by a single set \(G_{i} .\) This contradiction completes the proof.]
Prove that if \(\left\{x_{m}\right\} \subseteq A \subseteq(S, \rho)\) and \(A\) is compact, then \(\left\{x_{m}\right\}\) converges iff it has a single cluster point.
[Hint: Proceed as in Problem 12 of Chapter 3, §16.]
Prove that if \(\emptyset \neq A \subseteq(S, \rho)\) and \(A\) is compact, there are two points \(p, q \in A\) such that \(d A=\rho(p, q) .\)
\(\text { [Hint: As } A \text { is bounded (Theorem } 3), d A<+\infty .\) By the properties of suprema,
\[
(\forall n)\left(\exists x_{n}, y_{n} \in A\right) \quad d A-\frac{1}{n}<\rho\left(x_{n}, y_{n}\right) \leq d A . \quad \text { (Explain!) }
\]
By compactness, \(\left\{x_{n}\right\}\) has a subsequence \(x_{n_{k}} \rightarrow p \in A .\) For brevity, put \(x_{k}^{\prime}=x_{n_{k}}\), \(y_{k}^{\prime}=y_{n_{k}} .\) Again, \(\left\{y_{k}^{\prime}\right\}\) has a subsequence \(y_{k_{m}}^{\prime} \rightarrow q \in A .\) Also,
\[
d A-\frac{1}{n_{k_{m}}}<\rho\left(x_{k_{m}}^{\prime}, y_{k_{m}}^{\prime}\right) \leq d A .
\]
Passing to the limit \((\text { as } m \rightarrow+\infty),\) obtain
\[
d A \leq \rho(p, q) \leq d A
\]
by Theorem 4 in Chapter 3, §15.]
Given nonvoid sets \(A, B \subseteq(S, \rho),\) define
\[
\rho(A, B)=\inf \{\rho(x, y) | x \in A, y \in B\} .
\]
Prove that if \(A\) and \(B\) are compact and nonempty, there are \(p \in A\) and \(q \in B\) such that \(\rho(p, q)=\rho(A, B) .\) Give an example to show that this \(\left.\text { may fail if } A \text { and } B \text { are not compact (even if they are closed in } E^{1}\right)\).
[Hint: For the first part, proceed as in Problem 12 .]
Prove that every compact set is complete. Disprove the converse by examples.