Just like with sequences of real numbers, we define Cauchy sequences.
Definition7.4.1.
Let \((X,d)\) be a metric space. A sequence \(\{ x_n \}_{n=1}^\infty\) in \(X\) is a Cauchy sequence if for every \(\epsilon > 0\text{,}\) there exists an \(M \in \N\) such that for all \(n \geq M\) and all \(k \geq M\text{,}\) we have
The definition is again simply a translation of the concept from the real numbers to metric spaces. A sequence of real numbers is Cauchy in the sense of Chapter 2 if and only if it is Cauchy in the sense above, provided we equip the real numbers with the standard metric \(d(x,y) = \abs{x-y}\text{.}\)
Proposition7.4.2.
A convergent sequence in a metric space is Cauchy.
Proof.
Suppose \(\{ x_n \}_{n=1}^\infty\) converges to \(p\text{.}\) Given \(\epsilon > 0\text{,}\) there is an \(M\) such that for all \(n \geq M\text{,}\) we have \(d(p,x_n) < \nicefrac{\epsilon}{2}\text{.}\) Hence for all \(n,k \geq M\text{,}\) we have \(d(x_n,x_k) \leq d(x_n,x) + d(x,x_k) < \nicefrac{\epsilon}{2} +
\nicefrac{\epsilon}{2} = \epsilon\text{.}\)
Definition7.4.3.
We say a metric space \((X,d)\) is complete or Cauchy-complete if every Cauchy sequence \(\{ x_n \}_{n=1}^\infty\) in \(X\) converges to a \(p \in X\text{.}\)
Proposition7.4.4.
The space \(\R^n\) with the standard metric is a complete metric space.
For \(\R = \R^1\text{,}\) completeness was proved in Chapter 2. The proof of completeness in \(\R^n\) is a reduction to the one-dimensional case.
Proof.
Let \(\{ x_m \}_{m=1}^\infty\) be a Cauchy sequence in \(\R^n\text{,}\) where \(x_m = \bigl(x_{m,1},x_{m,2},\ldots,x_{m,n}\bigr) \in \R^n\text{.}\) As the sequence is Cauchy, given \(\epsilon > 0\text{,}\) there exists an \(M\) such that for all \(i,j \geq M\text{,}\)
Hence the sequence \(\{ x_{m,k} \}_{m=1}^\infty\) is Cauchy. As \(\R\) is complete the sequence converges; there exists a \(y_k \in \R\) such that \(y_k = \lim_{m\to\infty} x_{m,k}\text{.}\) Write \(y = (y_1,y_2,\ldots,y_n) \in \R^n\text{.}\) By Proposition 7.3.9, \(\{ x_m \}_{m=1}^\infty\) converges to \(y \in \R^n\text{,}\) and hence \(\R^n\) is complete.
In the language of metric spaces, the results on continuity of section Section 6.2, say that the metric space \(C\bigl([a,b],\R\bigr)\) of Example 7.1.8 is complete. The proof follows by “unrolling the definitions,” and is left as Exercise 7.4.7.
Proposition7.4.5.
The space of continuous functions \(C\bigl([a,b],\R\bigr)\) with the uniform norm as metric is a complete metric space.
A subset of a complete metric space such as \(\R^n\) with the subspace metric need not be complete. For example, \((0,1]\) with the subspace metric is not complete, as \(\{ \nicefrac{1}{n} \}_{n=1}^\infty\) is a Cauchy sequence in \((0,1]\) with no limit in \((0,1]\text{.}\) However, a closed subspace of a complete metric space is complete. After all, one way to think of a closed set is that it contains all points reachable from the set via a sequence. The proof is Exercise 7.4.16.
Proposition7.4.6.
Suppose \((X,d)\) is a complete metric space and \(E \subset X\) is closed. Then \(E\) is a complete metric space with the subspace metric.
Subsection7.4.2Compactness
Definition7.4.7.
Let \((X,d)\) be a metric space and \(K \subset X\text{.}\) The set \(K\) is said to be compact if for every collection of open sets \(\{ U_{\lambda} \}_{\lambda \in I}\) such that
\begin{equation*}
K \subset \bigcup_{\lambda \in I} U_\lambda ,
\end{equation*}
there exists a finite subset \(\{ \lambda_1, \lambda_2,\ldots,\lambda_m \} \subset I\) such that
\begin{equation*}
K \subset \bigcup_{j=1}^m U_{\lambda_j} .
\end{equation*}
A collection of open sets \(\{ U_{\lambda} \}_{\lambda \in I}\) as above is said to be an open cover of \(K\text{.}\) A way to say that \(K\) is compact is to say that every open cover of \(K\) has a finite subcover.
Example7.4.8.
Let \(\R\) be the metric space with the standard metric.
The set \(\R\) is not compact. Proof: For \(j \in \N\text{,}\) let \(U_j \coloneqq (-j,j)\text{.}\) Any \(x \in \R\) is in some \(U_j\) (by the Archimedean property), so we have an open cover. Suppose we have a finite subcover \(\R \subset U_{j_1} \cup U_{j_2} \cup \cdots \cup U_{j_m}\text{,}\) and suppose \(j_1 < j_2 < \cdots < j_m\text{.}\) Then \(\R \subset U_{j_m}\text{,}\) but that is a contradiction as \(j_m \in \R\) on one hand and \(j_m \notin U_{j_m} =
(-j_m,j_m)\) on the other.
The set \((0,1) \subset \R\) is also not compact. Proof: Take the sets \(U_{j} \coloneqq (\nicefrac{1}{j},1-\nicefrac{1}{j})\) for \(j=3,4,5,\ldots\text{.}\) As above \((0,1) = \bigcup_{j=3}^\infty U_j\text{.}\) And similarly as above, if there exists a finite subcover, then there is one \(U_j\) such that \((0,1)
\subset U_j\text{,}\) which again leads to a contradiction.
The set \(\{ 0 \} \subset \R\) is compact. Proof: Given an open cover \(\{
U_{\lambda} \}_{\lambda \in I}\text{,}\) there must exist a \(\lambda_0\) such that \(0
\in U_{\lambda_0}\) as it is a cover. But then \(U_{\lambda_0}\) gives a finite subcover.
We will prove below that \([0,1]\text{,}\) and in fact every closed and bounded interval \([a,b]\text{,}\) is compact.
Proposition7.4.9.
Let \((X,d)\) be a metric space. If \(K \subset X\) is compact, then \(K\) is closed and bounded.
Proof.
First, we prove that a compact set is bounded. Fix \(p \in X\text{.}\) We have the open cover
\begin{equation*}
K \subset \bigcup_{n=1}^\infty B(p,n) = X .
\end{equation*}
If \(K\) is compact, then there exists some set of indices \(n_1 < n_2 < \ldots < n_m\) such that
\begin{equation*}
K \subset \bigcup_{j=1}^m B(p,n_j) = B(p,n_m) .
\end{equation*}
As \(K\) is contained in a ball, \(K\) is bounded. See the left-hand side of Figure 7.11.
Next, we show a set that is not closed is not compact. Suppose \(\widebar{K} \not= K\text{,}\) that is, there is a point \(x \in \widebar{K}
\setminus K\text{.}\) If \(y \not= x\text{,}\) then \(y \notin C(x,\nicefrac{1}{n})\) for \(n \in \N\) such that \(\nicefrac{1}{n} < d(x,y)\text{.}\) Furthermore, \(x \notin K\text{,}\) so
\begin{equation*}
K \subset \bigcup_{n=1}^\infty {C(x,\nicefrac{1}{n})}^c .
\end{equation*}
A closed ball is closed, so its complement \({C(x,\nicefrac{1}{n})}^c\) is open, and we have an open cover. If we take any finite collection of indices \(n_1 < n_2 < \ldots < n_m\text{,}\) then
As \(x\) is in the closure of \(K\text{,}\) then \(C(x,\nicefrac{1}{n_m}) \cap K \not= \emptyset\text{.}\) So there is no finite subcover and \(K\) is not compact. See the right-hand side of Figure 7.11.
Figure7.11.Proving compact set is bounded (left) and closed (right).
We prove below that in a finite-dimensional euclidean space, every closed bounded set is compact. So closed bounded sets of \(\R^n\) are examples of compact sets. It is not true that in every metric space, closed and bounded is equivalent to compact. A simple example is an incomplete metric space such as \((0,1)\) with the subspace metric from \(\R\text{.}\) There are many complete and very useful metric spaces where closed and bounded is not enough to give compactness: \(C\bigl([a,b],\R\bigr)\) is a complete metric space, but the closed unit ball \(C(0,1)\) is not compact, see Exercise 7.4.8. However, see also Exercise 7.4.12.
Not worrying about the boundedness for a moment, note further the difference between being closed and being compact. Being closed depends on the ambient metric space: The set \((0,1]\) is not closed in \(\R\text{,}\) but it is closed in the subspace \((0,\infty)\text{.}\) However, a set \(K\) is compact in some metric space \((X,d)\) if and only if it is compact in the subspace metric on \(K\text{.}\) So for a compact set, we do not have to ask what metric space it lives in. On the other hand, every set is always closed in the subspace metric as a subset of itself. See also Exercise 7.4.6.
A useful property of compact sets in a metric space is that every sequence in the set has a convergent subsequence converging to a point in the set. Such sets are called sequentially compact. We will prove that in the context of metric spaces, a set is compact if and only if it is sequentially compact. First we prove a lemma.
Lemma7.4.10.Lebesgue covering lemma.
1
Named after the French mathematician Henri Léon Lebesgue (1875–1941). The number \(\delta\) is sometimes called the Lebesgue number of the cover.
Let \((X,d)\) be a metric space and \(K \subset X\text{.}\) Suppose every sequence in \(K\) has a subsequence convergent in \(K\text{.}\) Given an open cover \(\{ U_\lambda \}_{\lambda \in I}\) of \(K\text{,}\) there exists a \(\delta > 0\) such that for every \(x \in K\text{,}\) there exists a \(\lambda \in I\) with \(B(x,\delta) \subset U_\lambda\text{.}\)
Proof.
We prove the lemma by contrapositive. If the conclusion is not true, then there is an open cover \(\{ U_\lambda \}_{\lambda \in I}\) of \(K\) with the following property. For every \(n \in \N\text{,}\) there exists an \(x_n \in K\) such that \(B(x_n,\nicefrac{1}{n})\) is not a subset of any \(U_\lambda\text{.}\) Take any \(x \in K\text{.}\) There is a \(\lambda \in I\) such that \(x \in U_\lambda\text{.}\) As \(U_\lambda\) is open, there is an \(\epsilon > 0\) such that \(B(x,\epsilon) \subset U_\lambda\text{.}\) Take \(M\) such that \(\nicefrac{1}{M} < \nicefrac{\epsilon}{2}\text{.}\) If \(y \in
B(x,\nicefrac{\epsilon}{2})\) and \(n \geq M\text{,}\) then
where \(B(y,\nicefrac{\epsilon}{2}) \subset B(x,\epsilon)\) follows by triangle inequality. See Figure 7.12. Thus \(y \not= x_n\text{.}\) In other words, for all \(n \geq M\text{,}\)\(x_n \notin B(x,\nicefrac{\epsilon}{2})\text{.}\) The sequence cannot have a subsequence converging to \(x\text{.}\) As \(x \in K\) was arbitrary we are done.
Figure7.12.Proof of Lebesgue covering lemma. Note that \(B(y,\nicefrac{\epsilon}{2}) \subset
B(x,\epsilon)\) by triangle inequality.
It is important to recognize what the lemma says. It says that if \(K\) is sequentially compact, then given any cover there is a single \(\delta > 0\text{.}\) The \(\delta\) depends on the cover, but, of course, it does not depend on \(x\text{.}\)
For example, let \(K \coloneqq [-10,10]\) and let \(U_n \coloneqq (n,n+2)\) for \(n \in \Z\) give an open cover. Consider \(x \in K\text{.}\) There is an \(n \in \Z\text{,}\) such that \(n \leq x < n+1\text{.}\) If \(n \leq x < n+\nicefrac{1}{2}\text{,}\) then \(B\bigl(x,\nicefrac{1}{2}\bigr) \subset U_{n-1}\text{.}\) If \(n+ \nicefrac{1}{2} \leq x < n+1\text{,}\) then \(B\bigl(x,\nicefrac{1}{2}\bigr) \subset U_{n}\text{.}\) So \(\delta =
\nicefrac{1}{2}\) will do. The sets \(U'_n \coloneqq \bigl(\frac{n}{2},\frac{n+2}{2} \bigr)\text{,}\) again give an open cover, but now the largest \(\delta\) that works is \(\nicefrac{1}{4}\text{.}\)
On the other hand, \(\N \subset \R\) is not sequentially compact. It is an exercise to find a cover for which no \(\delta > 0\) works.
Theorem7.4.11.
Let \((X,d)\) be a metric space. Then \(K \subset X\) is compact if and only if every sequence in \(K\) has a subsequence converging to a point in \(K\text{.}\)
Proof.
Claim: Let \(K \subset X\) be a subset of \(X\) and \(\{ x_n \}_{n=1}^\infty\) a sequence in \(K\text{.}\) Suppose that for each \(x \in K\text{,}\) there is a ball \(B(x,\alpha_x)\) for some \(\alpha_x > 0\) such that \(x_n \in B(x,\alpha_x)\) for only finitely many \(n \in \N\text{.}\) Then \(K\) is not compact.
Proof of the claim: Notice
\begin{equation*}
K \subset \bigcup_{x \in K} B(x,\alpha_x) .
\end{equation*}
Any finite collection of these balls contains at most finitely many elements of \(\{ x_n \}_{n=1}^\infty\text{,}\) and so there must be an \(x_n \in K\) not in their union. Hence, \(K\) is not compact and the claim is proved.
So suppose that \(K\) is compact and \(\{ x_n \}_{n=1}^\infty\) is a sequence in \(K\text{.}\) Then there exists an \(x \in K\) such that for all \(\delta > 0\text{,}\)\(B(x,\delta)\) contains \(x_n\) for infinitely many \(n \in \N\text{.}\) We define the subsequence inductively. The ball \(B(x,1)\) contains some \(x_k\text{,}\) so let \(n_1 \coloneqq k\text{.}\) Suppose \(n_{j-1}\) is defined. There must exist a \(k > n_{j-1}\) such that \(x_k \in B(x,\nicefrac{1}{j})\text{.}\) Define \(n_j \coloneqq k\text{.}\) We now posses a subsequence \(\{ x_{n_j} \}_{j=1}^\infty\text{.}\) Since \(d(x,x_{n_j}) < \nicefrac{1}{j}\text{,}\)Proposition 7.3.5 says \(\lim_{j\to\infty} x_{n_j} = x\text{.}\)
For the other direction, suppose every sequence in \(K\) has a subsequence converging in \(K\text{.}\) Take an open cover \(\{ U_\lambda \}_{\lambda \in I}\) of \(K\text{.}\) Using the Lebesgue covering lemma above, find a \(\delta > 0\) such that for every \(x \in K\text{,}\) there is a \(\lambda \in I\) with \(B(x,\delta) \subset U_\lambda\text{.}\)
Pick \(x_1 \in K\) and find \(\lambda_1 \in I\) such that \(B(x_1,\delta) \subset
U_{\lambda_1}\text{.}\) If \(K \subset U_{\lambda_1}\text{,}\) we stop as we have found a finite subcover. Otherwise, there must be a point \(x_2 \in K \setminus U_{\lambda_1}\text{.}\) Note that \(d(x_2,x_1) \geq \delta\text{.}\) There must exist some \(\lambda_2 \in I\) such that \(B(x_2,\delta) \subset U_{\lambda_2}\text{.}\) We work inductively. Suppose \(\lambda_{n-1}\) is defined. Either \(U_{\lambda_1} \cup
U_{\lambda_2} \cup \cdots \cup
U_{\lambda_{n-1}}\) is a finite cover of \(K\text{,}\) in which case we stop, or there must be a point \(x_n \in K \setminus \bigl( U_{\lambda_1} \cup
U_{\lambda_2} \cup \cdots \cup
U_{\lambda_{n-1}}\bigr)\text{.}\) Note that \(d(x_n,x_j) \geq \delta\) for all \(j = 1,2,\ldots,n-1\text{.}\) Next, there must be some \(\lambda_n \in I\) such that \(B(x_n,\delta) \subset U_{\lambda_n}\text{.}\) See Figure 7.13.
Figure7.13.Covering \(K\) by \(U_{\lambda}\text{.}\) The points \(x_1,x_2,x_3,x_4\text{,}\) the three sets \(U_{\lambda_1}\text{,}\)\(U_{\lambda_2}\text{,}\)\(U_{\lambda_2}\text{,}\) and the first three balls of radius \(\delta\) are drawn.
Either at some point we obtain a finite subcover of \(K\text{,}\) or we obtain an infinite sequence \(\{ x_n \}_{n=1}^\infty\) as above. For contradiction, suppose that there is no finite subcover and we have the sequence \(\{ x_n \}_{n=1}^\infty\text{.}\) For all \(n\) and \(k\text{,}\)\(n \not= k\text{,}\) we have \(d(x_n,x_k) \geq \delta\text{.}\) So no subsequence of \(\{ x_n \}_{n=1}^\infty\) is Cauchy. Hence, no subsequence of \(\{ x_n \}_{n=1}^\infty\) is convergent, which is a contradiction.
Example7.4.12.
Theorem 2.3.8, the Bolzano–Weierstrass theorem for sequences of real numbers, says that every bounded sequence in \(\R\) has a convergent subsequence. Therefore, every sequence in a closed interval \([a,b] \subset \R\) has a convergent subsequence. The limit is also in \([a,b]\) as limits preserve non-strict inequalities. Hence a closed bounded interval \([a,b]
\subset \R\) is (sequentially) compact.
Proposition7.4.13.
Let \((X,d)\) be a metric space and let \(K \subset X\) be compact. If \(E \subset K\) is a closed set, then \(E\) is compact.
Because \(K\) is closed, \(E\) is closed in \(K\) if and only if it is closed in \(X\text{.}\) See Proposition 7.2.12.
Proof.
Let \(\{ x_n \}_{n=1}^\infty\) be a sequence in \(E\text{.}\) It is also a sequence in \(K\text{.}\) Therefore, it has a convergent subsequence \(\{ x_{n_j} \}_{j=1}^\infty\) that converges to some \(x \in K\text{.}\) As \(E\) is closed the limit of a sequence in \(E\) is also in \(E\) and so \(x \in E\text{.}\) Thus \(E\) must be compact.
A closed bounded subset \(K \subset \R^n\) is compact.
So subsets of \(\R^n\) are compact if and only if they are closed and bounded, a condition that is much easier to check. Let us reiterate that the Heine–Borel theorem only holds for \(\R^n\) and not for metric spaces in general. The theorem does not hold even for subspaces of \(\R^n\text{,}\) just in \(\R^n\) itself. In general, compact implies closed and bounded, but not vice versa.
Proof.
For \(\R = \R^1\text{,}\) suppose \(K \subset \R\) is closed and bounded. Then \(K \subset [a,b]\) for some closed and bounded interval, which is compact by Example 7.4.12. As \(K\) is a closed subset of a compact set, it is compact by Proposition 7.4.13.
We carry out the proof for \(n=2\) and leave arbitrary \(n\) as an exercise. As \(K \subset \R^2\) is bounded, there exists a set \(B=[a,b]\times[c,d] \subset \R^2\) such that \(K \subset B\text{.}\) We will show that \(B\) is compact. Then \(K\text{,}\) being a closed subset of a compact \(B\text{,}\) is also compact.
Let \(\bigl\{ (x_k,y_k) \bigr\}_{k=1}^\infty\) be a sequence in \(B\text{.}\) That is, \(a \leq x_k \leq b\) and \(c \leq y_k \leq d\) for all \(k\text{.}\) A bounded sequence of real numbers has a convergent subsequence so there is a subsequence \(\{ x_{k_j} \}_{j=1}^\infty\) that is convergent. The subsequence \(\{ y_{k_j} \}_{j=1}^\infty\) is also a bounded sequence so there exists a subsequence \(\{ y_{k_{j_i}} \}_{i=1}^\infty\) that is convergent. A subsequence of a convergent sequence is still convergent, so \(\{ x_{k_{j_i}} \}_{i=1}^\infty\) is convergent. Let
\begin{equation*}
x \coloneqq \lim_{i\to\infty} x_{k_{j_i}}
\qquad \text{and} \qquad
y \coloneqq \lim_{i\to\infty} y_{k_{j_i}} .
\end{equation*}
By Proposition 7.3.9, \(\bigl\{ (x_{k_{j_i}},y_{k_{j_i}}) \bigr\}_{i=1}^\infty\) converges to \((x,y)\text{.}\) Furthermore, as \(a \leq x_k \leq b\) and \(c \leq y_k \leq d\) for all \(k\text{,}\) we know that \((x,y) \in B\text{.}\)
Example7.4.15.
The discrete metric provides interesting counterexamples again. Let \((X,d)\) be a metric space with the discrete metric, that is, \(d(x,y) = 1\) if \(x \not= y\text{.}\) Suppose \(X\) is an infinite set. Then
\((X,d)\) is a complete metric space.
Any subset \(K \subset X\) is closed and bounded.
A subset \(K \subset X\) is compact if and only if it is a finite set.
The conclusion of the Lebesgue covering lemma is always satisfied, e.g. with \(\delta = \nicefrac{1}{2}\text{,}\) even for noncompact \(K \subset X\text{.}\)
The proofs of the statements above are either trivial or are relegated to the exercises below.
Remark7.4.16.
A subtle issue with Cauchy sequences, completeness, compactness, and convergence is that compactness and convergence only depend on the topology, that is, on which sets are the open sets. On the other hand, Cauchy sequences and completeness depend on the actual metric. See Exercise 7.4.19.
Exercises7.4.3Exercises
7.4.1.
Let \((X,d)\) be a metric space and \(A\) a finite subset of \(X\text{.}\) Show that \(A\) is compact.
7.4.2.
Let \(A \coloneqq \{ \nicefrac{1}{n} : n \in \N \} \subset \R\text{.}\)
Show that \(A\) is not compact directly using the definition.
Show that \(A \cup \{ 0 \}\) is compact directly using the definition.
7.4.3.
Let \((X,d)\) be a metric space with the discrete metric.
Prove that \(X\) is complete.
Prove that \(X\) is compact if and only if \(X\) is a finite set.
7.4.4.
Show that the union of finitely many compact sets is a compact set.
Find an example where the union of infinitely many compact sets is not compact.
7.4.5.
Prove Theorem 7.4.14 for arbitrary dimension. Hint: The trick is to use the correct notation.
7.4.6.
Show that a compact set \(K\) (in any metric space) is itself a complete metric space (using the subspace metric).
7.4.7.
Let \(C\bigl([a,b],\R\bigr)\) be the metric space as in Example 7.1.8. Show that \(C\bigl([a,b],\R\bigr)\) is a complete metric space.
7.4.8.
(Challenging) Let \(C\bigl([0,1],\R\bigr)\) be the metric space of Example 7.1.8. Let \(0\) denote the zero function. Then show that the closed ball \(C(0,1)\) is not compact (even though it is closed and bounded). Hints: Construct a sequence of distinct continuous functions \(\{ f_n \}_{n=1}^\infty\) such that \(d(f_n,0) = 1\) and \(d(f_n,f_k) = 1\) for all \(n \not= k\text{.}\) Show that the set \(\{ f_n : n \in \N \} \subset C(0,1)\) is closed but not compact. See Chapter 6 for inspiration.
7.4.9.
(Challenging) Show that there exists a metric on \(\R\) that makes \(\R\) into a compact set.
7.4.10.
Suppose \((X,d)\) is complete and suppose we have a countably infinite collection of nonempty compact sets \(E_1 \supset E_2 \supset E_3 \supset
\cdots\text{.}\) Prove \(\bigcap_{j=1}^\infty E_j \not= \emptyset\text{.}\)
7.4.11.
(Challenging) Let \(C\bigl([0,1],\R\bigr)\) be the metric space of Example 7.1.8. Let \(K\) be the set of \(f \in C\bigl([0,1],\R\bigr)\) such that \(f\) is equal to a quadratic polynomial, i.e. \(f(x) = a+bx+cx^2\text{,}\) and such that \(\abs{f(x)} \leq 1\) for all \(x \in [0,1]\text{,}\) that is \(f \in C(0,1)\text{.}\) Show that \(K\) is compact.
7.4.12.
(Challenging) Let \((X,d)\) be a complete metric space. Show that \(K \subset X\) is compact if and only if \(K\) is closed and such that for every \(\epsilon > 0\) there exists a finite set of points \(x_1,x_2,\ldots,x_n\) with \(K \subset \bigcup_{j=1}^n B(x_j,\epsilon)\text{.}\) Note: Such a set \(K\) is said to be totally bounded, so in a complete metric space a set is compact if and only if it is closed and totally bounded.
7.4.13.
Take \(\N \subset \R\) using the standard metric. Find an open cover of \(\N\) such that the conclusion of the Lebesgue covering lemma does not hold.
7.4.14.
Prove the general Bolzano–Weierstrass theorem: Any bounded sequence \(\{ x_k \}_{k=1}^\infty\) in \(\R^n\) has a convergent subsequence.
7.4.15.
Let \(X\) be a metric space and \(C \subset \sP(X)\) the set of nonempty compact subsets of \(X\text{.}\) Using the Hausdorff metric from Exercise 7.1.8, show that \((C,d_H)\) is a metric space. That is, show that if \(L\) and \(K\) are nonempty compact subsets, then \(d_H(L,K) = 0\) if and only if \(L=K\text{.}\)
7.4.16.
Prove Proposition 7.4.6. That is, let \((X,d)\) be a complete metric space and \(E \subset X\) a closed set. Show that \(E\) with the subspace metric is a complete metric space.
7.4.17.
Let \((X,d)\) be an incomplete metric space. Show that there exists a closed and bounded set \(E \subset X\) that is not compact.
7.4.18.
Let \((X,d)\) be a metric space and \(K \subset X\text{.}\) Prove that \(K\) is compact as a subset of \((X,d)\) if and only if \(K\) is compact as a subset of itself with the subspace metric.
7.4.19.
Consider two metrics on \(\R\text{.}\) Let \(d(x,y) \coloneqq \abs{x-y}\) be the standard metric, and let \(d'(x,y) \coloneqq \bigl\lvert \frac{x}{1+\abs{x}} - \frac{y}{1+\abs{y}}
\bigr\rvert\text{.}\)
Show that \((\R,d')\) is a metric space (if you have done Exercise 7.3.10, the computation is the same).
Show that the topology is the same, that is, a set is open in \((\R,d)\) if and only if it is open in \((\R,d')\text{.}\)
Show that a set is compact in \((\R,d)\) if and only if it is compact in \((\R,d')\text{.}\)
Show that a sequence converges in \((\R,d)\) if and only if it converges in \((\R,d')\text{.}\)
Find a sequence of real numbers that is Cauchy in \((\R,d')\) but not Cauchy in \((\R,d)\text{.}\)
While \((\R,d)\) is complete, show that \((\R,d')\) is not complete.
7.4.20.
Let \((X,d)\) be a complete metric space. We say a set \(S \subset X\) is relatively compact if the closure \(\widebar{S}\) is compact. Prove that \(S \subset X\) is relatively compact if and only if given any sequence \(\{ x_n \}_{n=1}^\infty\) in \(S\text{,}\) there exists a subsequence \(\{ x_{n_k} \}_{k=1}^\infty\) that converges (in \(X\)).
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