15. Complete spaces; and Compact sets PDF
Proposition
Proposition. Let \((X,\rho)\) be a metric space, \(E \subseteq X\) is closed iff for every sequence \((x_n)_{n \in \mathbb{N}}\subseteq E\) such that \(x_n \ _{\overrightarrow{n \to \infty}}\ x \in X\) we have \(x \in E.\)
Proof. (\(\Longrightarrow\)) Suppose that \(E\) is closed and consider \((x_n)_{n \in \mathbb{N}}\subseteq E\) such that \(x_n \ _{\overrightarrow{n \to \infty}}\ x \in X\). We have to show that \(x \in E\). Observe that \[B(x,r) \cap E \neq \varnothing \quad \text{ for any }\quad r>0.\] But \(x_n \ _{\overrightarrow{n \to \infty}}\ x\) iff \(x_n \in B(x,r)\) for all but finitely many \(n \in \mathbb{N}\), and consequently we conclude that \(x \in {\rm cl\;}(E)\), hence \(x\in E\).
(\(\Longleftarrow\)) Conversely, if \(x \in {\rm cl\;}(E)\) then there is a sequence \((x_n)_{n \in \mathbb{N}}\subseteq E\) so that \(x_n \ _{\overrightarrow{n \to \infty}}\ x \in {\rm cl\;}(E)\subseteq X\) thus by our assumption \(x \in E\). $$\tag*{$\blacksquare$}$$
Theorem
Theorem. The subsequential limits of a sequence \((x_n)_{n \in \mathbb{N}}\) in a metric space \((X,\rho)\) form a closed subset of \(X\).
Proof. Let \(E^{*}\) be the set of subsequential limits of \((x_n)_{n \in \mathbb{N}}\) and let \(q\) be an accumulation point of \(E^{*}\). We will show that \(q \in E^{*}\).
Choose \(n_1 \in \mathbb{N}\) so that \(E^*\ni x_{n_1} \neq q\) (if no such point exists then \(E^{*}\) has only one point and there is nothing to prove). Set \[\delta=\rho(x_{n_1},q)>0.\]
Suppose that \(n_1,\ldots,n_{i-1}\) have been chosen. Since \(q\) is an accumulation point of \(E^{*}\) there is \(x \in E^{*}\) so that \[{\color{red}\rho(x,q)<\delta2^{-i-1}}.\]
Since \(x \in E^{*}\) there is \(n_i>n_{i-1}\) such that \({\color{blue}\rho(x,x_{n_i})<\delta 2^{-i-1}}\).
Hence, by the triangle inequality \[\rho(q,x_{n_i}) \leq {\color{red}\rho(q,x)}+{\color{blue}\rho(x,x_{n_i})}<{\color{red}\delta 2^{-i-1}}+{\color{blue}\delta 2^{-i-1}}=\delta 2^{-i}.\]
This means that \((x_{n_i})_{i \in \mathbb{N}}\) converges to \(q\), i.e. \[\lim_{i\to \infty}x_{n_i}=q \quad \iff \quad \lim_{i\to \infty}\rho(x_{n_i}, q)=0\] thus \(q \in E^{*}\).
In fact, we have shown that \[{\rm acc\;}(E^*) \subseteq E^{*},\] which means that \(E^{*}\) is closed. $$\tag*{$\blacksquare$}$$
Cauchy sequences and complete spaces
Cauchy sequences
Cauchy sequence. A sequence \((x_n)_{n \in \mathbb{N}}\) is a metric space \((X,\rho)\) is said to be a Cauchy sequence if for every \(\varepsilon>0\) there is \(N_{\varepsilon} \in \mathbb{N}\) such that \[m,n \geq N_{\varepsilon}\quad \text{ implies } \quad \rho(x_{m},x_{n})<\varepsilon.\]
Complete spaces. A subset of a metric space \((X,\rho)\) is called complete if every Cauchy sequence in \(E\) converges and its limit is in \(E\).
Complete spaces - examples
Example 1. The set of real numbers \(\mathbb{R}\) is complete.
Example 2. The open unit interval \((0,1)\) is not complete space in \(\mathbb{R}\).
Indeed, let \(x_n=\frac{1}{n}\) for \(n \in \mathbb{N}\), then \(x_n \in (0,1)\) and \((x_n)_{n \in \mathbb{N}}\) is Cauchy in \((0,1)\), but \(0\), which is the limit of \((x_n)_{n \in \mathbb{N}}\) is not in \((0,1)\).
Example 3. \([0,1]\) is complete space in \(\mathbb{R}\).
Some facts
Fact 1. If \((x_n)_{n \in \mathbb{N}}\) is Cauchy in a metric space \((X,\rho)\) then \((x_n)_{n \in \mathbb{N}}\) is bounded.
Fact 2. If \((x_n)_{n \in \mathbb{N}}\) is a Cauchy sequence in a metric space \((X,\rho)\) and \(\lim_{k \to \infty}\rho(x_{n_k},x)=0\) for some \((x_{n _k})_{k \in \mathbb{N}}\), then \[\lim_{n \to \infty} \rho(x_n,x)=0.\]
Proposition. A closed subset of a complete metric space is complete and a complete subset of an arbitrary metric space is closed.
Proof. If \((X,\rho)\) is complete, \(E \subseteq X\) is closed and \((x_n)_{n \in \mathbb{N}}\) is a Cauchy sequence in \(E\), then \((x_n)_{n \in \mathbb{N}}\) has a limit in \(X\). But \({\rm cl\;}(E)=E\), thus \(x \in {\rm cl\;}(E)\), so \(x \in E\).
If \(E \subseteq X\) is complete and \(x \in {\rm cl\;}(E)\) then we know that there exists \((x_n)_{n \in \mathbb{N}}\subseteq E\) converging to \(x\). But \((x_n)_{n \in \mathbb{N}}\) is Cauchy so its limit lies in \(E\), thus \({\rm cl\;}(E)=E\) as desired.$$\tag*{$\blacksquare$}$$
Remark. In the second part of the proof we have used the fact that if \((x_n)_{n \in \mathbb{N}}\) converges (say to \(x\) in a metric space \((X,\rho)\)) then \((x_n)_{n \in \mathbb{N}}\) is Cauchy.
Cantor intersection theorem
Cantor intersection theorem. A metric space is complete iff for every decreasing sequence \[F_1 \supseteq F_2 \supseteq F_3 \supseteq \ldots\] of nonempty closed sets in \(X\) with \({\rm diam\;}(F_n) \ _{\overrightarrow{n \to \infty}}\ 0\), one has \[{\color{blue}\bigcap_{n \in \mathbb{N}}F_n=\{x_0\} \quad \text{ for some }\quad x_0 \in X}.\]
Assume that \((X,\rho)\) is complete.
Let \((F_n)_{n \in \mathbb{N}}\) be such that \(F_1 \supseteq F_2 \supseteq F_3 \supseteq \ldots\) and \({\rm diam\;}(F_n) \ _{\overrightarrow{n \to \infty}}\ 0\).
Choose \(x_n \in F_n\), let \(\varepsilon>0\) and pick \(N_{\varepsilon} \in \mathbb{N}\) such that \({\rm diam\;}(F_n)<\varepsilon\) for all \(n \geq N_{\varepsilon}\). Note that for \(n \geq m \geq N_{\varepsilon}\) we have
\[x_n \in F_n \subseteq F_m,\]
so \[\rho(x_n,x_m) \leq {\rm diam\;}(F_m)<\varepsilon.\]
This ensures that \((x_n)_{n \in \mathbb{N}}\) is a Cauchy sequence and, consequently, converges to some \(x_0 \in X\). Since each \(F_n\) is closed then \(x_0 \in F_n\) for all \(n \in \mathbb{N}\), thus \(x_0 \in \bigcap_{n \in \mathbb{N}}F_n\).
Suppose there is \(y \neq x_0\) so that \(y \in \bigcap_{n \in \mathbb{N}}F_n\), then
\[0<\rho(x_0,y) \leq {\rm diam\;}(F_n) \ _{\overrightarrow{n \to \infty}}\ 0,\] contradiction. Thus \(\bigcap_{n \in \mathbb{N}}F_n =\{x_0\}\).
To prove the converse implication assume that \((x_n)_{n \in \mathbb{N}}\) is a Cauchy sequence.
Let \[{\color{red}F_n={\rm cl\;}\left(\{x_m\;:\; m \geq n\}\right).}\]
Fact. \[{\rm diam\;}(E)={\rm diam\;}({\rm cl\;}(E)).\]
We see that \(F_1 \supseteq F_2 \supseteq F_3 \supseteq \ldots\) and
\[{\rm diam\;}(F_n)={\rm diam\;}\left(\{x_m\;:\;m \geq n\}\right) \ _{\overrightarrow{n \to \infty}}\ 0.\]
Thus \(\bigcap_{n \in \mathbb{N}}F_n=\{x_0\}\) for some \(x_0 \in X\).
Finally, we conclude \(\lim_{n \to \infty}\rho(x_n,x_0)=0\) as desired. $$\tag*{$\blacksquare$}$$
Compact sets
Coverings and compact sets
Coverings. Let \((X,\rho)\) be a metric space.
If \(E \subseteq X\) and \((V_{\alpha})_{\alpha \in A}\) is a family of sets such that \(E \subseteq \bigcup_{\alpha \in A}V_{\alpha}\), then \((V_{\alpha})_{\alpha \in A}\) is called a cover of \(E\) and \(E\) is said to be covered by the \(V_{\alpha}\)’s.
If additionally each \(V_{\alpha}\) is open \((V_{\alpha})_{\alpha \in A}\) is called an open cover of \(E\).
Heine–Borel property. A subset \(K\) of a metric space \((X,\rho)\) is said to be compact if every open cover of \(K\) contains a finite subcover. More explicitly, if \((V_{\alpha})_{\alpha \in A}\) is an open cover of \(K\) then there are finitely many \(\alpha_1,\alpha_2,\ldots,\alpha_n \in A\) such that \[K \subseteq \bigcup_{j=1}^{n}V_{\alpha_j}.\]
Compact sets - examples
Example 1. Every finite subset of \(\mathbb{R}\) is compact.
Example 2. \(K=\{\frac{1}{n}\;:\;n \in \mathbb{N}\} \cup \{0\}\) is compact in \(\mathbb{R}\).
Indeed, let \((V_{\alpha})_{\alpha \in A}\) be an open cover of \(K\), then there is \(\alpha_0 \in A\) such that \(0 \in V_{\alpha_0}\) since \(\lim_{n \to \infty}\frac{1}{n}=0\) and \(V_{\alpha_0}\) is open thus it contains all but finitely many \(\frac{1}{n}\)’s. In other words, there is \(n_0 \in \mathbb{N}\) such that \(n \geq n_0\) implies \(\frac{1}{n} \in V_{\alpha_0}\). Then, for each \(j \in \{1,2,\ldots,n_0-1\}\) we can pick \(\alpha_j \in A\) so that \(\frac{1}{j}\in V_{\alpha_j}\) and we see
\[K \subseteq \bigcup_{j=0}^{{\color{red}n_0}}V_{\alpha_j}.\]
Theorem
Theorem. Compact subsets of metric spaces are closed.
Proof. Let \(K\) be compact subset of a metric space \(X\).
We shall prove that \(K^c\) is open in \(X\). Let \(x \in X \setminus K\). If \(y \in K\), let \[V_{y}=B(x,r_y) \quad \text{ and }\quad W_y=B(y,r_y),\] where \(r_y<\frac{1}{2}\rho(x,y)\), then \({\color{red}V_{y} \cap W_y = \varnothing}\).
Since \(K\) is compact \(K \subseteq \bigcup_{y \in K}W_y\), then we can find \(y_1,\ldots,y_n \in K\) so that \[{\color{red}K \subseteq \bigcup_{j=1}^nW_{y_j}=W}.\]
If \(V=V_{y_1} \cap \ldots \cap V_{y_n}\) then \(V\) is an open set containing \(x\) and
\[V \cap W = \varnothing.\]
Hence \(x \in V \subseteq W^c \subseteq K^c\) thus \(x\) is an interior point of \(K^c\). $$\tag*{$\blacksquare$}$$
Theorem. Closed subsets of compact sets are compact.
Proof. Suppose that \(F \subseteq K \subseteq X\) and \(F\) is closed in \(X\) and \(K\) is compact.
Let \((V_{\alpha})_{\alpha \in A}\) be an open cover of \(F\). Observe that \[F \subseteq K \subseteq \underbrace{\bigcup_{\alpha \in A}V_{\alpha}}_{{\color{red} F\subseteq }} \cup \underbrace{F^c}_{\text{open}}.\]
The set \(K\) is compact thus there is a finite subcover of \[(V_{\alpha})_{\alpha \in A} \cup \{F^c\}\] that covers \(K\).
But \(F \subseteq K\) hence this is also a finite subcover of \(F\) upon removing \(F^c\) as desired.$$\tag*{$\blacksquare$}$$
Theorem. If \((K_{\alpha})_{\alpha \in A}\) is a collection of compact sets of a metric space \((X,\rho)\) such that the intersection of every finite subcollection of \((K_{\alpha})_{\alpha \in A}\) is non-empty then \[\bigcap_{\alpha \in A}K_{\alpha} \neq \varnothing.\]
Proof. Fix a member \(K_{\alpha_0}\) of \((K_{\alpha})_{\alpha \in A}\) and set \(G_{\alpha}=K_{\alpha}^c\).
Suppose that \[\bigcap_{\alpha \in A}K_{\alpha}=K_{\alpha_0} \cap \left(\bigcap_{\alpha \in A \setminus \{\alpha_0\}}K_{\alpha}\right)=\varnothing.\]
Then \[K_{\alpha_0} \subseteq \bigcup_{\alpha \in A \setminus \{\alpha_0\}}G_{\alpha}.\]
Since \(K_{\alpha_0}\) is compact there are \(\alpha_1,\ldots ,\alpha_n \in A\) so that \[K_{\alpha_0} \subseteq \bigcup_{j=1}^n G_{\alpha_j}.\]
Hence \[K_{\alpha_0} \cap K_{\alpha_1} \cap \ldots \cap K_{\alpha_n}=\varnothing,\] which is a contradiction. So we must have \[\bigcap_{\alpha \in A}K_{\alpha} \neq \varnothing.\] as desired. $$\tag*{$\blacksquare$}$$
Totally bounded sets
Totally bounded set. Let \((X,\rho)\) be a metric space, \(E \subseteq X\) is called totally bounded if for every \(\varepsilon>0\), the set \(E\) can be covered by finitely many balls of radius \(\varepsilon\).
It means that there is \(N_{\varepsilon} \in \mathbb{N}\) so that \[E \subseteq \bigcup_{j=1}^{N_{\varepsilon}}B(x_j,\varepsilon)\quad \text{ for some }\quad x_1,x_2,\ldots,x_{N_{\varepsilon}} \in X.\]
Remark 1. If \(E\) is totally bounded so is \({\rm cl\;}(E)\). Indeed, \[E \subseteq \bigcup_{j=1}^{N_{\varepsilon}}B(x_j,\varepsilon) \Longrightarrow {\rm cl\;}(E) \subseteq \bigcup_{j=1}^{N_{\varepsilon}}B(x_j,2\varepsilon).\]
Remark
Remark 2. Every totally bounded set \(E\) is bounded. If \[x,y \in E \subseteq \bigcup_{j=1}^{N_{\varepsilon}}B(x_j,\varepsilon),\] then say \(x \in B(x_1,\varepsilon)\), \(y \in B(x_2,\varepsilon)\) and \[\begin{aligned} \rho(x,y) &\leq \rho(x,x_1)+\rho(x_1,x_2)+\rho(x_2,y) \\&\leq \varepsilon +\max\{\rho(x_i,x_j)\;:\;1 \leq i,j \leq N_{\varepsilon}\}+\varepsilon. \end{aligned}\]
The converse is false in general.
Characterization of compactness
Theorem. If \(E\) is a subset of a metric space \((X,\rho)\) the following are equivalent.
\(E\) is complete and totally bounded.
(The Bolzano–Weierstrass property) Every sequence in \(E\) has a subsequence that converges to a point of \(E\).
(The Heine–Borel property) If \((V_{\alpha})_{\alpha \in A}\) is an open cover of \(E\) then there is finite \(F \subseteq A\) such that \((V_{\alpha})_{\alpha \in F}\) covers \(E\).
Remark. This theorem can be thought of as a characterization of compactness in metric spaces.
Suppose that (a) holds and \((x_n)_{n \in \mathbb{N}}\subseteq E\). We find \((x_{n_k})_{k \in \mathbb{N}}\) such that \(\rho(x_{n_k},x_0) \ _{\overrightarrow{k \to \infty}}\ 0\) for some \(x_0 \in E\).
\(E\) can be covered by finitely many balls of radius \(1 / 2\). At least one of them must contain \(x_n\) for infinitely many \(n \in \mathbb{N}\):
say \(x_n \in B_1\) for \(n \in \mathbb{N}_1 \subseteq \mathbb{N}\) and \({\rm card\;}(\mathbb{N}_1)={\rm card\;}(\mathbb{N})\).
Now \(E \cap B_1\) can be covered by finitely many balls of radius \(1 / 4\). At least one of them must contain \(x_n\) for infinitely many \(n \in \mathbb{N}\):
say \(x_n \in B_2\) for \(n \in \mathbb{N}_2 \subseteq \mathbb{N}_1\) and \({\rm card\;}(\mathbb{N}_2)={\rm card\;}(\mathbb{N})\).
Continuing inductively we obtain a sequence of balls \(B_j\) of radius \(2^{-j}\) and decreasing sequence of subsets \(\mathbb{N}_j\) of \(\mathbb{N}\) such that
\(x_n \in B_j\) for \(n \in \mathbb{N}_j\), \(\mathbb{N}_{j+1} \subseteq \mathbb{N}_j \subseteq \mathbb{N}\), \({\rm card\;}(\mathbb{N}_j)={\rm card\;}(\mathbb{N})\).
Pick \(n_1 \in \mathbb{N}_1\), \(n_2 \in \mathbb{N}_2, \ldots\) such that \[n_1<n_2<n_3<\ldots.\]
Then \((x_{n_j})_{j \in \mathbb{N}}\) is a Cauchy sequence for \[\rho(x_{n_j},x_{n_k})<2^{1-j} \quad \text{ if } \quad k \geq j,\] since \(x_{n_j},x_{n_k} \in B_j\) and \[{\rm diam\;}(B_j) \leq 2^{1-j}.\]
Since \(E\) is complete the sequence \((x_{n_k})_{k \in \mathbb{N}}\) has a limit in \(E\) and the implication (a) \(\Rightarrow\) (b) is proved.$$\tag*{$\blacksquare$}$$
We show that of either condition in (a) fails then so does (b).
If \(E\) is not complete there is a Cauchy sequence \((x_n)_{n \in \mathbb{N}}\subseteq E\), with no limit in \(E\). No subsequence of \((x_n)_{n \in \mathbb{N}}\) can converge in \(E\), for otherwise the whole sequence would converge to the same limit.
On the other hand if \(E\) is not totally bounded, let \(\varepsilon>0\) be such that \(E\) cannot be covered by finitely many balls of radius \(\varepsilon>0\). Choose \(x_n \in E\) inductively as follows. Let \(x_1 \in E\), and having chosen \(x_1,\ldots,x_n\) pick \[x_{n+1} \in E \setminus \bigcup_{j=1}^nB(x_j,\varepsilon),\] then \(\rho(x_n,x_m) \geq \varepsilon\) for all \(n \neq m\), so \((x_n)_{n \in \mathbb{N}}\) has no convergent subsequence. Thus (b) \(\Rightarrow\) (a). $$\tag*{$\blacksquare$}$$
It suffices to show that if (b) holds and \((V_{\alpha})_{\alpha \in A}\) is an open cover of \(E\) then the following claim holds:
Claim. There exists \(\varepsilon>0\) such that every ball of radius \(\varepsilon>0\) that intersects \(E\) is contained in some \(V_{\alpha}\).
Then \(E\) can be covered by finitely many such balls by (a) this allows us to find a finite subcover of \((V_{\alpha})_{\alpha \in A}\).
Suppose for a contradiction that the claim is not true.
For each \(n \in \mathbb{N}\) there is a ball \(B_n\) of radius \(2^{-n}\) such that \(B_n \cap E \neq \varnothing\) and \(B_n\) is contained in no \(V_{\alpha}\).
Pick \(x_n \in B_n \cap E\). Using (b), (by passing to a subsequence if necessary) we may assume \(\lim_{n \to \infty}\rho(x_n,x)=0\) for some \(x \in E\).
We have \(x \in V_{\alpha}\) for some \(\alpha \in A\) and since \(V_{\alpha}\) is open there is \(\varepsilon>0\) so that \(B(x,\varepsilon) \subseteq V_{\alpha}\).
If \(n\) is large enough so that \(\rho(x_n,x)<\frac{\varepsilon}{3}\) and \(2^{-n}<\frac{\varepsilon}{3}\), then \(B_n \subseteq B(x,\varepsilon) \subseteq V_{\alpha}\), which is contradiction.
Indeed, pick \(y \in B_n\), then \[\rho(y,x) \leq \rho(x_n,y)+\rho(x_n,x)<2^{1-n}+\frac{\varepsilon}{3} \leq \varepsilon.\]
This completes the proof of the implication (a) and (b) \(\Rightarrow\) (c). $$\tag*{$\blacksquare$}$$
If \((x_n)_{n \in \mathbb{N}}\subseteq E\), with no convergent sequence, for each \(x \in E\) there is a ball \(B_x\) centered at \(x\) that contains \(x_n\) for only finitely many \(n\).
Otherwise, some sequence would converge to \(x\). Then \[{\color{blue}(B_{x})_{x \in E}}\] is a cover of \(E\) by open sets with no finite subcover.$$\tag*{$\blacksquare$}$$