Section 4.1 Sequential Compactness
Definition 4.1.
A subset \(K\) of a metric space \((X,d)\) is called sequentially compact if every sequence in \(K\) has subsequence which converges to a limit in \(K\text{.}\) If \(K=X\) is sequentially compact, we say the metric space \((X,d)\) is sequentially compact.
Example 4.2.
An interval of the form \([a,b]\) with \(a \lt b\text{,}\) considered as a subset of \(\R\) with the usual metric, is sequentially compact by the theorem of Bolzano–Weierstrass.
The following are some basic properties of sequentially compact sets.
Lemma 4.3. Sequential compactness and metric subspaces.
A subset \(K\) of a metric space \((X,d)\) is sequentially compact if and only if the associated metric subspace \((K,d')\) is sequentially compact.
Proof.
Let \(\seq xn\) be a sequence of points in \(K\text{,}\) and let \(x_0 \in
K\text{.}\) Then a subsequence \(\subseq xnk\) converges to \(x_0\) as a sequence in \((X,d)\) if and only if it converges to \(x_0\) as a sequence in \((K,d')\text{.}\) The result follows.
Theorem 4.4.
Any sequentially compact metric space is complete.
Proof.
In a sequentially compact space \((X,d)\text{,}\) let \((x_n)_{n \in \N}\) be a Cauchy sequence. Then there exists a convergent subsequence \((x_{n_k})_{k \in \N}\text{.}\) Let \(x_0 \in X\) be its limit. We claim that \(x_n \to x_0\) as \(n \to \infty\text{.}\)
Let \(\varepsilon \gt 0\text{.}\) Then there exists \(N \in \N\) such that \(d(x_m,x_n) \lt \frac{\varepsilon}{2}\) when \(m,n \ge N\text{.}\) Moreover, as \(x_{n_k} \to x_0\text{,}\) there exists \(K \in \N\) such that \(n_K \ge N\) and \(d(x_{n_K},x_0) \lt \frac{\varepsilon}{2}\text{.}\) Now
\begin{equation*}
d(x_n,x_0) \le d(x_n,x_{n_K}) + d(x_{n_K},x_0) \lt \varepsilon
\end{equation*}
for \(n \ge N\text{.}\)
Theorem 4.5.
If \(K\) is a sequentially compact subset of a metric space \((X,d)\text{,}\) then \(K\) is closed.
Proof.
Consider a sequence \((x_n)_{n \in \N}\) in \(K\) that converges in \(X\) to a limit \(x_0 \in X\text{.}\) By Theorem 1.35, it suffices to show that \(x_0 \in K\text{.}\)
As \(K\) is sequentially compact, there exists a subsequence \((x_{n_k})_{k \in \N}\) that converges to a limit \(y_0 \in K\text{.}\) The same subsequence also converges to \(x_0\text{,}\) and by the uniqueness of limits, we infer that \(x_0 = y_0 \in K\text{.}\)
Theorem 4.6.
Let \((X,d)\) be a sequentially compact metric space. Let \(S \subset X\) be closed. Then \(S\) is sequentially compact as well.
Proof.
Let \((x_n)_{n \in \N}\) be a sequence in \(S\text{.}\) As it is also a sequence in \(X\text{,}\) there exists a subsequence \((x_{n_k})_{k \in \N}\) with a limit \(x_0 \in X\text{.}\) Since \(S\) is closed, Theorem 1.35 then implies that \(x_0 \in S\text{,}\) and the proof is complete.
Exercises Exercises
1. (PS8) Compactness and minimisation.
Let \((X,d)\) be a non-empty metric space and \(f \maps X \to \R\) a function. A minimising sequence for \(f\) is a sequence \((x_n)_{n \in \N}\) in \(X\) with
\begin{equation*}
\inf_{x \in X} f(x) = \lim_{n \to \infty} f(x_n),
\end{equation*}
where we allow \(\lim_{n \to \infty} f(x_n) = -\infty\) if necessary.
(a)
Show that every function \(f \maps X \to \R\) has a minimising sequence.
Hint 1.
Exercise B.2 is useful here.
Hint 2.
Note that question asks you to consider the possibility that \(\inf_{x \in X} f(x) = -\infty\text{.}\)
Solution.
If
\begin{equation*}
\inf_{x \in X} f(x) \gt -\infty\text{,}
\end{equation*}
then this follows from Exercise B.2. Indeed, the definition of the infimum implies that for every \(n \in \N\text{,}\) there exists a point \(x_n \in X\) such that
\begin{equation*}
f(x_n) - \frac{1}{n} \le \inf_{x \in X} f(x) \le f(x_n).
\end{equation*}
Thus we obtain a sequence \((x_n)_{n \in \N}\) with the required property. If
\begin{equation*}
\inf_{x \in X} f(x) = -\infty,
\end{equation*}
then for any \(n \in \N\) there exists \(x_n \in X\) with \(f(x_n) \le -n\text{.}\) Then
\begin{equation*}
\lim_{n \to \infty} f(x_n) = -\infty
\end{equation*}
as well. (Note that \(\inf_{x \in X} f(x) = + \infty\) is not possible as \(X \ne \varnothing\text{.}\))
Comment.
Note that the question explicitly mentions the possibility that \(\inf_{x \in X} f(x) = - \infty\text{.}\) Many students did not treat this case; on an exam this would have cost marks.
(b)
Show that the infimum of \(f\) is attained, i.e., there exists a point \(x_0 \in X\) with
\begin{equation*}
f(x_0) = \inf_{x \in X} f(x),
\end{equation*}
if \((X,d)\) is sequentially compact and \(f\) is continuous.
Hint.
Argue that the minimising sequence from the previous part has a convergent subsequence.
Solution.
Let \((x_n)_{n \in \N}\) be a minimising sequence. If \((X,d)\) is sequentially compact, then there exists a convergent subsequence \((x_{n_k})_{k \in \N}\text{.}\) Let \(x_0\) be its limit. If \(f\) is continuous, then it follows (from Theorem 1.34 and Theorem 1.46) that
\begin{equation*}
f(x_0) = \lim_{k \to \infty} f(x_{n_k}) = \lim_{n \to \infty} f(x_n) = \inf_{x \in X} f(x).
\end{equation*}
Comment.
It’s easy to see to find counterexamples if we weaken the requirement that \(X\) is compact. For instance, the function \(x \mapsto x\) does not attains its minimum over either \(X = (0,1)\) or \(X = \R\text{.}\)
2. Bounding functions on compact spaces.
Let \((X,d)\) be a metric space and \(K\subseteq X\) be sequentially compact. Suppose \(f\maps X\to\R\) is continuous. Show that \(f\) is bounded on \(K\text{.}\) That is, show that there exists a real number \(M\gt 0\) such that \(| f(x)| \le M\) for every \(x \in K\text{.}\)
Hint.
Try an argument by contradiction.
3. Weierstrass extreme value theorem.
Let \(X\) be a metric space and \(K\subseteq X\) be compact. Let \(f
\in C_\bdd(K)\text{.}\) Show that there exists \(x_{\sup} \in K\) and \(x_{\inf}\in K\) such that \(\sup_{x \in K} f(x)= f(x_{\sup})\) and \(\inf_{x \in K} f(x) = f(x_{\inf})\text{.}\)
Hint.
Use Exercise 4.1.1.
4. Nested compact sets.
In a metric space \((X,d)\text{,}\) consider a sequence of sets \(C_n \subseteq
X\) for \(n \in \N\text{.}\) Suppose that \(C_n\) is non-empty and sequentially compact, and \(C_{n + 1} \subseteq C_n\) for every \(n \in
\N\text{.}\) Show that the intersection \(\bigcap_{n \in \N} C_n\) is not empty.
Hint 1.
For each \(n \in \N\text{,}\) pick a point \(x_n \in C_n\text{.}\) Argue that \(\seq xn\) is then a sequence in the sequentially compact set \(C_1\text{.}\)
Hint 2.
Theorem 4.5 is useful.
