Section 4.4 Relative Compactness
The following is a useful notion concerning sets in a metric space that are not necessarily compact, but share some of the properties of compact sets.
Definition 4.26.
A set \(S\) in a metric space is relatively compact if its closure \(\overline S\) is compact.
By Theorem 4.22, this is equivalent to all sequences in \(S\) having a subsequence converging in \(X\text{;}\) see Exercise 4.4.1.
Theorem 4.27.
- Any relatively compact subset of a metric space is totally bounded.
- Any totally bounded subset of a complete metric space is relatively compact.
Proof.
Let \(S\) be a subset of a metric space \((X,d)\text{.}\)
Part i.
Suppose that \(S\) is relatively compact. Then \(\overline S\) is compact. Theorem 4.22 implies that \(\overline S\) is totally bounded, and Lemma 4.11 implies that \(S\) is totally bounded as well.
Part ii.
Suppose that \(X\) is complete and \(S\) is totally bounded. We first show that \(\overline S\) is totally bounded as well. To this end, let \(r \gt 0\text{.}\) Then there exist finitely many balls of radius \(r/2\) that cover \(S\text{,}\) say \(B_{r/2}(s_1), \dotsc,
B_{r/2}(s_N)\text{.}\) Now for every \(x \in \overline S\text{,}\) there exists \(s \in S\) with \(d(x, s) \lt \frac{r}{2}\text{.}\) Furthermore, there exists \(n \in \{1, \dotsc, N\}\) with \(s \in B_{r/2}(s_n)\text{.}\) Hence \(x \in B_r(s_n)\text{,}\) which shows that the balls \(B_r(s_1), \dotsc, B_r(s_N)\) cover \(\overline S\text{.}\)
Since \(X\) is complete and \(\overline S\) is closed, the metric subspace \(\overline S\) is complete as well by Theorem 1.41. Theorem 4.22 then implies that \(\overline S\) is compact, which means that \(S\) is relatively compact.
Exercises Exercises
1. (PS9) Relative compactness and subsequences.
Let \((X,d)\) be a metric space and \(S \subseteq X\text{.}\) Show that the following are equivalent:
- \(S\) is relatively compact.
- For all sequences \(\seq sn\) in \(S\text{,}\) there is a subsequence \(\subseq snk\) which converges in \(X\text{.}\)
Hint.
Showing that Part i implies Part ii is not so bad, but the reverse implication requires a bit more ingenuity. One way to proceed, given a sequence \(\seq xn\) in \(\overline S\text{,}\) is to use the definition of the closure to argue that there is another sequence \(\seq sn\) in \(S\) with \(d(x_n,s_n) \lt 1/n\text{.}\)
Solution.
First suppose that Part i holds, and let \(\seq sn\) be a sequence in \(S\text{.}\) Then \(\seq sn\) is also a sequence in \(\overline S\text{,}\) which is compact and hence sequentially compact. Thus there exists a subsequence \(\subseq snk\) which converges to some \(x \in \overline S \subseteq X\text{.}\)
Now suppose that Part ii holds. To show that \(S\) is relatively compact, it suffices to show that \(\overline
S\) is sequentially compact. So let \(\seq xn\) be a sequence in \(\overline S\text{.}\) We must find a subsequence \(\subseq xnk\) which converges to some \(x_0 \in \overline S\text{.}\) Using the definition of the closure, for each \(n \in \N\) we can find \(s_n \in S\) with \(d(s_n,x_n) \lt 1/n\text{.}\) By Part ii, the sequence \(\seq sn\) has a subsequence \(\subseq snk\) which converges to some \(x_0 \in X\text{,}\) and by Theorem 1.35 we have \(x_0 \in \overline S\text{.}\) Finally, by the triangle inequality
\begin{equation*}
d(x_{n_k},x_0) \le
d(x_{n_k},s_{n_k})
+ d(s_{n_k},x_0)
\to 0
\end{equation*}
as \(k \to \infty\text{,}\) so that \(x_{n_k} \to x_0\) as desired.
Comment.
Note that in order to show that \(\overline S\) is sequentially compact, we must show that any sequence \(\overline S\) has a subsequence converging to a limit lying in the same set \(\overline S\text{.}\)
