Chapter 10: Compactness and Sequential Compactness in Metric Spaces
In pointβset topology, metric spaces are a class of topological spaces with excellent properties. In this chapter we focus specifically on compactness in metric spaces, introduce the Lebesgue number lemma, and prove that compactness and sequential compactness are equivalent in metric spaces.
1. Diameter of a Set and the Lebesgue Number Lemma
First we define the diameter of a set in a metric space.
Definition 10.1 (Diameter of a Set)
For a metric space \((X, d)\) and a subset \(A \subseteq X\), the diameter of \(A\) is defined as the supremum of distances between any two points in \(A\):
Clearly, if \(x \in A\), then
An extremely important property of compactness in metric spaces is the existence of a Lebesgue number, which links open covers to concrete distances.
Lemma 10.1 (Lebesgue Number Lemma)
Let \(\mathcal{A} = \{A_i\}_{i \in I}\) be an open cover of a compact metric space \((X, d)\). Then there exists a real number \(\delta = \delta(\mathcal{A}) > 0\) (called a Lebesgue number of the cover) such that for every subset \(S \subseteq X\) with \(d(S) < \delta\), there exists some index \(i \in I\) with
Proof of Lemma 10.1 (Click to expand)
(1) If the whole space \(X\) itself belongs to the cover \(\mathcal{A}\), i.e., \(X \in \mathcal{A}\), then the conclusion is trivially true (just take any positive \(\delta\); every subset is contained in \(X\)).
(2) Assume \(X \notin \mathcal{A}\). Since \(X\) is compact, the open cover \(\mathcal{A}\) has a finite subcover. Denote this finite subcover by \(\{A_1, A_2, \dots, A_k\}\). Define a realβvalued function \(f: X \to \mathbb{R}_{\ge 0}\) as follows:
where \(d(x, X \setminus A_j)\) is the distance from the point \(x\) to the closed set \(X \setminus A_j\). Note that \(x \in A_j\) if and only if \(d(x, X \setminus A_j) > 0\).
Since distance functions are continuous and a sum of finitely many continuous functions is continuous, \(f(x)\) is continuous on \(X\). Because \(\{A_1, \dots, A_k\}\) covers \(X\), for every \(x \in X\) there is at least one \(A_j\) with \(x \in A_j\), hence \(d(x, X \setminus A_j) > 0\). Therefore \(f(x) > 0\) for all \(x \in X\).
As the continuous function \(f(x)\) is defined on the compact space \(X\), by the Extreme Value Theorem it attains a strictly positive minimum. Denote this minimum by \(\delta\):
This means that for every \(x \in X\), \(f(x) \ge \delta\), and consequently there exists some \(i \in \{1, \dots, k\}\) such that the largest distance term is at least the average, i.e., \(d(x, X \setminus A_i) \ge f(x) \ge \delta\).
(3) Now take any subset \(S \subseteq X\) with diameter \(d(S) < \delta\). Choose any point \(s_0 \in S\). Since \(d(S) < \delta\), we have
By the conclusion from (2), for the point \(s_0\) there exists an index \(i\) such that
This means that the open ball of radius \(\delta\) centred at \(s_0\) does not touch \(X \setminus A_i\). In other words,
Combining the inclusions, we obtain
Hence \(S \subseteq A_i\), which completes the proof. \(\square\)
2. Sequential Compactness
In analysis, sequential convergence is often used to characterise properties of spaces. We introduce the concept of sequential compactness.
Definition 10.2 (Sequential Compactness)
A metric space \((X, d)\) is called sequentially compact if every sequence \(\{x_n\}_{n \in \mathbb{N}}\) in \(X\) has a convergent subsequence whose limit lies in \(X\).
3. Equivalence of Compactness and Sequential Compactness in Metric Spaces
In general topological spaces, compactness and sequential compactness are not necessarily equivalent. However, in metric spaces the two notions coincide completely.
Theorem 10.1 (Equivalent Characterisations of Compactness in Metric Spaces)
For a metric space \((X, d)\) and any subset \(S \subseteq X\), the following two statements are equivalent:
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(1) \(S\) is compact.
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(2) \(S\) is sequentially compact.
Proof of Theorem 10.1 (Click to expand)
Sufficiency: (1) \(\implies\) (2) (Compactness implies sequential compactness)
Suppose \(S\) is compact. Let \(A = \{x_n\}_{n \in \mathbb{N}} \subseteq S\) be an arbitrary sequence in \(S\). We need to show that \(A\) has a cluster point in \(S\) (i.e., the limit of a convergent subsequence).
We argue by contradiction. Assume \(A\) has no cluster point in \(S\); denote the set of cluster points of \(A\) by \(\text{Lps}(A) = \emptyset\).
Because \(A\) has no cluster points, \(A\) is closed, and for every point \(x \in S\) there exists an open neighbourhood \(U_x\) such that \(U_x\) contains at most finitely many points of \(A\) (specifically, it can contain at most \(x\) itself).
Clearly, the family \(\{U_x\}_{x \in S}\) is an open cover of the compact set \(S\). By compactness, there exist finitely many points \(x_1, x_2, \dots, x_k \in S\) such that their corresponding open neighbourhoods cover \(S\):
However, since \(A \subseteq S\), \(A\) is also covered by these finitely many open sets. By construction each \(U_{x_m}\) contains only finitely many points of the sequence \(A\), so the union of these finitely many open sets can contain at most finitely many points. This directly contradicts the fact that \(A\) is an infinite sequence!
Therefore our assumption is false; \(A\) must have a cluster point in \(S\), and we can extract a subsequence converging to that cluster point. This proves \(S\) is sequentially compact.
Necessity: (2) \(\implies\) (1) (Sequential compactness implies compactness)
Assume \(S\) is sequentially compact. Let \(\mathcal{A} = \{A_i\}_{i \in I}\) be an arbitrary open cover of \(S\). We will prove that it has a finite subcover through several claims.
Claim 1: Existence of a Lebesgue number \(\delta > 0\)
We first claim that there exists \(\delta > 0\) such that for any subset \(B \subseteq S\) with diameter \(d(B) < \delta\), we have \(B \subseteq A_i\) for some \(i\).
Suppose, for contradiction, that no such \(\delta\) exists. Then for every \(n \in \mathbb{N}\), there exists a subset \(B_n \subseteq S\) with \(d(B_n) < \frac{1}{n}\) but \(B_n\) is not contained in any \(A_i\). Choose a point \(b_n \in B_n\) arbitrarily; we obtain a sequence \(\{b_n\}_{n \in \mathbb{N}} \subseteq S\).
By sequential compactness, \(\{b_n\}\) has a convergent subsequence \(\{b_{n_k}\}\); let its limit be \(b \in S\). Since \(b \in S\) and \(\mathcal{A}\) covers \(S\), there exists an open set \(A_i \in \mathcal{A}\) with \(b \in A_i\). Because \(A_i\) is open, there exists \(\epsilon > 0\) such that
Since \(b_{n_k} \to b\) and the radii satisfy \(d(B_{n_k}) < \frac{1}{n_k} \to 0\), for sufficiently large \(k\) the whole set \(B_{n_k}\) is contained in \(B_\epsilon(b)\):
This contradicts the assumption that \(B_{n_k}\) is not contained in any \(A_i\). Therefore a global Lebesgue number \(\delta > 0\) must exist.
Claim 2: \(S\) is totally bounded
We claim that for the \(\delta > 0\) found above, the open cover \(\{B_{\delta/3}(x) : x \in S\}\) has a finite subcover.
Suppose, to the contrary, that it has no finite subcover. We can then construct a sequence recursively: pick \(x_1 \in S\); pick \(x_2 \in S \setminus B_{\delta/3}(x_1)\); pick \(x_3 \in S \setminus (B_{\delta/3}(x_1) \cup B_{\delta/3}(x_2))\); and so on. In general, choose
The set \(P = \{x_1, x_2, \dots, x_n, \dots\} \subseteq S\) thus constructed satisfies
Such a sequence with a fixed lower bound on pairwise distances cannot have a Cauchy subsequence, and therefore cannot have a convergent subsequence. This directly violates the assumption that \(S\) is sequentially compact! Hence our assumption is false; \(S\) must be covered by finitely many open balls of radius \(\delta/3\).
Claim 3: Extracting a finite subcover
By Claim 2, there exist finitely many points \(x_1, x_2, \dots, x_r \in S\) such that
Notice that each ball \(B_{\delta/3}(x_m)\) has diameter strictly less than \(\delta\):
By Claim 1 (existence of the Lebesgue number), each such ball is contained in some open set \(A_{i_m}\) from the original cover \(\mathcal{A}\):
Consequently,
Thus we have extracted a finite subcover \(\{A_{i_1}, A_{i_2}, \dots, A_{i_r}\}\) from the arbitrary open cover \(\mathcal{A}\). The proof is complete. \(\square\)