§ 28. Limit Point Compactness

Definition 28.1 : Limit Point Compact

A space \(X\) is said to be limit point compact if every infinite subset of \(X\) has a limit point.

Theorem 28.1

Compactness implies limit point compactness, but not conversely.

Proof

Let \(X\) be a compact space. Given a subset \(A\) of \(X\), we wish to prove that if \(A\) is infinite, then \(A\) has a limit point. We prove the contrapositive-if \(A\) has no limit point, then \(A\) must be finite.

So suppose \(A\) has no limit point. Then \(A\) contains all its limit points, so that \(A\) is closed. Furthermore, for each \(a \in A\) we can choose a neighborhood \(U_{a}\) of \(a\) such that \(U_{a}\) intersects \(A\) in the point \(a\) alone. The space \(X\) is covered by the open set \(X-A\) and the open sets \(U_{a}\); being compact, it can be covered by finitely many of these sets. Since \(X-A\) does not intersect \(A\), and each set \(U_{a}\) contains only one point of \(A\), the set \(A\) must be finite.

Definition 28.2 : Subsequence / Sequentially Compact

Let \(X\) be a topological space. If \(\left(x_{n}\right)\) is a sequence of points of \(X\), and if

\[ n_{1}<n_{2}<\cdots<n_{i}<\cdots \]

is an increasing sequence of positive integers, then the sequence \(\left(y_{i}\right)\) defined by setting \(y_{i}=x_{n_{i}}\) is called a subsequence of the sequence \(\left(x_{n}\right)\). The space \(X\) is said to be sequentially compact if every sequence of points of \(X\) has a convergent subsequence.

Theorem 28.2

Let \(X\) be a metrizable space. Then the following are equivalent:

1. \(X\) is compact.
2. \(X\) is limit point compact.
3. \(X\) is sequentially compact.

Proof

We have already proved that (1) \(\Rightarrow\) (2). To show that (2) \(\Rightarrow\) (3), assume that \(X\) is limit point compact. Given a sequence \(\left(x_{n}\right)\) of points of \(X\), consider the set \(A=\left\{x_{n} \mid n \in \mathbb{Z}_{+}\right\}\). If the set \(A\) is finite, then there is a point \(x\) such that \(x=x_{n}\) for infinitely many values of \(n\). In this case, the sequence \(\left(x_{n}\right)\) has a subsequence that is constant, and therefore converges trivially. On the other hand, if \(A\) is infinite, then \(A\) has a limit point \(x\). We define a subsequence of \(\left(x_{n}\right)\) converging to \(x\) as follows: First choose \(n_{1}\) so that

\[ x_{n_{1}} \in B(x, 1) \text {. } \]

Then suppose that the positive integer \(n_{i-1}\) is given. Because the ball \(B(x, 1 / i)\) intersects \(A\) in infinitely many points, we can choose an index \(n_{i}>n_{i-1}\) such that

\[ x_{n_{i}} \in B(x, 1 / i) . \]

Then the subsequence \(x_{n_{1}}, x_{n_{2}}, \ldots\) converges to \(x\).

Finally, we show that \((3) \Rightarrow\) (1). This is the hardest part of the proof.

First, we show that if \(X\) is sequentially compact, then the Lebesgue number lemma holds for \(X\). (This would follow from compactness, but compactness is what we are trying to prove!) Let \(\mathcal{A}\) be an open covering of \(X\). We assume that there is no \(\delta>0\) such that each set of diameter less than \(\delta\) has an element of \(\mathcal{A}\) containing it, and derive a contradiction.

Our assumption implies in particular that for each positive integer \(n\), there exists a set of diameter less than \(1 / n\) that is not contained in any element of \(\mathcal{A}\); let \(C_{n}\) be such a set. Choose a point \(x_{n} \in C_{n}\), for each \(n\). By hypothesis, some subsequence \(\left(x_{n_{i}}\right)\) of the sequence \(\left(x_{n}\right)\) converges, say to the point \(a\). Now \(a\) belongs to some element \(A\) of the collection \(\mathcal{A}\); because \(A\) is open, we may choose an \(\epsilon>0\) such that \(B(a, \epsilon) \subset A\). If \(i\) is large enough that \(1 / n_{i}<\epsilon / 2\), then the set \(C_{n_{i}}\) lies in the \(\epsilon / 2\)-neighborhood of \(x_{n_{i}}\); if \(i\) is also chosen large enough that \(d\left(x_{n_{i}}, a\right)<\epsilon / 2\), then \(C_{n_{i}}\) lies in the \(\epsilon\)-neighborhood of \(a\). But this means that \(C_{n_{i}} \subset A\), contrary to hypothesis.

Second, we show that if \(X\) is sequentially compact, then given \(\epsilon>0\), there exists a finite covering of \(X\) by open \(\epsilon\)-balls. Once again, we proceed by contradiction. Assume that there exists an \(\epsilon>0\) such that \(X\) cannot be covered by finitely many \(\epsilon\)-balls. Construct a sequence of points \(x_{n}\) of \(X\) as follows: First, choose \(x_{1}\) to be any point of \(X\). Noting that the ball \(B\left(x_{1}, \epsilon\right)\) is not all of \(X\) (otherwise \(X\) could be covered by a single \(\epsilon\)-ball), choose \(x_{2}\) to be a point of \(X\) not in \(B\left(x_{1}, \epsilon\right)\). In general, given \(x_{1}, \ldots, x_{n}\), choose \(x_{n+1}\) to be a point not in the union

\[ B\left(x_{1}, \epsilon\right) \cup \cdots \cup B\left(x_{n}, \epsilon\right) \]

using the fact that these balls do not cover \(X\). Note that by construction \(d\left(x_{n+1}, x_{i}\right) \geq\) \(\epsilon\) for \(i=1, \ldots, n\). Therefore, the sequence \(\left(x_{n}\right)\) can have no convergent subsequence; in fact, any ball of radius \(\epsilon / 2\) can contain \(x_{n}\) for at most one value of \(n\).

Finally, we show that if \(X\) is sequentially compact, then \(X\) is compact. Let \(\mathcal{A}\) be an open covering of \(X\). Because \(X\) is sequentially compact, the open covering \(\mathcal{A}\) has a Lebesgue number \(\delta\). Let \(\epsilon=\delta / 3\); use sequential compactness of \(X\) to find a finite covering of \(X\) by open \(\epsilon\)-balls. Each of these balls has diameter at most \(2 \delta / 3\), so it lies in an element of \(\mathcal{A}\). Choosing one such element of \(\mathcal{A}\) for each of these \(\epsilon\)-balls, we obtain a finite subcollection of \(\mathcal{A}\) that covers \(X\).