(Theorem) Compactness, limit point compactness, and sequential compactness are equivalent on metric spaces

Theorem (Munkres 28.1): Compactness implies limit point compactness in any topological space.

Proof.

We will show the contrapositive: If $X$ is a compact topological space and $A\subseteq X$ is any subset, then if $A$ does not have a limit point, then $A$ is finite.

If $A$ has no limit points, then every point $a\in A$ has a neighborhood $U_a$ that intersects $A$ at that that point alone, i.e., $U\cap A=\{a\}$. Then these neighborhoods ${\bigcup }_{a\in A}{U}_a$ cover $A$. Further, $A$ is trivially closed, so the complement $X\backslash A$ is open, and the union ${\bigcup }_{a\in A}{U}_a\cup X\backslash A$ covers $X$. Since $X$ is compact, this cover must have a finite subcover, so $A$ is covered by finitely many sets $U_a$. Each $U_a$ contains only one point $a\in A$, so we conclude that $A$ has finitely many points.

Theorem (Munkres 28.2): Equivalent forms of compactness in metric spaces

Let $(X,d)$ be a metrizable space. Then the following are equivalent:

• (i) $X$ is compact;
• (ii) $X$ is limit point compact;
• (iii) $X$ is sequentially compact

Note that (i) $⟹$ (ii) in any topological space.

Proof that (ii) $⟹$ (iii); also see Compactness, Proof Appendix.

Let $(x_n{)}_{n\in \mathbb{N}}$ be a sequence in $X$, and let $A\subseteq X$ be the set of all points in the sequence. We have two possible cases:

1. The sequence (or $A$) is finite. Then there exists some $N\in \mathbb{N}$ such that for all $n\ge N$, the value $x_n$ is constant. Constant sequences converge trivially.
2. The sequence (or $A$) is infinite. By limit point compactness, $A$ has a limit point $x$. Then we can construct a convergent subsequence as follows: fix a basic open neighborhood $B(x,1)$ about $x$; since $x\in A$ is a limit point, there exists some $x_{n_1}\in A$ that lies in $B(x,1)$. For each $i$, continue fixing points $x_{n_i}$ that lie in the neighborhood $B(x,1/i)$. Since each ball intersects $A$ at infinitely many points, such an element $x_{n_i}$ always exists, so $x_{n_i}\to x$.