Sequential compactness $\rightarrow$ limit point compactness and axiom of choice

Question

A space $S$ is sequentially compact if every sequence has a convergent subsequence. A space $S$ is limit point compact is every infinite subset has a limit point in $S$.

Proving sequential compactness implies limit point compactness more or less involves letting $E \subset S$ be infinite and extracting a countably infinite subset from $E$, which we can then view as a sequence. What the convergent subsequence converges to will be the limit point.

However, the claim that an arbitrary infinite set has a countably infinite subset is dependent on the axiom of choice. Does sequentially compact $\to $ limit point compact still hold without choice? Can my proof above be modified so that choice isn't needed?

Answer 1

No, this is in fact equivalent (in ZF) to the statement that every infinite set has a countably infinite subset. Indeed, suppose $S$ is an infinite set with no countably infinite subset. Then any sequence in $S$ can take only finitely many values (otherwise its image would be a countably infinite subset of $S$), and so has a convergent subsequence with respect to any topology (just pick a subsequence that is constant). In particular, giving $S$ the discrete topology, $S$ is sequentially compact but not limit point compact.