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

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?

## 1 Answer

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.