3
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.