In the literature on measure theory, probability, and functional analysis the definition of a subset A⊆X of a topological space X to be relatively sequentially compact is not unique:
- A is relatively sequentially compact in X if its closure cl(A) in X is sequentially compact, i.e. if every sequence in cl(A) has a convergent subsequence (with limit in cl(A))
- A is relatively sequentially compact in X if every sequence in A has a convergent subsequence with limit in cl(A).
Clearly, 1 ⇒ 2. In this question's answer the writer claims that if X is first-countable it is not hard to prove that definition (2) implies definition (1).
Can someone enlighten me on how this (not hard) proof would go?
[I can see that (2) implies (1) if X is a metric space, just not sure how to generalize the argument to the case where X is merely first-countable.]
Mrówka’s space $\Psi$, which is described in this answer, is a counterexample. It is clearly first countable. The set $\omega$ is dense in $\Psi$, and I claim that it is relatively sequentially compact in sense (2).
Let $\sigma=\langle x_n:n\in\omega\rangle$ be a sequence in $\omega$. If $\sigma$ has a constant subsequence, we’re done. If not, the maximality of the almost disjoint family $\mathscr{A}$ ensures that there is an $A\in\mathscr{A}$ that meets $\{x_n:n\in\omega\}$ in an infinite set $\{x_{n_k}:k\in\omega\}$, and $\langle x_{n_k}:k\in\omega\rangle$ is then a subsequence of $\sigma$ that converges to the point $A\in\Psi$.
However, $\mathscr{A}$ is an infinite (in fact uncountable) closed, discrete subset of $\Psi=\operatorname{cl}_\Psi\omega$, so $\omega$ is not relatively sequentially compact in sense (1).