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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:

  1. 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))
  2. 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.]

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Selrach Dunbar
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1 Answers1

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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).

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Brian M. Scott
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  • So ... if I take your point, the bottom line is: $\omega\subseteq\bar{\omega}=\Psi$. $\Psi$ is first countable. $\omega$ is relatively sequentially compact in sense (2), but not in sense (1) (since $\Psi$ is not sequentially compact). Can you confirm your last sentence is meant to imply $\Psi$ is not sequentially compact? ... is $\mathscr{A}$ a family of subsets of $\Psi$ or a subset of $\Psi$) – Selrach Dunbar Nov 07 '16 at 01:09
  • @SelrachDunbar: Yes, that’s exactly right. – Brian M. Scott Nov 07 '16 at 01:11
  • @SelrachDunbar: $\mathscr{A}$ is a family of subsets of $\omega$. Each of those subsets is also a point of $\Psi$. It may be more comfortable to introduce a new set $P={p_A:A\in\mathscr{A}}$ of points not in $\omega$ and let $\Psi=\omega\cup P$. Then basic open nbhds of $p_A$ are sets of the form ${p_A}\cup(A\setminus F)$, where $F$ is any finite subset of $A$. That way $A$ doesn’t have to do double duty as an element of $\Psi\setminus\omega$ and a subset of $\omega$. That’s the approach used ... – Brian M. Scott Nov 07 '16 at 01:19
  • ... in this answer. See also this page in Dan Ma’s Topology Blog. – Brian M. Scott Nov 07 '16 at 01:22
  • @BrianMScott: Awesome. Thanks! – Selrach Dunbar Nov 07 '16 at 09:51
  • @ BrianMScott: P.S. I always take a little (perverse) pleasure in discovering that some allegedly not hard claim, upon further inspection, is actually impossible (since it is actually not true) :) – Selrach Dunbar Nov 07 '16 at 09:51
  • @SelrachDunbar: You're welcome! (It all depends on which side of the claim one is on. :-) I've been on both sides.) – Brian M. Scott Nov 07 '16 at 13:50