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I was reviewing my Topology notes today and came across the proof that "if $(X,\tau)$ is first countable and $A\subset X$ is sequentially closed, then it is closed". I will do a brief recap of (our) proof:

Let $x\in \overline{A}$ and open $U_1\supset U_2\supset \dots$ be a nested local base at $x$. Then $U_n\cap A\neq \emptyset $ for all $n$. So $\textbf{choose}$ $x_n\in U_n\cap A$. Define $f: \mathbb{N}\cup \{\infty\}\to X$ as $f(n)=x_n$ and $f(\infty)=x$. To show $f$ is continuous, let $U\subset X$ be open. If $x\not\in U$, then $f^{-1}(U)\subset\mathbb{N}$ is open and if $x\in U$, then there exists $N$ such that $U_n\subset U$ and therefore, $f^{-1}(U)\supset\{N,N+1, \dots\}$ so it is open in this case as well. We conclude that $x_n\to x$ and thus that $x\in A$.

Now we very clearly invoke (at least a weak version of) the axiom of choice here. As indicated in the title, I am wondering if this result still is true without this assumption?

Adam Martens
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  • I mean, metric spaces are first countable. So the example in that answer is also a counterexample for your question. – Asaf Karagila Nov 26 '19 at 02:24
  • @Eric: The example with a Dedekind-finite set of reals is an example of a set which is sequentially closed and sequentially compact in a metric space (which is first countable), but it is neither closed nor compact. – Asaf Karagila Nov 26 '19 at 02:27
  • @Eric: Possibly. Let me find another link, because this is definitely a dup. – Asaf Karagila Nov 26 '19 at 02:29
  • @Eric: This one should be better. – Asaf Karagila Nov 26 '19 at 02:38
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    For the benefit of OP: Although Asaf's answer in the question he linked does not say so explicitly, the example he describes also shows that "first countable and sequentially closed implies closed" is equivalent to countable choice, using the set $A=X\setminus{\infty}$. – Eric Wofsey Nov 26 '19 at 02:47
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    In particular, I point this out because it is not obvious that condition (2) of Proposition 5 is equivalent to "metric and sequentially closed implies closed", because a space can satisfy "sequentially closed implies closed" without having accumulation points of subsets always be limits of sequences. – Eric Wofsey Nov 26 '19 at 02:48

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