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It is known that in a first countable space, any limit point $c$ of a sequence $x_n$ is the limit of some subsequence of $x_n$. This is not true in general as noted in this previous question Accumulation points of sequences as limits of subsequences?.

My question is, does the converse hold? I.e., if a space $X$ has the property that for any sequence $x_n \in X$, every limit point of the sequence is the limit of some subsequence, does it follow that $X$ is first countable?

If not, does this property coincide with some other known notion? If it does coincide with some other known notion, I'd expect that it would be Urysohn-Frechet or sequential spaces, but I haven't been able to prove that.

Fernando Chu
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The converse does not hold. Let $X=\omega_1+1=\omega_1\cup\{\omega_1\}$, where $\omega_1$ is the first uncountable ordinal, equipped with the order topology. We claim this has the property described. Indeed, if $x$ is a limit point of $x_n$ and $x\neq \omega_1$, then $\{(y,x]\mid y<x\}$ forms a countable base at $x$ and we can deduce the property from that.

On the other hand if $\omega_1$ is a limit point for $x_n$ then $x_n$ must frequently equal $\omega_1$. To see this, observe that otherwise there is some $N$ for which $\{x_n\mid n> N\}\cap \{\omega_1\}=\emptyset$, but then $\alpha=\bigcup_{n>N}x_n$ is a countable ordinal, hence less than $\omega$, and so $$\{x_n\mid n> N\}\cap (\alpha,\omega_1]=\emptyset\text{,}$$ contradicting the assumption that $\omega_1$ is a limit point.

Since $x_n$ is frequently equal to $\omega_1$, we certainly have a subsequence that is constantly equal to $\omega_1$ and thus converges.

Note that $X$ is not first countable, as $\omega_1$ has no countable base. Moreover, $X$ is not a Fréchet–Urysohn space, nor a sequential space of any type at all, since $\{x\in X\mid x<\omega_1\}$ is sequentially closed but not closed.

Thus while I do not know if there is some other characterization for this property, it is not one of the ones mentioned.

M W
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    I'd change notation to use $\omega_1$ as $\omega$ is more typically the first infinite (countable) ordinal. – Steven Clontz Sep 08 '23 at 20:43
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    @StevenClontz thanks, my ordinals were a little rusty and I'd forgotten some of the standard conventions, I fixed it. – M W Sep 08 '23 at 20:48
  • It's a good answer. I'm a little surprised that after some light searching I didn't find a name for this property, but I'd be more surprised if it isn't out there somewhere. – Steven Clontz Sep 09 '23 at 00:53