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Let $X$ be a Fréchet-Urysohn space (i.e., for all $A\subseteq X$, the closure of $A$ coicides with the sequential closure of $A$). Let $(x_n)_{n\in\mathbb N}$ be a sequence in $X$ converging to $x$. For each $n\in\mathbb N$, let $(x_{n,m})_{m\in\mathbb N}$ be a sequence in $X$ converging to $x_n$. Make $\mathbb N\times\mathbb N$ into a directed set by letting $(n,m)\leq(n',m')$ if and only if [$n<n'$ or [$n=n'$ and $m\leq m'$]]. Then does there exist a subnet $(a_n)_{n\in\mathbb N}$ (required to be a sequence) of $(x_{n,m})_{(n,m)\in\mathbb N\times\mathbb N}$ such that $(a_n)_{n\in\mathbb N}$ converges to $x$?

I know that it is true if $X$ is first-countable, but I wonder if it is still true if I weaken the assumption to Fréchet-Urysohn.

zxcv
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1 Answers1

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Let $X=\mathbb{N}^2\cup\{\infty\}$ with the topology such that every subset of $\mathbb{N}^2$ is open and a set containing $\infty$ is open iff for each $n\in\mathbb{N}$, it contains $(n,m)$ for all but finitely many $m\in\mathbb{N}$. Then $X$ is Fréchet-Urysohn: the only nontrivial thing to check is that if $A\subseteq X$ and $\infty\in\overline{A}$, then $\infty$ is a limit of a sequence in $A$. Note that for any fixed $n$ the sequence $(n,m)_{m\in\mathbb{N}}$ converges to $\infty$, so if $A$ contains no sequence converging to $\infty$, then $A\subset\mathbb{N}^2$ contains only finitely many points of the form $(n,m)$ for any fixed $n$. But then the complement of $A$ is a neighborhood of $\infty$, so $\infty\not\in\overline{A}$.

However, consider $x_n=\infty$ for each $n$ and $x_{n,m}=(n,m)$. Then $(x_n)$ converges to $\infty$ and $(x_{n,m})$ converges to $x_n$ for each fixed $n$. However, no subsequence of the net $(x_{n,m})$ converges to $\infty$, since such a subsequence would have to have only finitely many terms of the form $(n,m)$ for each $n$ and so as above $\infty$ would not be in its closure.

On the other hand, it is true if $x_n\neq x$ for all $n$ and $X$ is sequentially Hausdorff (i.e., limits of sequences in $X$ are unique). In that case, for each $n$ only finitely many of the terms $x_{n,m}$ can be equal to $x$, so we may remove those terms from each of these sequences to assume that $x_{n,m}\neq x$ for all $n,m$ as well. Now since $x$ is in the closure of the set $A=\{x_{n,m}:n,m\in\mathbb{N}\}$, there is a sequence in $A$ that converges to $x$; since $x\not\in A$ and $X$ is sequentially Hausdorff we may assume the terms of this sequence are distinct. If this sequence has infinitely many terms of the form $x_{n,m}$ for some fixed value of $n$, then the subsequence formed by those terms converges to $x_n$, contradicting sequential Hausdorffness. So the sequence has only finitely many terms of the form $x_{n,m}$ for each fixed $n$, and thus is a subnet of $(x_{n,m})$.

Eric Wofsey
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  • Lol, within seconds, and I think these are the same example? – M W Jan 30 '24 at 03:25
  • Indeed, looks like it. – Eric Wofsey Jan 30 '24 at 03:26
  • This second part is pretty cool, shows in some sense this is the only counterexample. – M W Jan 30 '24 at 03:34
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    Yes, I first was trying to prove it and realized I seemed to need certain extra hypotheses, and then tried to build a counterexample violating those hypotheses. (You can similarly get a counterexample where $x_n\neq x$ for each $n$ but $X$ fails to be sequentially Hausdorff. Basically take the example above and add an extra point $x_n$ that is a second limit of each sequence $(x_{n,m})$ and make $(x_n)$ converge to $\infty$ as well.) – Eric Wofsey Jan 30 '24 at 03:37