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A sequential space is a space which the closed (open) sets are precisely the sequentially closed (open) sets. These spaces are often described as spaces in which "sequences suffice" to describe the topology. This phrase certainly make sense to a certain degree.

In a first countable space, every convergent net admits a cofinal subsequence (i.e. subnet which is a sequence). I'd like to know if something like this can be done in a sequential space which is not first countable.

Question: Suppose $X$ is sequential and $T_1$. Let $(J,\leq)$ be a directed set and $f:J\to X$, $f(j)=x_j$ be a net converging to $x\in X$ where all $x_j$ are distinct from $x$. If the image of $f$ is countable, is it possible to find a sequence converging to $x$ whose image lies in the image of $f$?

If not, I wonder if there is a property weaker than 1st countable which admits a positive answer.

Jeremy Brazas
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  • What is a "cofinal subsequence"? If it means we want a cofinal map $g \colon \mathbb{N}\to J$ and look at the sequence $y_k = f(g(k))$, then the existence of such has nothing to do with $X$ and only with $J$ [and for e.g. $J = \omega_1$, none exist]. So something else. A sequence $(y_k)_{k \in \mathbb{N}}$ such that for every $n$ there is a $j_n \in J$ with $x_j \in { y_k : k \geqslant n}$ for all $j \geqslant j_n$? – Daniel Fischer Sep 04 '16 at 18:46
  • There are (at least) three different notions of subnet and hence of subsequence of a net; which one did you have in mind? – Brian M. Scott Sep 04 '16 at 18:49
  • Right, sorry for the ambiguity. I would like to find a sequence converging to $x$ whose image is in the image of the net. – Jeremy Brazas Sep 04 '16 at 19:05

1 Answers1

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It is not always possible in a sequential space.

Let $X$ be the Arens space, which I’ll describe in the following form. Let $S=\left\{\frac1n:n\in\Bbb Z^+\right\}$, $K=\{0\}\cup S$, and $V_n=\left\{\frac1n\right\}\times K$ for $n\in\Bbb Z^+$; then

$$X=\{\langle 0,0\rangle\}\cup\bigcup_{n\in\Bbb Z^+}V_n\;.$$

For brevity let $p=\langle 0,0\rangle$. Points of $S\times S$ are isolated. For each $n,m\in\Bbb Z^+$ let

$$B(n,m)=\left\{\left\langle\frac1n,\frac1k\right\rangle:k\ge m\right\}\;;$$

$\{B(n,m):m\in\Bbb Z^+\}$ is a local base at $\left\langle\frac1n,0\right\rangle$. For each $\varphi:\Bbb Z^+\to\Bbb Z^+$ and $m\in\Bbb Z^+$ let

$$B(\varphi,m)=\{p\}\cup\bigcup_{k\ge m}B\big(k,\varphi(k)\big)\;;$$

then

$$\left\{B(\varphi,m):\varphi\in{^{\Bbb Z^+}\Bbb Z^+}\text{ and }m\in\Bbb Z^+\right\}$$

is a local base at $p$. $X$ is the canonical example of a sequential space that isn’t Fréchet (and hence isn’t first countable).

Let $\mathscr{D}={^{\Bbb Z^+}\Bbb Z^+}\times\Bbb Z^+$, and for $\langle\varphi,m\rangle,\langle\psi,n\rangle\in\mathscr{D}$ set $\langle\varphi,m\rangle\preceq\langle\psi,n\rangle$ iff $\varphi(k)\le\psi(k)$ for all $k\in\Bbb Z^+$ and $m\le n$; $\langle\mathscr{D},\preceq\rangle$ is a directed set. Let

$$\nu:\mathscr{D}\to X:\langle\varphi,n\rangle\mapsto\left\langle\frac1n,\frac1{\varphi(n)}\right\rangle\;;$$

if $B=B(\varphi,m)$ is any basic open nbhd of $p$, then $\nu(\langle\psi,n\rangle)\in B$ whenever $\langle\varphi,m\rangle\preceq\langle\psi,n\rangle$, so $\nu\to p$. However, the subspace $\{p\}\cup(S\times S)$ is not sequential: no sequence in $S\times S$ converges to $p$.

Brian M. Scott
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