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Let $(X,\tau)$ be a topological space. The collection of all sequentially open subsets of $X$ (i.e. the complements of the sequentially closed subsets) is itself a topology $\tau_\text{seq}$, equal to $\tau$ if $X$ is a sequential space, and a strictly finer topology otherwise. (See the wikipedia article for details.)

Question: Is $(X,\tau_\text{seq})$ a sequential space?

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  • $\begingroup$ Isn't it immediate from the first characterization on the wiki page? $\tau_{seq}$ consists of all sequential open subsets of X. $\endgroup$ Jun 27 '20 at 21:53
  • $\begingroup$ It is not obvious. Working equivalently in terms of closed subsets, what needs to be shown is that a subset that is sequentially closed in $(X,\tau_\text{seq})$ is closed with respect to $\tau_\text{seq}$, i.e., is sequentially closed in $(X,\tau)$. $\endgroup$
    – PatrickR
    Jun 27 '20 at 21:59
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Suppose that $U$ is sequentially open with respect to $\tau_{\text{seq}}$, i.e., that every sequence converging in $\tau_{\text{seq}}$ to a point of $U$ is eventually in $U$. $\langle X,\tau_{\text{seq}}\rangle$ and $\langle X,\tau\rangle$ have the same convergent sequences, so $U$ is sequentially open with respect to $\tau$ and is therefore in $\tau_{\text{seq}}$. Thus, $\langle X,\tau_{\text{seq}}\rangle$ is sequential.

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  • $\begingroup$ Nice. Thank you. $\endgroup$
    – PatrickR
    Jun 27 '20 at 22:06
  • $\begingroup$ @PatrickR: You’re welcome. $\endgroup$ Jun 27 '20 at 22:07
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Preliminaries:

Define, for a sequence $(x_n)$ in $X$ the tail filter of the sequence by $$\textrm{tail}(x_n) = \{A \subseteq X\mid \exists m\in \Bbb N: \{x_n: n \ge m\} \subseteq A\}$$

The definition of convergence says in essence that $x_n \to x$ for any topology $\tau'$ on $X$ iff $$\{O \in \tau'\mid x \in O\} \subseteq \text{tail}(x_n)\tag{0}$$

It is standard that $\tau_{\text{seq}}$ is defined by all sets $O$ such that

$$\forall x \in O: \forall (x_n) \subseteq X: (x_n \to_\tau x) \to O \in \textrm{tail}(x_n)\tag{1}$$

which are called the sequentially open sets w.r.t. the topology $\tau$. The wikipedia page has a proof that this actually forms a topology on $X$ which clearly obeys $$\tau \subseteq \tau_{\text{seq}}$$ basically by the definition of convergence under $\tau$.


Proof

Now suppose that $O$ is sequentially open for $\tau_{\text{seq}}$. We have to show $O$ is actually in $\tau_{\text{seq}}$. So let $x \in O$ be arbitrary and $x_n \to_{\tau} x$.

Fact: $x_n \to x$ for $\tau_s$ as well: let $U$ be any open set in $\tau_{\text{seq}}$ containing $x$. Because $U \in \tau_{\text{seq}}$ and $x_n \to_\tau x$ by definition $U \in \textrm{tail}(x_n)$. But as $U \in \tau_{\text{seq}}$ was arbitrary containing $x$, indeed $x_n \to x$ for $\tau_{\text{seq}}$ ( we apply $(0)$ to $\tau_{\text{seq}}$) as required.

But as $O$ is sequentially open for $\tau_{\text{seq}}$, by definition again $O \in \textrm{tail}(x_n)$ and so $O$ (by $(1)$) is sequentially open for $\tau$, i.e. $O \in \tau_{\text{seq}}$.

Which shows that $(X, \tau_{\text{seq}})$ is a sequential space by definition.

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  • $\begingroup$ I realise this is essentially Brian's proof. But he does not show (the page he refers to also gives no proof, just a claim) why the topologies have the same convergent sequences and I do show all the necessary details. So I think this answer adds something. $\endgroup$ Jun 28 '20 at 21:53
  • $\begingroup$ Thank you. Using the tail filter of sequences is a good way to organize the argument. $\endgroup$
    – PatrickR
    Jun 28 '20 at 22:16
  • $\begingroup$ @PatrickR I hope it helps you. $\endgroup$ Jun 29 '20 at 4:35

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