# Topology of sequentially open sets is sequential?

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?

• Isn't it immediate from the first characterization on the wiki page? $\tau_{seq}$ consists of all sequential open subsets of X. Jun 27 '20 at 21:53
• 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 $(X,\tau)$. Jun 27 '20 at 21:59

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.

• Nice. Thank you. Jun 27 '20 at 22:06
• @PatrickR: You’re welcome. Jun 27 '20 at 22:07

## 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.

• 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. Jun 28 '20 at 21:53
• Thank you. Using the tail filter of sequences is a good way to organize the argument. Jun 28 '20 at 22:16
• @PatrickR I hope it helps you. Jun 29 '20 at 4:35