What separation is required to ensure extremally disconnected spaces are sequentially discrete?

Question

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In https://github.com/pi-base/data/issues/387 it is noted that Tychonoff extremally disconnected spaces are sequentially discrete (see Encyclopedia of General Topology by Hart, Nagata, Vaughan).

It seems from The only convergent sequences in an extremally disconnected $T_2$ space are those which are ultimately constant. that in fact Hausdorff is sufficient (Willard exercise 15G.3).

Can Hausdorff be weakened? (Note that an indiscrete space with multiple points is extremally disconnected but not sequentially discrete, so some separation seems to be necessary.)

Answer 1

Here is a proof for T2:

$X$ is a topological space.

Lemma. Let X be T2, $(x_n)_{n \in \mathbb N}$ a sequence in $X$ of pairwise distinct elements, converging to $x \in X, x \neq x_n$ for all $n \in \mathbb N$. Then there exists a pairwise disjoint family $(U_n)_{n \in \mathbb N}$ of open sets, such that $x_n \in U_n$ for all $n \in \mathbb N$.

PROOF. Step 1: For $n \in \mathbb N$ there is an open set $U$ containing $x_n$, such that $x_m \notin \overline{U}$ for all $m > n$. [Consider disjoint neighborhoods of $x_n$ and $x$. Then only finitely many $x_m$ are not in the latter, which can easily be taken care of by T2.]
Step 2: By induction, using step 1, it is now easy to define $(U_n)_{n \in \mathbb N}$ as required.

Now let $X$ be extremally disconnected, T2. Then each converging sequence is eventually constant:

Assume not. Then it is easy to see that by T1, there is a sequence $(x_n)_{n \in \mathbb N}$ in $X$ of pairwise distinct elements, converging to $x \in X, x \neq x_n$ for all $n \in \mathbb N$. Let $(U_n)_{n \in \mathbb N}$ be as in the lemma. Define $U := \bigcup_{n \in \mathbb N} U_{2n}$ and $V := \bigcup_{n \in \mathbb N} U_{2n+1}$. Then $U, V$ are open and disjoint, $x \in \overline{U} \cap \overline{V}$ contradicting extremal disconnectness of $X$.

Answer 2

Potentially helpful to find a full answer:

Note that $T_2$ implies US. The Countable complement topology on the real numbers is the only example of a US Λ ¬T2 Λ Extremally disconnected space known to pi-Base as of this post. Every countable set in this space is closed.

Now, for every space, if countable sets are closed, then the space is sequentially discrete. Likewise, if the space is sequentially discrete, it is US.

To see this, if all countable sets are closed, a sequence cannot converge to a point outside itself, and thus only eventually constant sequences converge. Likewise, if every convergent sequence is eventually constant, that tail gives the unique limit of the sequence.

In general these do not reverse. The Stone-Cech compactification $\beta \omega$ is sequentially discrete but $\omega$ is a countable non-closed set. Likewise, the real line with its usual topology is US but not sequentially discrete.

As it turns out, $\beta\omega$ is also extremally disconnected, giving an example of a space which is extremally disconnected and sequentially discrete, but not all countable sets are closed.

This still leaves open the possibility that all extremally disconnected $US$ spaces are sequentially discrete.

Answer 3

KP Hart provides an example of a strongly KC extremally disconnected space which is not sequentially discrete on MathOverflow: https://mathoverflow.net/a/453013/

In short: let $X$ be the co-countable topology on an uncountable set, and $Y$ be a converging sequence. Then give the disjoint union $X\cup Y$ the weakest topology preserving these subspaces such that $Y$ is closed.