*Based upon discussion at [Math.SE](https://math.stackexchange.com/questions/4751580/what-separation-is-required-to-ensure-extremally-disconnected-spaces-are-sequent)*


Consider the property [extremally disconnected](https://topology.pi-base.org/properties/P000049), for which the closure of any open set remains open.


Frequently, this property is paired with the assumption of Hausdorff. This allows nice results like [all extremally disconnected spaces are totally separated](https://topology.pi-base.org/theorems/T000045) and disallows silliness like [all hyperconnected spaces are extremally disconnected](https://topology.pi-base.org/theorems/T000096).


But I like silliness, so how much of the theory can we recover without Hausdorff? In [this Math.SE post](https://math.stackexchange.com/questions/4751580/what-separation-is-required-to-ensure-extremally-disconnected-spaces-are-sequent) it was pointed out that while all Hausdorff extremally disconnected spaces are sequentially discrete (only trivial sequences converge), the cofinite topology on an infinite set is a $T\_1$ extremally disconnected space that is not sequentially discrete.


On the other hand, the only example I know of a sequentially discrete, $T\_1$, extremally disconnected space is US (limits of sequences are unique): the cocountable topology on an uncountable set. In fact, this is pretty immediate: all spaces where countable sets are closed are sequentially discrete, and all sequentially discrete spaces are US.


Might it be the case that the theorem that all Hausdorff extremally disconnected spaces are sequentially discrete can be improved to only assume US? If not, what counterexample can be constructed?



Must US extremally disconnected spaces be sequentially discrete?
Here's an example, I think.
Let $X=(\omega\_1+1)\times(\omega\_0+1)$ and let $Y$ denote the subset $\omega\_1\times(\omega\_0+1)$.
Define a topology on $X$ by declaring every set of the form $Y\setminus C$ with $C$ countable open. So $Y$ will be an open subspace. We ensure that $Y$ has the co-countable topology by specifying local bases at the other points.


At $\langle\omega\_1,n\rangle$ (with $n\in\omega\_0$) the basic neighbourhoods are of the form $\{\langle\omega\_1,n\rangle\}\cup(Y\setminus C)$ with $C$ countable.


At $\langle\omega\_1,\omega\_0\rangle$ the basic neighbourhoods are of the form $\{\{\omega\_1\}\times(n,\omega\_0]\cup(Y\setminus C)$ with $n\in\omega\_0$ and $C$ countable.


If $A$ is a countable subset of $X$ then $A\cap Y$ is closed in $X$ and so if $A\cap Y$ is infinite then $A$ cannot form a convergent sequence under any enumeration. Hence if $A$ is an infinite convergent sequence then $A\cap Y$ must be finite and upon deleting that finite set $A\subseteq \{\omega\_1\}\times(\omega\_0+1)$; the latter has the natural topology of $\omega\_0+1$ so $A$ converges to $\omega\_0$ (and no other point).


Finally: every nonempty open set is co-countable, and so every nonempty open set is dense and that makes $X$ extremally disconnected.


