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Based upon discussion at Math.SE

Consider the property extremally disconnected, 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 and disallows silliness like all hyperconnected spaces are extremally disconnected.

But I like silliness, so how much of the theory can we recover without Hausdorff? In this Math.SE post 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?

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    $\begingroup$ You use the abbreviation US several times, and seem to indicate what it means I think, but it isn't so clear. Can you tell us what the letters stand for? This is difficult to Google. $\endgroup$
    – Joel David Hamkins
    Aug 18 at 12:05
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    $\begingroup$ Ah, evidently it goes back to Wilansky (1967), doi.org/10.2307/2316017. But he defines US space without saying what U and S stand for. I suppose "unique" and "sequence", but it would be silly to call them unique sequence spaces, since they often have many sequences—what they have is unique limits for their convergent sequences. Do people just say US space, pronouncing the letters? I find this to be a regrettable style. $\endgroup$
    – Joel David Hamkins
    Aug 18 at 12:33
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    $\begingroup$ Oh, I had seen your parenthetical remark, but I hadn't been sure whether that was intended as a definition of the term. And to clarify: I was not objecting to your style, but rather to Wilansky's. An abbreviation that doesn't actually stand for something seems ridiculous to me, and I think mathematics is better without such jargon. Probably too late to change this terminology, though... $\endgroup$
    – Joel David Hamkins
    Aug 18 at 15:14
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    $\begingroup$ Wilansky doesn't seem to say that, although it would have been better if he had. But then, still, it is a strange abbreviation without the L. Although I suppose people also use US similarly instead of USA, dropping the A. But in that case, one would say it stands for "United States" rather than "United States of America." But I don't think we'll refer to unique sequential spaces. $\endgroup$
    – Joel David Hamkins
    Aug 18 at 22:16
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    $\begingroup$ Yeah, Wilanski should have mentioned that explicitly. Same for KC = "compacts are closed". $\endgroup$
    – PatrickR
    Aug 18 at 22:34

2 Answers 2

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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 we can assume that $A\subseteq \{\omega_1\}\times(\omega_0+1)$; 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.

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    $\begingroup$ Cool. So this is the cocountable topology on an uncountable set, with a copy of a convergent sequence which is attached so that it's closed but not open. This is a useful technique. $\endgroup$
    – Steven Clontz
    Aug 18 at 15:18
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KP answered the question, and in fact provided the answer to a stronger question: his space is strongly KC, showing that Hausdorff is quite necessary to show extremally disconnected spaces are both totally separated and sequentially discrete.

To see this, consider a countably compact subset. It must have finite intersection with the cocountable topology subspace, as otherwise it would contain an infinite closed discrete set. So that piece is closed.

Then its intersection with the converging sequence subspace is closed in that subspace, since that subspace is strongly KC. Since that subspace is closed, the intersection there is closed.

Since the union of two closed sets is closed, the arbitrary countably compact set is closed.

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