Extremally disconnected without Hausdorff

Question

In Theorem T000045 of pi-Base, a proof is given to defend the assertion from Counterexamples in Topology that all Extremally disconnected ($T_2$ where the closure of open is open) spaces are Totally separated (for every two points there is a disconnection of the entire space that separates them). A current pull request to pi-Base removes the $T_2$ requirement for the extremally disconnected property; however, does T45 still hold without such an assumption?

Answer 1

You are right that the Hausdorff assumption is needed to show that extremally disconnected implies totally separated.

If we take extremally disconnected to mean the closure of any open set is open (without Hausdorff), then the following hold.

(1) Any hyperconnected space is extremally disconnected. (Reason: any nonempty open set $U\subseteq X$ is dense in $X$, so its closure is the whole space, which is open.)

(2) Any space that is at the same time extremally disconnected and connected is hyperconnected. (Reason: Any nonempty open set $U\subseteq X$ has a nonempty clopen closure. Since the space is connected, that closure must be the whole space, that is, $U$ is dense in $X$.)

Answer 2

Some additional assumption is required for this to work. For example, a two-point indiscrete space has the property that the closure of open is open, but is connected. A particular point topology provides a connected $T_0$ example: the closure of any non-empty open set is the whole space (and thus is open).

In fact, the countable complement topology on the real numbers is a $T_1$ (in fact, strongly KC) not $T_2$ space. This space is connected, and the closure of the any non-empty open set is the whole space (and thus is open).