I want to prove that if a topological space $(X,\tau)$ is $T_1$, compact and extremally disconnected, then it's 0-dimensional.
My attempt of proof:
Let $x\in X$, and let $U_x$ be an open neighborhood of $x$.
Since $X$ is extremally disconnected, $\overline{U_x}$ is still open, and it's compact since it's a closed subset of a compact space.
Let $\mathcal{B}$ be a base for $X$. Then $\overline{U_x}=\bigcup\limits_{i\in I} V_i$ for $V_i\in \mathcal{B}$, and we can refine this open cover to a finite subcover of $\overline{U_x}=\bigcup\limits_{i\in \overline{I}}V_i$. Then $ \bigcup\limits_{i\in \overline{I}}\overline{V}_i$ is a covering of $U_x$ made of clopen sets, and $\overline{\mathcal{B}}=\{\overline{V}\hspace{1mm}|\hspace{1mm} V\in\mathcal{B}\}$ is a base for $X$.
I think that something's wrong with this. First of all, I didn't use the fact that $X$ is $T_1$, and I don't know if this is actually necessary. Second of all, is it necessary to refine the cover of $\overline{U_x}$ to a finite one?
Thank you in advance.
Best Answer
If you define extremal disconnectness and compactness without T2, then the statement is false: The cofinite topology is then extremally disconnected (and connected), T1, compact, but not zero-dimensional.
In order to have a zero-dimensional space, you need that the space is regular. You can arrange this, if you require in your definition of extremal disconnectness or of compactness that the space is T2. Since then, by compactness, the space is regular.
On the other hand, compactness is much too strong for the statement: An extremally disconnected, regular space is already zero-dimensional:
Let $x \in U \subset X$, $U$ open in $X$. By regularity, pick an open $V$ such that $x \in V \subset \overline{V} \subset U$. Since $\overline{V}$ is clopen, we are done.