Recall that a topological space is called extremally disconnected if the closure of every open subset is still open. Every discrete space is of course extremally disconnected, and the standard non-trivial examples are the Stone-Čech compactifications $\beta X$ of discrete spaces $X$.


In a paper I have seen now the claim that if $X$ is infinite and discrete, then the Stone-Čech boundary $\partial\_\beta X := \beta X \setminus X$ is never extremally disconnected. But I have a hard time figuring out why ... ?



Stone-Čech boundary is not extremally disconnected
We can suppose $X=\omega$. Let $(X\_i)\_{i\in I}$ be a continuum family of infinite subsets of $\omega$ with pairwise finite intersection. Define $Y\_i=\bar{X\_i}-X\_i$. So the $Y\_i$ are pairwise disjoint non-empty clopen subsets in $\beta\omega-\omega$.


For $J\subset I$, define $Y\_J$ as the closure of $\bigcup\_{j\in J}Y\_j$. So $Y\_J\cap Y\_i=\emptyset$ for each $i\notin J$.


If by contradiction $\beta\omega-\omega$ is extremally disconnected, $Y\_J$ is clopen for each $J$. These are pairwise distinct, by the previous paragraph. So we obtain $2^c$ distinct clopen subsets in $\beta\omega-\omega$. But there a surjection from $2^\omega$ to the set of clopen subsets of $\beta\omega-\omega$ (namely $Y\mapsto\bar{Y}-Y$), so we get a surjection from $2^\omega$ onto some set in which $2^c$ injects. Contradiction.




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Added: Here's a possibly more intuitive and explicit answer. Write $A\subset^\* B$ to mean $A-B$ is finite. Write $X=\bigsqcup X\_n$ with each $X\_n$ infinite. Define $Y\_n=\bar{X\_n}-X\_n$. Then $Y\_n$ is clopen, so $\bigcup\_n Y\_n$ is open. I claim that $\bar{Y}$ is not open.


Otherwise, it's clopen, hence has the form $\bar{Z}-Z$ for some subset $Z$ of $X$. Then $Z$ has the following property [which expresses being the sup of $(X\_n)$ in the Boolean algebra $2^X/\mathrm{fin}$]: $Y\_n\subset^\* Z$ is finite, and for every $Z'\subset X$ having the property that $Y\_n\subset^\* Z'$ for every $n$, we have $Z\subset^\* Z'$. But now choose $z\_n\in Z\cap Y\_n$ and define $Z'=Z-\{z\_n:n\in\omega\}$. Then $Y\_n\subset^\* Z'$ for all $n$, but $Z-Z'$ is infinite, contradiction.


