Stone-Čech boundary is not extremally disconnected 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 ... ?
 A: 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.

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.
