Stone-Čech boundary is not extremally disconnected

Question

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 ... ?

Answer 1

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.

Answer 2

For an explicit pair of disjoint open sets with intersecting closures work on the binary tree $2^{<\omega}$ of finite sequences of $0$s and $1$s. For every $x\in2^\omega$ let $A_x=\{x\mathbin{\upharpoonright}n:n\in\omega\}$, the branch that consists of the initial segments of $x$. Let $Q$ be the (countable) set of $q\in2^\omega$ that are eventually constant, and let $P=2^\omega\setminus Q$ be its complement. Let $U=\bigcup_{q\in Q}A_q^*$ and $V=\bigcup_{p\in P}A_p^*$. ($A^*$ is short for $\operatorname{cl}(A)\setminus A$.) Then $U$ and $V$ are open and disjoint in the boundary. If their closures were disjoint then there would be a subset $C$ of $2^{<\omega}$ such that $V\subseteq C^*$ and $C^*\cap U=\emptyset$. However, an application of the Baire Category theorem will yield many $q\in Q$ for which $A_q^*\cap C^*\neq\emptyset$.

Answer 3

The question has already been answered satisfactorily, but I think the following presentation, which is implicit in the second part of YCor's answer, will clarify things. The fact that $\beta\mathbb{N} \setminus \mathbb{N}$ isn't extremally disconnected follows from three observations which are each interesting in their own right:

1. The Boolean algebra $\operatorname{Clop}(X)$ of clopen subsets of an extremally disconnected space $X$ is complete. Indeed, the infinite join of a family $(F_i)_{i\in I}$ of clopen sets is explicitly given by $\bigvee_{i\in I} F_i = \operatorname{closure}(\bigcup_{i\in I} F_i)$ (that the space is extremally disconnected ensures that this closure is, in fact, clopen, and it is clearly the smallest clopen set containing all the $F_i$). Actually, we have a sort of converse: the Stone space of a complete Boolean algebra is extremally disconnected (such spaces are called “Stonean”).

2. The Boolean algebra $\operatorname{Clop}(\mathbb{N}^*)$ of clopen subsets of $\mathbb{N}^* := \beta\mathbb{N} \setminus \mathbb{N}$ is that $\mathscr{P}(\mathbb{N})/\textrm{fin}$ of subsets of $\mathbb{N}$ modulo finite differences. (In fact, $\mathbb{N}^*$ is the Stone space of $\mathscr{P}(\mathbb{N})/\textrm{fin}$.) This is explained in a comment to YCor's answer for the particular situation at hand; but more generally: if $X$ is a Stone space, and $Y\subseteq X$ is closed, then $\operatorname{Clop}(Y) = \{F\cap Y : F\in \operatorname{Clop}(X)\}$: see Monk & Bonnet eds. (Koppelberg), Handbook of Boolean Algebras (1989), chap. 3, lemma 7.6(b); so $F \mapsto F\cap Y$ defines a surjective homomorphism of Boolean algebras $\operatorname{Clop}(X) \to \operatorname{Clop}(Y)$, so that $\operatorname{Clop}(Y)$ can be seen as the quotient of $\operatorname{Clop}(X)$ by the equivalence relation of “coinciding on $Y$”, and in the special case of $X=\beta\mathbb{N}$ and $Y=\mathbb{N}^*$ it is clear that the kernel of $\mathscr{P}(\mathbb{N}) \to \operatorname{Clop}(\mathbb{N}^*)$ consists of finite subsets of $\mathbb{N}$ because they are the ones that belong to no free ultrafilter. (Side note: analogously, the ring $C^*(Y)$ of bounded continuous real-valued functions on $Y$ consists of restrictions to $Y$ of the functions in $C^*(X)$ (see, e.g., Gilman & Jerrison, Rings of Continuous Functions, problem 3D), so that here $C^*(\mathbb{N}^*)$ consists of bounded sequences of real numbers up to finite differences. The Boolean algebras under discussion are the Boolean algebras of idempotents of these rings of bounded continuous real-valued functions.)

3. The Boolean algebra $\mathscr{P}(\mathbb{N})/\textrm{fin}$ of subsets of $\mathbb{N}$ modulo finite differences is not complete. Here I can only repeat YCor's argument: if $\mathbb{N}$ is partitioned into countably many infinite sets $S_n$, and if $Z$ represents their putative upper bound $[Z]$ in $\mathscr{P}(\mathbb{N})/\textrm{fin}$, then we can pick an element $z_n \in Z \cap S_n$ (because each $S_n \setminus Z$ is finite) for each $n$ and let $Z' := Z \setminus \{z_n : n\in\mathbb{N}\}$: then $[Z'] < [Z]$ in $\mathscr{P}(\mathbb{N})/\textrm{fin}$ yet $[S_n] \leq [Z']$ for each $n$ since each $S_n \setminus Z'$ is finite, and this contradicts the assumption that $[Z]$ is the upper bound of the $S_n$. (Here I write $[P] \in \mathscr{P}(\mathbb{N})/\textrm{fin}$ for the class mod finite sets of an element $P \in \mathscr{P}(\mathbb{N})$.)

Putting these three facts together we get the desired conclusion, and, again, the proof is precisely that given in the second part of YCor's answer, but I think the above helps clarify what is going on.

Answer 4

A Banach space $X$ is said to be $\lambda$-injective provided for all Banach spaces $Y$, all subspaces $W$ of $Y$, and all bounded linear operators $T$ from $W$ into $X$, there is an extension $\tilde{T}$ of $T$ to a bounded linear operator from $Y$ into $X$ so that $\|\tilde{T}\| \le \lambda \|T\|$. The relevance of this definition to the OP's question is that the Kelley-Goodner-Nachbin Theorem states that a Banach space $X$ is $1$-injective if and only if $X$ is isometrically isomorphic to $C(K)$ for some compact extremally disconnected Hausdorff space $K$. In functional analytic terms, the OP is asking whether $\ell_\infty/c_0$ is $1$-injective. Actually, $\ell_\infty/c_0$ is not $\lambda$-injective for any $\lambda < \infty$. $c_0(2^{\aleph_0})$ embeds isomorphically into $\ell_\infty/c_0$ (e.g., Yves' answer implies that), and H.P. Rosenthal proved in 1970 that if a bounded linear operator from $\ell_\infty(2^{\aleph_0})$ into a Banach space $X$ acts isomorphically on $c_0(2^{\aleph_0})$, then it acts isomorphically on some subspace of $\ell_\infty(2^{\aleph_0})$ that is isomorphic to $\ell_\infty(2^{\aleph_0})$. But $\ell_\infty(2^{\aleph_0})$ is not isomorphic to any subspace of $\ell_\infty/c_0$ because the former space has density character strictly larger than the continuum.