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Space $X$ is basically disconnected if every cozero-set has an open closure.

Every extremally disconnected space is basically disconnected But i think the converse fails. The one-point compactification $\alpha D(\tau)$ of an uncountable discrete space is a counterexample?

Eric Wofsey
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TXC
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3 Answers3

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That doesn’t quite work: if $C$ is a countably infinite subset of $D(\tau)$, $C$ is a cozero-set in $\alpha D(\tau)$ whose closure isn’t open.

Take $X$ to be the one-point Lindelöfization of an uncountable discrete space, with non-isolated point $p$: nbhds of $p$ are the sets $X\setminus C$ for countable $C\subseteq(X\setminus\{p\})$. $X$ is a $P$-space, meaning that $G_\delta$-sets in $X$ are open, so every zero-set in $X$ is open, and therefore every cozero-set in $X$ is closed, and $X$ is basically disconnected.

Now let $X\setminus\{p\}=U\cup V$, where $U$ and $V$ are uncountable and disjoint. $U$ and $V$ are open, but $\operatorname{cl}U=U\cup\{p\}$ and $\operatorname{cl}V=V\cup\{p\}$ are not disjoint, so $X$ is not extremally disconnected.

Brian M. Scott
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  • $\omega_1+1$ does'nt work as a counterexample too? one-point Lindelöfization of an uncountable discrete space is a moscow space? – TXC May 06 '13 at 16:51
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    @TXC: No, my $X$ is not Moscow: the closure of the $U$ in the last paragraph is not a union of $G_\delta$-sets, because every $G_\delta$ that contains $p$ is co-countable. $\omega_1+1$ isn’t basically disconnected. – Brian M. Scott May 06 '13 at 17:10
  • This is the first time that i learned about Lindelöfization, please recommend me some references. $X$ is a tychonoff space? – TXC May 08 '13 at 04:58
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    @TXC: $X$ is even hereditarily $T_4$. I don’t know of any references for Lindelöfization; it isn’t very useful in general. Someone once asked here about it, and this was my answer. – Brian M. Scott May 08 '13 at 13:03
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One can use Stone duality to construct many counterexamples. It is not to hard to show that a Boolean algebra $B$ is complete if and only if the Stone space $S(B)$ is extremally disconnected. Furthermore, a Boolean algebra $B$ is $\sigma$-complete (recall that a Boolean algebra is $\sigma$-complete iff every countable subset has a least upper bound) if and only if the Stone space $S(B)$ is basically disconnected. Therefore, if $B$ is a $\sigma$-complete but not complete, then $S(B)$ is basically disconnected but not extremally disconnected. Moreover, a zero-dimensional space is extremally disconnected if and only if the Boolean algebra of clopen sets is a complete Boolean algebra. One can easily show that your proposed counterexample is not a true counterexample by looking at that Boolean algebra of all clopen sets.

quid
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Joseph Van Name
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you can find examples in the Gillman and Jerison's book: Ring of continuous functions

Lei Li
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