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Suppose $X$ is a zero-dimensional completely regular space (clopen sets form a base) such that the Boolean algebra of clopen sets is a $\sigma$-complete Boolean algebra. Must $X$ be basically disconnected? That is, must every cozero set in $X$ have open closure?

I thought this was known, but I can't find it. The point is to show that such $X$ is strongly zero-dimensional, i.e., $\beta(X)$ is zero-dimensional, so that every cozero set is a countable union of clopens.

It should be noted that the analogous statement for extremally disconnected spaces is true: A zero-dimensional space is extremally disconnected (open sets have open closures) if and only if the Boolean algebra of all its clopen sets is a complete Boolean algebra. The proof is an easy exercise.

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  • $\begingroup$ Have you looked on M.H. Stone, Boundedness properties in function-lattices, Canadian Math. Soc. 1 (1949), 176-186? cms.math.ca/cjm/v1/cjm1949v01.0176-0186.pdf $\endgroup$ Commented Mar 11, 2013 at 11:03
  • $\begingroup$ Yes. I did not see an answer to my question in that article. I could have missed somethig, however. $\endgroup$ Commented Mar 13, 2013 at 6:20
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    $\begingroup$ Problem 6K in the book Extensions and Absolutes of Hausdorff Spaces by Porter and Woods claims that a Tychonoff space $X$ is Basically disconnected if and only if $X$ is zero-dimensional and the Boolean algebra of clopen sets is $\sigma$-complete. Unfortunately, I have not yet solved this problem, and problems in books are only authoritative if you solve them (i.e. I do not believe this problem yet). $\endgroup$ Commented Sep 29, 2013 at 20:20
  • $\begingroup$ I also found that the paper Frolik's theorem for basically disconnected spaces by Johannes Vermeer it is also claimed that the basically disconnected spaces are precisely the zero-dimensional spaces $X$ where the Boolean algebra of all clopen sets is $\sigma$-complete as a Boolean algebra. I however have not seen a proof yet, so I do not believe this claim. $\endgroup$ Commented Jan 29, 2015 at 2:32
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    $\begingroup$ Vermeer's paper assumes compact spaces, in which the claim is true. See p. 136, line 3. $\endgroup$ Commented Jan 30, 2015 at 18:10

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The answer to the posted question is probably no. Thanks to the above comment by @Joseph Van Name, the claim "yes" is posed as problem 6K in the Porter-Woods book. Grant Woods has told me in a recent email that this matter came to his attention some time ago, and he concluded that the problem statement in the book should have stated "strongly zero-dimensional" in the assumption.

A complete answer to the posted question will require a counterexample. This may not be so well-known. This will require a zero-dimensional space which is not strongly zero-dimensional, whose Boolean algebra of all clopen sets is $\sigma$-complete. Perhaps a modification of the Dowker construction would work (see Porter-Woods Problem 4V, or Gillman-Jerison Problem 16M, or Engelking Example 6.2.20).

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