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Recall the definition of basically disconnected:

A space is basically disconnected if every cozero-set has an open closure.

There exists a Basically disconnected space which is not extremally disconnected; the one-point Lindelöfization of an uncountable discrete space is such a space.

But with what conditions basically disconnectedness does imply extremally disconnectedness?

Thanks.

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TXC
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1 Answers1

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Completely regular plus basically disconnected gives extremely disconnectedness, since then all open sets are cozero sets. From the counterexamples given, it sounds like second countability and some separtion axiom may be a requirement (which would imply being normal), so I'm not sure that you can get anything sharper than that while being as general.

Edit: I realized I was wrong in the comments. I wanted to find conditions for all open sets to be cozero sets; however, if there are functions $f,g$ that are zero precisely on $A$ and $B$, then $f/(f+g)$ gives a function that is $0$ precisely on $A$ and 1 precisely on $B$, so we have to have normality. Also, such sets must be $G_{\delta}$'s, since they are the intersection of preimages of open sets in the interval. So a space must in fact be $T_6$ for all open sets to be cozero sets, so my approach won't work.

Brian Rushton
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  • In particular, the counterexample is regular and Lindelof, although it is not first-countable. So perhaps regularity and first-countability is enough, but I'm not sure. – Brian Rushton Jun 01 '13 at 01:01
  • Every $T_6$ basically disconnected space is extremally disconnected (since by vedenissoff's theorem, Every open set in $T_6$ space is cozero-set), But $T_6$ is a very stronger than i need. – TXC Jun 01 '13 at 03:39
  • Right, but completely regular is only $T_{3.5}$. – Brian Rushton Jun 01 '13 at 03:41
  • As Brian said in comments, the one-point Lindelöfization of an uncountable discrete space is $T_4$ basically disconnected wich is not extremally disconnected, Therefore i think completely regularity is not enough, But perhaps some cardinal functions (like tightness or $ko$-tightness) are useful. – TXC Jun 01 '13 at 03:55
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    He may use a different definition than I do; in a completely regular space, every closed set is a zero set...ah, I see the problem; you need something to be sharply completely regular, meaning you can separate closed sets from any point by a function that is 0 precisely on the closed set. I believe this is possible iff every closed set is a $G_{delta}$ and your space is completely regular. – Brian Rushton Jun 01 '13 at 04:06
  • @Brian: You need more. For a $T_1$ space $X$, all closed sets being zero-sets is equivalent to $X$ being perfectly normal. The Alexandroff double circle is hereditarily $T_4$ but not perfectly normal. – Brian M. Scott Jun 02 '13 at 04:48
  • @BrianM.Scott & BrianRushton: Thank you for taking time to the question :). – TXC Jun 02 '13 at 06:28