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In my recent research, I defined a topological space $X$ to be an $EZ$-space if for every open subset $A$ of $X$, there exists a collection $\{A_{\alpha}: \alpha\in S\}$ of clopen subsets of $X$ such that $\newcommand{\cl}{\operatorname{cl}} \cl_{X}{A} = \cl_{X}({\bigcup_{\alpha\in S}A_{\alpha}})$.

Also, I defined $X$ to be an $EB$-space if for every cozero-set $H$ of $COZ[X]$, there exists a collection $\{H_{\alpha}: \alpha\in S\}$ of clopen subsets of $X$ such that $\cl_{X}{H} = \cl_{X}({\bigcup_{\alpha\in S}H_{\alpha}})$.

For example, any zero-dimensional space is an $EZ$-space, and any basically disconnected space is an $EB$-space.

Problem: Give an example of a completely regular $EB$-space which is not an $EZ$-space.

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    $\begingroup$ Am I the only one to notice that EZ-space sounds a lot like easy-space? $\endgroup$
    – Joseph Van Name
    Commented Jan 18, 2013 at 17:24

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Not possible, in a completely regular space every open set is a union of cozero sets. Just collect the clopen sets that you use for these subordinate cozero sets.

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  • $\begingroup$ Hi Nik, you dont understand my definitins. On the other hand a cozeroset is not clopen. $Coz(f)=\{x\in X: f(x)\neq 0\}$. Please again see question. $\endgroup$
    – Ali
    Commented Jan 18, 2013 at 12:59
  • $\begingroup$ you notice in $EB$-space the closure of any open set is the closure of a union of a collection of clopen subsets. Ok, in completely regular, any open set is a union of cozerosets. and in $EZ$-space the closure of any cozerset is the closure of a union of a collection of clopen subsets. But does $cl(\bigcup_{\alpha}coz_{\alpha}=\bigcup_{\alpha}cl coz$. $\endgroup$
    – Ali
    Commented Jan 18, 2013 at 13:05
  • $\begingroup$ Please dont negative vote for question. This is a god question. $\endgroup$
    – Ali
    Commented Jan 18, 2013 at 13:07
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    $\begingroup$ $\def\cl{\operatorname{cl}}$@Ali: If $A=\bigcup_{i\in I}A_i$, and for each $i$, $\cl A_i=\cl(\bigcup_{\alpha\in S_i}A_{i,\alpha})$, then $\cl A=\cl(\bigcup_{i,\alpha}A_{i,\alpha})$. This is a trivial exercise. $\endgroup$
    – Emil Jeřábek
    Commented Jan 18, 2013 at 13:34
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    $\begingroup$ @Ali: I stand by my answer. $\endgroup$
    – Nik Weaver
    Commented Jan 18, 2013 at 13:52

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