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In my recent research, I defined a topological space $X$ to be $EZ$-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 $cl_{X}{A}=cl_{X}({\bigcup_{\alpha\in S}A_{\alpha}})$.

Also I defined $X$ to be $EB$-space if 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 zerodimensinal space is an $EZ$-space. and any basically disconnected space is an $EB$-space.

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

edited Jan 18 at 12:19

Emil Jeřábek
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asked Jan 18 at 12:14

Ali
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Dear emil, thank you very much. – Ali Jan 18 at 12:21

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Am I the only one to notice that EZ-space sounds a lot like easy-space? – Joseph Van Name Jan 18 at 17:24

Nik Weaver · Accepted Answer · 2013-01-18 12:27:20Z UTC

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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.

answered Jan 18 at 12:27

Nik Weaver
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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. – Ali Jan 18 at 12:59

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 $cl(\bigcup_{\alpha}coz_{\alpha}=\bigcup_{\alpha}cl coz$. – Ali Jan 18 at 13:05

Please dont negative vote for question. This is a god question. – Ali Jan 18 at 13:07

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$\def\cl{\operatorname{cl}}$@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. – Emil Jeřábek Jan 18 at 13:34

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@Ali: I stand by my answer. – Nik Weaver Jan 18 at 13:52

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