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A Hausdorff topological space $X$ is called strongly zero-dimensional whenever for every closed subset $A$ of $X$ and every open subset $U$ of $X$ such that $A \subseteq U$, there exists a clopen subset $V$ of $X$ such that $A \subseteq V \subseteq U$.

Now, let $X'$ be a Hausdorff topological space such that for every open subset $O$ of $X'$ and every closed subset $C$ of $X'$ such that $O \subseteq C$, there exists a clopen subset $V$ of $X'$ such that $O \subseteq V \subseteq C$. In this case, we say $X'$ is a $*$-space.

It is clear that a $*$-space is strongly zero-dimensional.

I am trying to answer the following questions:

1) Is it true that a strongly zero-dimensional is $*$-space? If not, is there an additional assumption $A $ on $X'$ such that $A $ + strongly zero-dimensional implies $X'$ is $*$-space?

2) Are $*$-spaces famous? ( For example, are $*$-spaces well-known cases of topological spaces that I do not know)

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    $\begingroup$ Are you the one who asked this question earlier? Because what is called "property $*$" there is what is called "strongly $0$-dimensional" here, and what is called a $*$-space here implies what is called "extremally disconnected" there, which is insanely confusing. Where did you get this terminology? At any rate, you should check the examples from Steen & Seebach which I mentioned in the comments there, maybe one of them will help in some way, or at least, clarify the question. $\endgroup$
    – Gro-Tsen
    Commented Mar 26, 2017 at 17:27

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(Having posted this, I saw that all of it is in the comment by Gro-Tsen)

Taking, in the definition of *-space, $C=\text{closure of $O$}$ shows that closure of an open set must be clopen. This is clearly also sufficient. So *-spaces are exactly extremally disconnected Hausdorff spaces - those with closures of open sets clopen. For an example of a strongly zero-dimensional space which is not extremally disconnected, take the Stone space of any incomplete Boolean algebra, e. g. one-point compactification $S\cup\{\infty\}$ of an infinite discrete space $S$. In there, take $O\subset S$ any infinite subset with infinite complement, and $C$ its closure $O\cup\{\infty\}$. Then $C$ is not clopen (the only clopens of $S\cup\{\infty\}$ are finite subsets of $S$ and their complements), so there is obviously no clopen $V$ with $O\subseteq V\subseteq C$.

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  • $\begingroup$ It's not entirely in my comment, because I hadn't noticed that "extremally disconnected" was also sufficient for what OP calls being a $*$-space. $\endgroup$
    – Gro-Tsen
    Commented Mar 26, 2017 at 17:40
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What you call a strongly zero-dimensional space is commonly known as an ultranormal space. An ultranormal space is a Hausdorff space where for every pair of disjoint closed sets $C,D$ there is a clopen set $Z$ with $C⊆Z⊆D^{c}$. A strongly zero-dimensional space is a space whose Stone-Cech compactification is zero-dimensional. The ultranormal spaces are precisely the strongly zero-dimensional normal spaces. See my expository for more details on these discussions. So it is true that the $*$-spaces are called extremally disconnected spaces. Finally, there are extremally disconnected spaces which are Hausdorff but not normal (nor even regular) such as this one. In fact, the absolute of a normal space is always completely regular and extremally disconnected but not necessarily normal.

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  • $\begingroup$ The π-Base link does not work with me - says "Loading..." indefinitely. $\endgroup$ Commented Mar 26, 2017 at 18:50
  • $\begingroup$ The pi-base link works for me on chrome. I wonder what may be going on with the link for others? $\endgroup$ Commented Mar 28, 2017 at 4:14
  • $\begingroup$ Weird. I am on chrome too; when going to the front page, there is an announcement There is a major data cleanup and standardization effort underway on Github. Data will be in-flux while that effort is in progress. If I click any link there, it says the same Loading... without any further results $\endgroup$ Commented Mar 28, 2017 at 6:09
  • $\begingroup$ I don't understand the containment $Z \subseteq D^c \subseteq Z$ … do you mean $Z = D^c$, or is one of those $Z$'s supposed to be something else? Reading the introduction to your paper suggests that maybe you just meant $C \subseteq Z \subseteq D^c$. $\endgroup$
    – LSpice
    Commented Apr 12, 2018 at 19:14

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