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Let $U$ be a regular open set in a Tychonoff space $X$ (regular means that it is an interior of a closed set).

[ In my specific situation $U$ is of the form $\operatorname{int} f^{-1}(0)$, where $f$ is a continuous real-valued function on $X$, and $X$ is a Baire space (a sequence of dense open sets has a dense intersection), but I am not sure if it helps. ]

Is there a sequence $\{A_n\}_{n\in\mathbb{N}}$ of closed (in $X$) subsets of $U$ such that $\bigcup_{n\in\mathbb{N}} A_n$ is dense in $U$?

Of course, this is the case if $X$ is perfectly normal (which is equivalent to every open set being $F_{\sigma}$), or separable, but I hope a less restrictive assumption will suffice, e.g. normality.

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  • $\begingroup$ I'd be curious in the case when $X=\beta\mathbf{N}-\mathbf{N}$ (Stone-Cech remainder) and $U$ is the interior of the intersection over $n$ of the set $F_n$ of ultrafilters containing $2^n\mathbf{N}$. $\endgroup$
    – YCor
    Sep 14, 2021 at 8:36
  • $\begingroup$ A $P$-space is a regular space where every $G_{\delta}$-set is open. You can produce any $P$-space from a regular space simply by declaring every union of $G_{\delta}$-sets to be open (this is the $P$-space coreflection). If $X$ is a $P$-space, then whenever $U$ is open and $F$ is an $F_{\sigma}$-set where $F\subseteq U$ and $F$ is dense in $U$, we know that $F$ is closed and therefore $U=F$, so $U$ must be clopen. However, in a $P$-space, if $U=\text{int}(f^{-1}[\{0\}])$ for continuous real-valued $f$, then $U$ must already be clopen. $\endgroup$ Sep 14, 2021 at 11:31
  • $\begingroup$ @JosephVanName so this is an example of a sufficient condition, because every zero set must be clopen, right? When I started reading your comment, i thought it will be a counterexample $\endgroup$
    – erz
    Sep 14, 2021 at 15:55
  • $\begingroup$ @erz. Yes. That is a counterexample. In a $P$-space, an open set $U$ has a dense $F_{\sigma}$-subset if and only if $U$ is clopen. However, in a $P$-space the clopen sets are precisely the sets of the form $f^{-1}[\{0\}]$ for some continuous real-valued function. Therefore, if $X$ is a non-extremally disconnected $P$-space, then there is a regular open set $U$ that is not clopen and that set is a counterexample (and every $P$-space of cardinality below the first measurable cardinal is non-extremally disconnected). $\endgroup$ Sep 14, 2021 at 17:13
  • $\begingroup$ @JosephVanName Ok, got it, thank you. I was confused, because in my specific situation $U$ is the interior of a zero-set, which in P-space has to be clopen $\endgroup$
    – erz
    Sep 14, 2021 at 17:18

1 Answer 1

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Not in $\beta\mathbb{N}\setminus\mathbb{N}$: if $A$ is an $F_\sigma$-subset of $\operatorname{int}f^{-1}(0)$ then there is even a clopen set $C$ such that $A\subseteq C\subseteq \operatorname{int}f^{-1}(0)$. Of course this is only an example if the zero-set is not clopen but you can get an example by working on the countable set $N=\mathbb{N}^2$ the clopen sets determined by the vertical lines $V_n=\{n\}\times\mathbb{N}$ union up to a cozero set: let $f$ have value $2^{-n}$ on $V_n$. That cozero set is not clopen and for every closed-set $F$ contained in $\operatorname{int}f^{-1}(0)$ there is a function $h:\mathbb{N}\to\mathbb{N}$ such that $F$ is in the clopen set determined by $L_h=\{(m,n):n\le h(m)\}$. If $A$ is an $F_\sigma$ then we get a sequence $\langle h_n:n\in\omega\rangle$ of such functions. Define $h(m)=1+\max\{h_n(m):n\le m\}$. Then the clopen set determined by $L_h$ contains $A$ and is a subset of $\operatorname{int}f^{-1}(0)$.

We consider the behaviour of the extension of $f$ to $\beta N$ and its restriction to $N^*=\beta N\setminus N$ (all called $f$). The important thing to note is that if $X$ determines a clopen set, denoted $X^*$, in $N^*$ then $X^*\subseteq f^{-1}(0)$ iff $X\cap V_n$ is finite for all $n$. In general $X^*\cap Y^*=\emptyset$ iff $X\cap Y$ is finite, and in this case if $X\cap V_n$ is finite for all $n$ then $f[X^*]=\{0\}$. Furthermore: $X\cap V_n$ is finite for all $n$ iff $X\subseteq L_h$ for some $h$ as above. Finally: $N^*$ is compact and zerodimensional, so if $F$ is closed and contained in $\operatorname{int}f^{-1}(0)$ then there is a clopen set $C$ such that $F\subseteq C\subseteq \operatorname{int}f^{-1}(0)$.

See this note for a short introduction.

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  • $\begingroup$ Sorry, but i have many questions: how do you know that $f^{-1}(0)$ is not clopen? How do you know that $h$ exists for any $F$? Why is the last set contained in $\in f^{-1}(0)$? $\endgroup$
    – erz
    Sep 14, 2021 at 17:29
  • $\begingroup$ I added a few more remarks and a reference for more information. $\endgroup$
    – KP Hart
    Sep 15, 2021 at 7:35

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