$\DeclareMathOperator\cl{cl}\DeclareMathOperator\int{int}$A subset $A$ of a topological space $X$ is called regular closed if $A=\cl _{X}\int_{X}A$.

The family of all regular closed sets of a topological space is denoted by $% \mathcal{R}\left( X\right) $.

An ultrafilter $\mathcal{U}$ on $\mathcal{R}\left( X\right) $ is said to converge to a point $p\in \beta X$ if $\left\{ p\right\} =\bigcap \left\{ \cl_{\beta X}U:U\in \mathcal{U}\right\} $.

Lemma: Let $D$ be a dense subspace of a space $X$. Then the map $% A\rightarrow \cl_{X}A$ is a Boolean algebra isomorphism from $\mathcal{R}% \left( D\right) $ onto $\mathcal{R}\left( X\right) $.

I think that the family $\mathcal{F}=\left\{ F\in \mathcal{R}\left( \beta X\right) :p\in \int_{\beta X}F\right\} $ is a filterbasis in $\mathcal{R}% \left( \beta X\right) $. Therefore $\mathcal{F}$ can be imbedded in an ultrafilter $\mathcal{U}$ in $\mathcal{R}\left( \beta X\right) $. Therefore $% X\cap \mathcal{U}=\left\{ X\cap U:U\in \mathcal{U}\right\} $ is an ultrafilter in $\mathcal{R}\left( X\right) $, and it converges to $p$.

My question is: for every $p\in \beta X$, does there exist a unique ultrafilter $% \mathcal{U}$ in $\mathcal{R}\left( X\right) $ such that $\left\{ p\right\} =\bigcap \left\{ cl_{\beta X}U:U\in \mathcal{U}\right\} $, that is $% \mathcal{U}$ converges to $p$?

Now, let $f:X\longrightarrow Y$ be a continuous map between Tychonoff spaces. Then the Stone extension $\beta f:\beta X\longrightarrow \beta Y$ is defined as follows: for $p\in \beta X$, there exists a unique $z$-ultrafilter $\mathcal{% A}^{p}$ on $X$ with $p$, so is defined by $\left( \beta f\right) \left( p\right) =\bigcap f^{\#}\mathcal{A}^{p}$, where $f^{\#}\mathcal{A}% ^{p}=\left\{ E\in Z\left( Y\right) :f^{-1}\left( E\right) \in \mathcal{A}% ^{p}\right\} $ (Gillman and Jerison, Rings of continuous functions, p.85). In another article (K. Srivastava. On the Stone–Čech compactification of an orbit space), it is defined by $\left( \beta f\right) \left( p\right) =\bigcap_{Z\in \mathcal{A}^{p}}\cl_{\beta Y}f\left( Z\right) $. I guess that's not quite right.

If, for every $p\in \beta X$, there exists a unique ultrafilter of regular closed sets of $X$, then how can I define $\left( \beta f\right) \left( p\right) $?

  • 3
    $\begingroup$ For the first question the answer is no. If $\beta X$ is compact Hausdorff, the Stone space of ultrafilters of $\mathscr R(X)$ is the Gleason cover $\gamma\beta X$ of $\beta X$, and the limit map $\gamma\beta X\to\beta X$ is bijective iff $\beta X$ is extremally disconnected. $\endgroup$ Aug 7 at 7:50
  • $\begingroup$ For the second, if this condition is satisfied, then $\gamma\beta X\approx\beta X$, so you can just define $\beta X\approx\gamma\beta X\to\gamma\beta Y\to\beta Y$, using that both $\beta$ and $\gamma$ are functorial. $\endgroup$ Aug 7 at 8:00
  • $\begingroup$ TeX note: \left and \righting all pairs of parentheses is harmless in terms of sizing if the contents are small enough (though it can be really weird looking when the contents are asymmetric about the midline: $\displaystyle\left(\sum_n n^{-s}\right)$), but it does weird things to the spacing. Compare $\mathcal R(X)$ \mathcal R(X) vs. $\mathcal R\left(X\right)$ \mathcal R\left(X\right). $\endgroup$
    – LSpice
    Aug 10 at 19:42
  • 2
    $\begingroup$ I thinks this other question and its references and answer are highly relevant to the present one. $\endgroup$
    – Gro-Tsen
    Aug 10 at 20:48

1 Answer 1


The family $\mathcal{F}=\{ F\in \mathcal{R}( \beta X) :p\in \operatorname{int}_{\beta X}F\}$ is indeed a filterbase; it is a base for the neighbourhood filter at $p$. As noted in the comments there need not be a unique ultrafilter that extends it; for example in $\beta\mathbb{R}$. Take a point $p$ in $\mathbb{R}$ then $\mathcal{F}$ is generated by $\{[p-2^{-n},p+2^{-n}]:n\in\mathbb{N}\}$; you can add $(-\infty,p]$ or $[p,\infty)$ to $\mathcal{F}$ and still have a filter base in $\mathcal{R}$. So we have at least two $\mathcal{R}$-ultrafilters that extend $\mathcal{F}$.

Srivastava's definition is identical to the one in Gillman and Jerison and hence correct: note that if $Z\in\mathcal{A}^p$ and $E\in f^\#\mathcal{A}^p$ then $Z\cap f^{-1}[E]\neq\emptyset$, hence $f[Z]\cap E\neq\emptyset$ as well. The latter implies that $\operatorname{cl}_{\beta Y}f[Z]\cap f^\#\mathcal{A}^p\neq \emptyset$ too. So $\beta f(p)\in\bigcap\{\operatorname{cl}_{\beta Y}f[Z]:z\in\mathcal{A}^p\}$. Next if $q\neq \beta f(p)$ then take $h:\beta Y\to[0,1]$ with $h(q)=1$ and $h(\beta f(p))=0$. Then $h\circ f$ is continuous and $Z=\{x:h(f(x))\le\frac12\}$ is a member of $\mathcal{A}^p$, but $q$ is not in the closure of $f[Z]$, hence $\bigcap\{\operatorname{cl}_{\beta Y}f[Z]:z\in\mathcal{A}^p\}=\{\beta f(p)\}$.

As an answer to the last question: there is already a definition, independent of the uniqueness of the $\mathcal{R}$-ultrafilters.


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