Stone–Čech compactification and an ultrafilter of regular closed sets

Question

$\DeclareMathOperator\cl{cl}\DeclareMathOperator\int{int}$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) $?

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.