Equivalent characterizations of Stone-Cech compactification and generalization to Wallman-Frink compactifications

Question

In the book Normal topological spaces by Alo and Shapiro the following theorem is present:

Theorem. If $X$ is a dense subspace of Tychonoff space $Y$ then the following are equivalent:

1. Every continuous function $f:X\to Z$ where $Z$ is compact Hausdorff has continuous extension $\tilde{f}:Y\to Z$.
2. $X$ is $C^*$-embedded in $Y$.
3. If $Z_1, Z_2$ are disjoint zero-sets of $X$, then $\overline{Z_1}$ and $\overline{Z_2}$ are disjoint.
4. If $Z_1, Z_2$ are zero-sets of $X$, then $\overline{Z_1\cap Z_2} = \overline{Z_1}\cap \overline{Z_2}$.
5. Every $y\in Y$ is a limit of unique $z$-ultrafilter on $X$ (that is, a zero-set ultrafilter).

This is a standard theorem that gives conditions for $X\subseteq Y\subseteq \beta X$, see e.g. Rings of continuous functions by Gillman and Jerison. The authors ask the reader to prove the analogue of above theorem for arbitrary normal base $\mathcal{Z}$ of $X$.

I don't know what the anaologue of this theorem would be, but I think they are asking about the equivalence of 3, 4, 5 in the sense of

3. If $Z_1, Z_2\in\mathcal{Z}$ are disjoint, then $\overline{Z_1}$ and $\overline{Z_2}$ are disjoint.
4. f $Z_1, Z_2\in\mathcal{Z}$ then $\overline{Z_1\cap Z_2} = \overline{Z_1}\cap\overline{Z_2}$.
5. Every $y\in Y$ is a limit of unique $\mathcal{Z}$-ultrafilter (that is, ultrafilter consisting of elements of $\mathcal{Z}$).

or some other analogues of thereof.

Anyhow we should obtain conditions for when $X\subseteq Y\subseteq \omega(\mathcal{Z})$ where $\omega(\mathcal{Z})$ is the Wallman-Frink compactification of $X$ with respect to the normal base $\mathcal{Z}$.

I don't know how to prove it however. For example, lets look at the proof that 3 implies 4 above. One has obvious inclusion $\overline{Z_1\cap Z_2}\subseteq \overline{Z_1}\cap \overline{Z_2}$ and so just needs to show the opposite inclusion. If $y\in \overline{Z_1}\cap \overline{Z_2}$ we can take a zero-set neighbourhood $V$ of $y$, and then $V\cap Z_1$ and $V\cap Z_2$ are zero-sets of $X$ with $y\in \overline{V\cap Z_1}\cap \overline{V\cap Z_2}$ and so $V\cap Z_1\cap Z_2\neq \emptyset$. And since $V$ was arbitrary, $y\in \overline{Z_1\cap Z_2}$.

The same procedure doesn't seem to be possible to do for the analogue, since we don't have any structure on $Y$ similar to the role that zero-sets played in above proof.

What did the authors mean by the analogue of above theorem?

Answer 1

For results of this type see the article Wallman compactifications and Wallman realcompactifications by Gagrat and Naimpally.

I wrote the following theorem based on the article A note on compactifications and semi-normal spaces by Alo and Shapiro.

Below by $\omega(\mathcal{Z})$ I denote the Wallman-Frink compactification corresponding to the normal base $\mathcal{Z}$, that is the set of all $\mathcal{Z}$-ultrafilters with topology given by taking $Z^\omega = \{\mathcal{F}\in\omega(\mathcal{Z}) : Z\in\mathcal{F}\}$ as the base of closed sets.

Theorem. Let $X\subseteq Y$ be Tychonoff, $\mathcal{Z}$ a normal base on $X$. Then the following are equivalent:

1. There is embedding $Y\hookrightarrow \omega(\mathcal{Z})$ which preserves $X$. That is, $X\subseteq Y\subseteq\omega(\mathcal{Z})$.

2. If $Z_1, Z_2\in\mathcal{Z}$ then $\text{cl}_Y(Z_1\cap Z_2) = \text{cl}_Y{Z_1}\cap\text{cl}_Y{Z_2}$ and sets of the form $\text{cl}_Y Z$, $Z\in\mathcal{Z}$ separate points from closed sets in the sense that if $y\notin F$ then there exists $Z_1, Z_2\in\mathcal{Z}$ such that $\text{cl}_Y Z_1 \cap \text{cl}_Y Z_2 = \emptyset, y\in \text{cl}_Y Z_1, F\subseteq \text{cl}_Y Z_2$.

3. If $Z_1, Z_2\in\mathcal{Z}$ are disjoint then $\text{cl}_Y Z_1 \cap \text{cl}_Y Z_2 = \emptyset$ and sets of the form $\text{cl}_Y Z$, $Z\in\mathcal{Z}$ strongly separate points from closed sets in the sense that if $y\notin F$ then there exists $Z_1, Z_2\in\mathcal{Z}$ such that $\text{cl}_Y Z_1 \cap \text{cl}_Y Z_2 = \emptyset, y\in \text{int}_Y( \text{cl}_Y Z_1), F\subseteq \text{int}_Y(\text{cl}_Y Z_2)$.

4. Every $y\in Y$ is a limit of unique $\mathcal{Z}$-ultrafilter and sets of the form $\text{cl}_Y Z$, $Z\in\mathcal{Z}$ strongly separate points from closed sets

Proof: (1$\implies$2, 3) Note that the map $Z\mapsto Z^\omega$ preserves finite intersections and unions, and that $Z^\omega = \text{cl}_{\omega(\mathcal{Z})}(Z)$, so we have $$\text{cl}_Y(Z_1\cap Z_2) = \text{cl}_{\omega(\mathcal{Z})}(Z_1\cap Z_2)\cap Y = \text{cl}_{\omega(\mathcal{Z})}Z_1\cap \text{cl}_{\omega(\mathcal{Z})}Z_2\cap Y = \text{cl}_Y Z_1\cap\text{cl}_Y Z_2.$$

Suppose $y\in Y$ and $F\subseteq Y$ is closed with $y\notin F$, $F = F_0\cap Y$ where $F_0\subseteq \omega(\mathcal{Z})$ is closed. Since $\mathcal{Z}^\omega$ is a normal base for $\omega(\mathcal{Z})$, we can find disjoint $B_1, B_2\in\mathcal{Z}$ with $y\in B_1^\omega$ and $F\subseteq B_2^\omega$. Then we can further find $A_1, A_2\in\mathcal{Z}$ with $B_i^\omega\subseteq \omega(\mathcal{Z})\setminus A_i^\omega$. Now its clear that $$F_0\subseteq \omega(\mathcal{Z})\setminus A_2^\omega \subseteq A_1^\omega\subseteq \omega(\mathcal{Z})\setminus \{y\}$$ and so by taking intersection with $Y$, $F\subseteq \text{int}_Y(\text{cl}_Y A_1)$. By taking $y\in C_1^\omega$ and $C_2 = A_1$ and again finding $D_i\in\mathcal{Z}$ such that $C_i^\omega\subseteq \omega(\mathcal{Z})\setminus D_i^\omega$ we see that $$y\in \omega(\mathcal{Z})\setminus D_1^\omega \subseteq D_2^\omega$$ and so $y\in\text{int}_Y(\text{cl}_Y D_2)$. Moreover, $\text{cl}_Y D_2\cap \text{cl}_Y A_1 = \emptyset$. So $\{\text{cl}_Y Z : Z\in\mathcal{Z}\}$ strongly separates points and closed sets of $Y$.

(2$\implies$1) For $y\in Y$ define $\mathcal{F}_y = \{Z\in\mathcal{Z} : y\in\text{cl}_Y(Z)\}$. If $Z_1, Z_2\in\mathcal{F}_y$ then $y\in \text{cl}_Y Z_1\cap \text{cl}_Y Z_2 = \text{cl}_Y(Z_1\cap Z_2)$ so that $Z_1\cap Z_2\in\mathcal{F}_y$, and so its easy to see that $\mathcal{F}_y$ is a $\mathcal{Z}$-filter. Suppose that $A\in \mathcal{Z}\setminus \mathcal{F}_y$ so that $y\notin\text{cl}_Y A$. By assumption we can find $Z\in\mathcal{Z}$ such that $y\in \text{cl}_Y Z\subseteq Y\setminus \text{cl}_Y A$, and so $Z$ and $A$ are disjoint. So for any $A\in\mathcal{Z}$, either $A\in \mathcal{F}_y$ or there exists $Z\in\mathcal{Z}$ such that $Z\subseteq Y\setminus A$, and so $\mathcal{F}_y$ is a $\mathcal{Z}$-ultrafilter.

Thus $f:Y\to\omega(\mathcal{Z})$ given by $f(y) = \mathcal{F}_y$ is a well-defined map, and clearly it sends $x\in X$ to its corresponding element $\mathcal{F}_x = \{Z\in \mathcal{Z} : x\in Z\}$ in $\omega(\mathcal{Z})$, so that $f$ preserves $X$.

Note that if $y_1\neq y_2$ then picking $Z\in\mathcal{Z}$ such that $y_1\in\text{cl}_Y Z, y_2\notin \text{cl}_Y Z$ we have $Z\in\mathcal{F}_{y_1}\setminus \mathcal{F}_{y_2}$ so that $f$ is injective.

Moreover $f^{-1}(Z^\omega) = \{y\in Y : \mathcal{F}_y\in Z^\omega\} = \{y\in Y : Z\in \mathcal{F}_y\} = \text{cl}_Y Z$, so that $f$ is continuous and $f(\text{cl}_Y Z) = f(Y)\cap Z^\omega$ so that because sets of the form $\text{cl}_Y Z, Z\in\mathcal{Z}$ form a base for closed sets of $Y$, $f$ is an embedding.

(3$\implies$2) Suppose $y\in\text{cl}_Y Z_1\cap \text{cl}_Y Z_2$ and $U$ is a neighbourhood of $y$. Take $Z\in\mathcal{Z}$ with $y\in\text{int}_Y(\text{cl}_Y Z)\subseteq U$ and define $V = \text{cl}_Y Z$. Then $y\in\text{cl}_Y(V\cap Z_i) = \text{cl}_Y (Z\cap Z_i)$ and so $(V\cap Z_1)\cap (V\cap Z_2) = V\cap Z_1\cap Z_2\neq \emptyset$, so $U\cap (Z_1\cap Z_2)\neq\emptyset$. So $y\in\text{cl}_Y (Z_1\cap Z_2)$.

(1$\implies$4) If $\mathcal{F}$ is a $\mathcal{Z}$-ultrafilter convergent to $\mathcal{F}_0\in\omega(\mathcal{Z})$ and if $\mathcal{F}_0\in \omega(\mathcal{Z})\setminus A^\omega$ then there exists $Z\in\mathcal{F}$ with $Z\subseteq \omega(\mathcal{Z})\setminus A^\omega$. That is if $x\in Z$ then $A\notin \mathcal{F}_x$ i.e. $x\notin A$. In other words $Z\subseteq X\setminus A$. So $A\notin \mathcal{F}$. Also, $\mathcal{F}_0\in \omega(\mathcal{Z})\setminus A^\omega$ is equivalent to $A\notin \mathcal{F}_0$. So $\mathcal{F}\subseteq \mathcal{F}_0$ which implies $\mathcal{F} = \mathcal{F}_0$ since those are $\mathcal{Z}$-ultrafilters. So if $\mathcal{F}_0\in\omega(\mathcal{Z})$ then $\mathcal{F}_0$ is the unique $\mathcal{Z}$-ultrafilter convergent to $\mathcal{F}_0$ in $\omega(\mathcal{Z})$.

(4$\implies$3) Suppose that $\mathcal{F}_y$ is the unique $\mathcal{Z}$-ultrafilter convergent to $y$. If $y\in \text{int}_Y(\text{cl}_Y Z_i)$ for $i = 1, 2$, find $Z\in\mathcal{Z}$ with $y\in\text{int}_Y(\text{cl}_Y Z)$ and $\text{cl}_Y Z\subseteq \text{int}_Y(\text{cl}_Y Z_1)\cap \text{int}_Y(\text{cl}_Y Z_2)$. Then $Z\subseteq Z_1\cap Z_2$ so that $$y\in \text{int}_Y(\text{cl}_Y Z)\subseteq \text{int}_Y(\text{cl}_Y(Z_1\cap Z_2))$$ and so $\mathcal{G} = \{Z\in\mathcal{Z} : y\in\text{int}_Y(\text{cl}_Y Z)\}$ is a $\mathcal{Z}$-filter. If $y\in\overline{Z}$ then $\{Z\}\cup\mathcal{G}$ extends to a $\mathcal{Z}$-ultrafilter convergent to $y$, and hence $Z\in\mathcal{F}_y$. It follows that if $\text{cl}_Y Z_1\cap \text{cl}_Y Z_2\neq\emptyset$ then $Z_1\cap Z_2\neq\emptyset$ since by picking any $y\in \text{cl}_Y Z_1\cap \text{cl}_Y Z_2$ one has $Z_1\cap Z_2\in\mathcal{F}_y$. $\square$

The separation of points and closed sets in the case of Wallman-Frink compactifications replaces the role of density for Stone-Cech compactification, which just happens to be equivalent in the particular case of zero-sets. Indeed:

Note that if $\mathcal{Z} = Z(X)$ is the family of zero-sets on $X$, and $X$ is dense in $Y$, then sets of the form $\text{cl}_Y Z$ where $Z\in Z(X)$ strongly separate points and closed sets. Indeed, if $y\notin F$ where $F$ is closed, take disjoint zero-set neighbourhoods $A_1, A_2$ in $Y$ of $y$ and $F$ respectively and let $Z_i = A_i\cap X\in Z(X)$. Then since $\text{cl}_Y (A_i\cap X) \supseteq \text{cl}_Y(\text{int}_Y(A_i)\cap X) = \text{cl}_Y(\text{int}_Y(A_i))$ one can see that if $y\in \text{int}_Y(A_1)\subseteq \text{cl}_Y Z_1$ and similarly $F\subseteq \text{int}_Y(A_2)\subseteq \text{cl}_Y Z_2$ with $\text{cl}_Y Z_1\cap \text{cl}_Y Z_2 = \emptyset$.

And conversely, if sets of the form $\text{cl}_Y Z$ with $Z\in Z(X)$ separate points and closed sets in $Y$, then for any non-empty open set $U$ of $Y$, if $y\in U$ then there is $Z\in Z(X)$ with $y\in \text{cl}_Y Z\subseteq U$ and so $X\cap U\neq \emptyset$.

So for the case of $\mathcal{Z} = Z(X)$ we recover:

Corollary. Let $X\subseteq Y$ be Tychonoff. Then the following are equivalent:

1. There is embedding $Y\hookrightarrow \beta X$ which preserves $X$. That is, $X\subseteq Y\subseteq\beta X$.

2. If $Z_1, Z_2$ are zero-sets then $\text{cl}_Y(Z_1\cap Z_2) = \text{cl}_Y{Z_1}\cap\text{cl}_Y{Z_2}$ and $X$ is dense in $Y$.

3. If $Z_1, Z_2$ are disjoint zero-sets then $\text{cl}_Y Z_1\cap \text{cl}_Y Z_2 = \emptyset$ and $X$ is dense in $Y$.

4. Every $y\in Y$ is a limit of unique $z$-ultrafilter on $X$.