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This is part of the proof that proximities on a topological space $X$ are order-isomorphic to compactifications of $X$. See wikipedia for definition of proximity space. Here I am defining proximity spaces using relation $<<$ instead of $\delta$, which are related to each other by $A << B \iff \lnot (A\delta (X\setminus B))$. My proximity spaces are separated i.e. $\{x\} << X\setminus \{y\}$ for $x\neq y$. The topology on proximity space $X$ is given by $O(X) = \{U\subseteq X : x\in U\implies \{x\} << U\}$. If $A\subseteq X$ then $\text{int}(A) = \{x\in A : \{x\} << A\}$. Every proximity space is Tychonoff. This is because if $A << B$ then $A$ and $X\setminus B$ are completely separated. Conversely if $X$ is Tychonoff and $Z$ is any compactification of $X$, then there is unique proximity $\lhd$ on $Z$, inducing a proximity $<<$ on $X$ by $A << B$ iff $A\lhd B\cup Z\setminus X$. For example $A << B$ iff $A$ and $X\setminus B$ are completely separated defines a proximity space corresponding to the Stone-Cech compactification $\beta X$ of $X$.

A $p$-filter on a proximity space $(X, <<)$ is a filter $\mathcal{F}$ such that if $B\in \mathcal{F}$ then there is $A\in \mathcal{F}$ with $A << B$. A $p$-ultrafilter is a maximal $p$-filter. Let $Y$ be the set of all $p$-ultrafilters on $X$. For $A\subseteq X$ define $o(A) = \{\mathcal{U}\in Y : A\in\mathcal{U}\}$. The topology on $Y$ is that given by $\{o(A) : A\subseteq X\}$ as a basis. Equivalently it can be given by proximity $\lhd$ on $Y$ defined as $C\lhd D$ iff there exist $A, B\subseteq X$ such that $A << B$, $C\subseteq o(A)$ and $Y\setminus D\subseteq o(X\setminus B)$. Then the map $\lambda:X\to Y$ sending $x\in X$ to the neighbourhood filter of $x$ is a dense embedding into $Y$ and so we can treat $X$ as a dense subspace of $Y$. Then $A << B$ iff $A\lhd B\cup Y\setminus X$ and so the proximity $<<$ on $X$ is induced from the proximity $\lhd$ on $Y$.

I want to show that $Y$ is compact. This is how my book wants to go about it. Since $Y$ is a proximity space, it's regular Hausdorff, and so it suffices to show that $Y$ is an $H$-closed space. To see that $Y$ is $H$-closed take a $O(Y)$-ultrafilter $\mathcal{U}$ (where $O(Y)$ is the lattice of open sets). Then it suffices to show that $\bigcap\{\overline{U} : U\in\mathcal{U}\}\neq\emptyset$. Then let $\mathcal{V}$ be the filter generated by $\{U\cap X : U\in\mathcal{U}\}$. Now they claim that $\mathcal{V}$ is a $p$-filter. So given $U\in\mathcal{U}$, the claim is that there exists $V\in\mathcal{U}$ with $V\cap X << U\cap X$. I'm not sure how am I supposed to show this from the information I've gathered so far.

Q. Why is $\mathcal{V}$ a $p$-filter?

Jakobian
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It doesn't have to be a $p$-filter. For example let $Y = X = [0, 1]$ and let $\mathcal{U} = \mathcal{V}$ be a $O(Y)$-ultrafilter extending $\{(0, 1/n) : n\in\mathbb{N}\}$. If $A << B$ and $A\in \mathcal{V}$ then $\text{cl}(A)\subseteq B$ so that $0\in B$, and so $\mathcal{V}$ is not a $p$-filter since if $B = (0, 1)\in\mathcal{V}$ there is no $A\in\mathcal{V}$ with $A << B$.

Instead consider $$\tilde{\mathcal{V}} = \{A\subseteq X : V << A\text{ for some }V\in\mathcal{V}\} = \{A\subseteq X : U\cap X << A\text{ for some }U\in\mathcal{U}\},$$ the largest $p$-filter contained in $\mathcal{V}$, and extend it to a $p$-ultrafilter $\mathcal{W}$. Let $A << B$ with $A\in\mathcal{W}$. Since $\mathcal{U}$ is a $O(Y)$-ultrafilter, either $o(B)\in \mathcal{U}$ or $Y\setminus \text{cl}_Y o(B)\in \mathcal{U}$. If it were $Y\setminus \text{cl}_Y o(B)\in\mathcal{U}$, then since $o(B)\cap X = \text{int}_X B$ and $A << \text{int}_X B$, we have $(Y\setminus \text{cl}_Y o(B))\cap X = X\setminus \text{cl}_X \text{int}_X B << X\setminus A$ so that $X\setminus A\in \mathcal{W}$, contradiction. So we have $o(B)\in \mathcal{U}$. Since any basic neighbourhood of $\mathcal{W}$ in $Y$ is of the form $o(B)$ for some $B\in\mathcal{W}$, it follows that $\mathcal{U}$ converges to $\mathcal{W}$. So $Y$ is $H$-closed, and hence compact.

Jakobian
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