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May be this question turns out trivial, but I can't figure out. I asked on StackExchange, but no one answers.

Let $S$ be a set, or equivalently the topological space with discrete topology and $2$ two point set with discrete tiopology. The $βS$ be the Stone–Čech compactification of $S$. By Tychonoff theorem the topology on $2^S$ is compact with respect to the product topology. Note that $P(S)$ the powerset of $S$ is isomorphic of $2^S$ as a set. Let $(P(S),\tau_{prod})$ be topologycal space on $P(S)$ with topology induced by product topology on $2^S$.

Now consider $X=\{(A,\Sigma)|A\in \Sigma\}\subseteq P(S)\times \beta S$. Where ultrafilter $\Sigma\in \beta S$ varies in $\beta S$. All possible $(A,\Sigma)$ s.t. $A\in P(S)$, $\Sigma\in \beta S$ and $A\in \Sigma$

My question is can $X$ be a clopen subset if $S$ is infinite?
Thanks in advance.

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    $\begingroup$ Did you mean to write $P(S)\times \beta S$ in the title? $\endgroup$ – Ramiro de la Vega Mar 2 '17 at 12:54
  • $\begingroup$ Oh sorry. I changed mistyped in a title $\endgroup$ – Evgeny Kuznetsov Mar 2 '17 at 12:59
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Your $X = \{(A,\Sigma) : A\in\Sigma\} \subseteq P(S) \times \beta S$ is never open.

Indeed, let $\Sigma_0$ be a nonprincipal ultrafilter on $S$. If $X$ were open, then in particular $\{A : A\in\Sigma_0\}$ would be open in $P(S)$ (as the inverse image of $X$ under the continuous map $P(S) \to P(S) \times \beta S$ given by $A \mapsto (A,\Sigma_0)$), in other words, $\Sigma_0$ would be open in $P(S)$. Let me assume $\Sigma_0$ open and derive a contradiction.

Let $A \in \Sigma_0$. Since $\Sigma_0$ is open, it contains a neighborhood of $A$ in $P(S)$, in other words, there exist (disjoint) finite subsets $P,Q\subseteq S$ with $P\subseteq A$ and $Q\cap A = \varnothing$ such that every $B \subseteq S$ satisfying $P\subseteq B$ and $Q\cap B = \varnothing$ also belongs to $\Sigma_0$. In particular, $P$ itself would belong to $\Sigma_0$. But an ultrafilter containing a finite set contains one of its singletons, so it is in fact principal, contradicting the assumption on $\Sigma_0$.

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  • $\begingroup$ Are X and S mistakenly placed some places here? $\endgroup$ – Evgeny Kuznetsov Mar 2 '17 at 13:43
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    $\begingroup$ @EvgenyKuznetsov Indeed, every $P(X)$ should have been $P(S)$. Fixed. $\endgroup$ – Gro-Tsen Mar 2 '17 at 14:01
  • $\begingroup$ What do you mean by $A \mapsto (A,\Sigma)$? $X=\{(A,\Sigma):A\in \Sigma\}$ for all possible ($A,\Sigma$), $A\in P(S)$ and $\Sigma \in \beta S$ $\endgroup$ – Evgeny Kuznetsov Mar 2 '17 at 14:19
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    $\begingroup$ @EvgenyKuznetsov I mean that one particular $\Sigma$ has been chosen. To make things clearer, I edited my answer so as to call it $\Sigma_0$. The set of $A$ such that $A\in\Sigma_0$ is the inverse image of $X=\{(A,\Sigma):A\in\Sigma\}$ under $A\mapsto(A,\Sigma_0)$. $\endgroup$ – Gro-Tsen Mar 2 '17 at 14:32

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