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.

  • 1
    $\begingroup$ 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

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 $\Sigma_0$.

  • $\begingroup$ Are X and S mistakenly placed some places here? $\endgroup$ – Evgeny Kuznetsov Mar 2 '17 at 13:43
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.