Vtyjkyuk

Let $X$ be a profinite space.(i.e. $X$ is the inverse limit of finite discrete space.) Suppose $X$ is 2nd countable. Then fix any open equivalence relation $R$ on $X$. Pick $xin X$. Then $Rxsubset X$ is a finite union of basic open sets.

$textbfQ:$ Why $Rx$ equivalence class is a finite union of basic open sets?

$textbfQ':$ It follows that there are only countably many open equivalence relation on $X$. What cardinality arithmetic is used here? I knew $X$ is compact and there are only finite covering required. How do I know there is a countably many open relations?

Ref. Profinite Groups Luis Ribes, Chpt 1

Q has been answered by the above comments. We know that there are only finitely many equivalence classes, each of these being the finite union of basic open sets.

$X$ has a countable base $mathcalB$, therefore the set of all finite sequences in $mathcalB$ is also countable and we conclude that the set $mathcalB^ast$ of all finite unions of basic open sets must be countable. Hence the set $mathcalB^ast ast$ of all finite sequences in $mathcalB^ast$ is countable. The set of open equivalence relations on $X$ can be identified with a subset of $mathcalB^ast ast$. This answers Q'.