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The Corollary 1.1.13 in my text book is the following statement.

$\textbf{Corollary 1.1.13}$ A profinite space $X$ is second countable if and only if

$$X\cong \varprojlim_{i\in I}X_i$$

where $(I,\leq)$ is a countable totally ordered set and each $X_i$ is finite discrete space.

And the proof starts as follows:

Proof

Suppose $X$ is profinite and second countable. Consider the set $\mathcal{R}$ of all open equivalence relations on $X$. For $R\in \mathcal{R}$, $xR$ is a finite union of basic open set. Hence $\mathcal{R}$ is countable.

But I can't understand why is $xR$ a finite union of basic open set and why does it imply $\mathcal{R}$ countable.

Please tell me if you know.

Hanul Jeon
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Combalge
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  • The equivalence classes of any one of your equivalence relations form an open cover of $X$, right? Then use compactness (in the form of profiniteness) to get a finite subcover with basic open sets. So you can identify each $R$ with a finite subset of your basis. There are only countably many finite subsets of a countable set, so $\mathcal R$ is countable. – Forever Mozart Feb 07 '15 at 10:30
  • How can I identify each $R$ with a finite subset of the basis? I think sometimes some $xR$ cannot be made by finite union. Is this impossible? – Combalge Feb 07 '15 at 12:24

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