Prove that the set of all isolated points of $A$ is countable Let $X$ be a second countable topology space. Prove that the set of all isolated points of $A$ is countable ($A$ is an arbitrary subset of X).
I tried to relate the base which is countable to set of isolated points but I didn't find solution. Can you help me ?

Yes, let $x ∈ X$ be an isolated point. Then $\{x\} ⊆ X$ is open, that is: an open neighbourhood of $x$. So for any topological base of $X$, you’ll find in that base a nonempty neighbourhood of $x$ within $\{x\}$, right? Meaning …?
So by $X$ having a countable topological base, you’ll find …?
