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I just want to be sure because I'm a little confused about something.

If I have a topological space $(X,\tau )$ and a subset $A\subseteq X$,

What does it mean that $A$ is separable?

I know that it means that there is a countable subset $S\subset A$ s.t $\overline{S} =A$ but I'm not sure if I need to take the closure with respect to $\tau$ or the closure with respect to the subspace topology.

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There is no distinct concept of a separable subset of a topological space. We can, however, speak about whether the subset is a separable space, in which case "... with the subspace topology" is implied.

So you should take the closure within $A$ with the subspace topology.

A bit of thought, however, should convince you that the condition in this case is the same as requiring $\overline S \supseteq A$ where the closure is taken in $X$ -- which may be conceptually easier to imagine.

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This is a good question and the answer is that the closure must be taken with respect to $\tau$, and not with respect to the subspace topology. Indeed, if $(X, \tau)$ is a topological space given any subset $A\subset X$, $A$ is both open and closed in the subspace topology $A$ inherits from $(X; \tau)$, hence the closure of $A$, in this topology, is $A$ itself.

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