# Proving $\overline{S}=S'\cup S$

As the topic, proving $\overline{S}=S'\cup S$. I think i know the definition of closure and derived sets and those def of limit points and adherent point but still don't know how to prove the equality.

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There are several different (though equivalent) definitions of the closure of a set; which one are you using? –  Brian M. Scott Sep 22 '12 at 10:48
closure of a set means the set of all adherent points of a set –  Mathematics Sep 22 '12 at 10:54
I’ll get you started. You have to prove two things: $\operatorname{cl}S\subseteq S\cup S'$, and $S\cup S'\subseteq\operatorname{cl}S$.
For the first, suppose that $x\in\operatorname{cl}S$. There are two possibilities: either $x\in S$, or $x\notin S$. If $x\in S$, then certainly $x\in S\cup S'$. What if $x\notin S$? We know that $x$ is an adherent point of $S$, so every open set $U$ containing $x$ contains at least one point of $S$, and that point can’t be $x$, because $x\notin S$; so what kind of point must $x$ be?
The second is even easier. If $x\in S\cup S'$, then either $x\in S$, or $x\in S'$. Take each case separately. If $x\in S$, is $x$ necessarily an adherent point of $S$? Why? If $x\in S'$, is $x$ necessarily an adherent point of $S$? Why?