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I'm trying to prove that the set of all limit points $S'$ of $S$ in a topological space $X$ is closed.

So assume $x$ is not a limit point, i.e. there is a point $x \not\in S'$ such that an open neighborhood of $x$, $U \cap S = x$ or $\emptyset$. Now I have to show that $U \cap S' = \emptyset$, so that I can claim that $U \subset X \setminus S'$, and so $X \setminus S'$ is open.

So how do I prove that $U \cap S' = \emptyset$? Because it doesn't seem like it's necessarily true. Take some point $y \in U \cap S'$. Then $y$ is a limit point of $S$, and there exists some open neighborhood $V$ of $y$ such that $V \cap S \neq \emptyset$. This doesn't contradict anything if $U \cap V = \{y\}$ and $y \not\in S$, so it seems like $U \cap S' \neq \emptyset$.

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If $y\in U\cap S'$, where $y\neq x$, then $y$ is a limit point of $S$ and $U\ \setminus \{x\}$ is an open neighborhood of $y$, and so $U\ \setminus \{x\} \cap S \neq \emptyset$, where it impossible, thus $U\cap S'= \emptyset$. Hint. If $X$ is not satisfying the $T_1$ axiom, then it dosen't imply that $S'$ is a closed set.

The condition that finite point sets be closed: is called $T_1$ axiom.

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Suppose that $y\in U$. Recall that $U\cap S$ is either $\varnothing$ or $\{x\}$. If $U\cap S$ is empty, then $U$ is an open nbhd of $y$ disjoint from $S$, so $y\notin S'$. If $U\cap S=\{x\}$, then either $y=x$, in which case $y\notin S'$, or $U\setminus\{x\}$ is an open nbhd of $y$ disjoint from $S$, and again $y\notin S'$. Thus, no point of $U$ belongs to $S'$.

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