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Let $A'$ denote the set of limit points of $A$ where $A$ is a subset of a metric space $X$

Is the claim in the title true? And if so, is the following proof correct (just the inclusion $\subset$)?

Let $x \in (A \cup B)'$. Then for every $B_r(x)$, there exists $y$ such that $x \ne y$ and $y \in B_r(x) \cap (A \cup B)$. Hence $y \in A$ or $y \in A$ or $y \in B$ or both. If $y \in A$, then $x \in A'$ and if $y \in B$, then $x \in B'$. Thus $x \in A' \cup B'$. So $$(A \cup B)' \subset A' \cup B'$$

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    Your proof is not very valid. If you want to show $x \in A' \cup B'$, you need to show that for a given $r$, you can always pick $y \in A \cap B_r(x)$, or always pick $y \in B \cap B_r(x)$. You cannot alternate between the two sets as $r$ varies.2012-09-10
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    It still needs a little work. For each $r>0$ you get a $y_r$ that’s in either $A$ or $B$. If they’re all in $A$, then certainly $x\in A'$, and similarly if they’re all in $B$, but some $y_r$’s may be in $A$ while others are in $B$. You have to argue a bit more to cover this possibility. It can be done with what you have, but it’s easier to suppose that $x\in(A\cup B)'$ and $x\notin A'$ and prove that this forces $x$ to be in $B\,'$.2012-09-10

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