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This question already has an answer here:

Definitions:

$A'$ is the set of all accumulation or limit points.

$\bar{A} = A \cup A'$ - this is known as the closure of $A$.

Let $A$ be a subset of $\mathbb{R}$. A point $p\in\mathbb{R}$ is an accumulation point of $A$ if and only if every open set $G$ containing $p$ contains a point of $A$ different from $p$.

Prove or disprove: $(\overline{A\bigcup B}) = \overline{A}\bigcup \overline{B}$

proof: let $p\in (\overline{A\bigcup B})$ be an accumulation of the union, then there exists some open set $S_{p}$ containing $p$ which also contains a point of $(\overline{A\bigcup B})$ different from $p$. Let's call this point $q$, where $p\neq q$ such that $q\in S_{p} \subset (\overline{A\bigcup B})$.

This is where I am lost, if anyone can help that would be greatly appreciated.

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marked as duplicate by Jonas Meyer, Eric Stucky, hardmath, Lord_Farin, N. F. Taussig Mar 6 '15 at 23:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @user149418 No. The linked question uses the notion of adherent point, which is different from limit point. $\endgroup$ – egreg Mar 6 '15 at 21:58
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Because $C \subseteq D$ implies $C' \subseteq D'$ we clearly have $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$. For the other implication: suppose $p$ is an accumulation point of $A \cup B$ but not of $A$ nor of $B$. Derive a contradiction.

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