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$A$ is a set of points in a metric space $M$ and $B$ is the set of all accumulations points of $A$. Prove that $B$ is closed.

Aim: Prove that $B^{\complement}$ is open.

Let $y \in B^{\complement}$. Then $y$ is not an accumulation point of $A$. Then there exists a $\epsilon>0$ such that the neighbourhood $N(y,\epsilon)$ does not contain a point of $A$ distinct from $y$. Hence, $N(y,\epsilon) \subset B^{\complement}$. This implies that $B^{\complement}$ is the union of a family of neighbourhood. Hence, $B^{\complement}$ is open. Thus, $B$ is closed.

Is my proof correct?

Idonknow
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  • Yes, it looks correct, but you can skip the last line; it is enough to show that every point x in the complement is contained in a ball lying in the complement. – DBFdalwayse Aug 24 '13 at 07:04

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It’s correct, though if you’re just beginning, an instructor might want you to say a little more about why $N(y,\epsilon)\subseteq M\setminus B$. Specifically, $B(y,\epsilon)\cap A\subseteq\{y\}$, so if $x\in B(y,\epsilon)\setminus\{y\}$, then $B\big(x,\epsilon-d(x,y)\big)$ is a nbhd of $x$ containing no point of $A$. Thus, no point of $B(y,\epsilon)$ is an accumulation point of $A$.

Brian M. Scott
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