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Show that a subset of $\mathbb{R}$ is closed iff it contains all its accumulation points.

Well, the definition of accumulation point for a set S is that I have is that for all $\epsilon>0$, $B_\epsilon(x)\cap S\neq \varnothing$. Also the definition of an open set is that for all $\epsilon>0$, $B_\epsilon(x)\subset S$. So it follows that the closed set contains all the acc. points, just not sure how to formalize.

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    @JonasMeyer: It's a bit of an abuse of the function notation (or maybe just wrong). I was trying to say that some sort of structure should be shown with $\mathbb{R}$ in the space where the dot was placed just so the community knows whether the OP is dealing in a metric space or a strictly topological space. Sorry for the confusion/bad notation.2012-01-24

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Suppose $C$ is closed (that is, it's complement is open). Let $x$ be an accumulation point of $C$. If $x\notin C$, then $x\in\mathbb{R}-C$, so there exists $\epsilon\gt 0$ such that $B_{\epsilon}(x)\subseteq \mathbb{R}-C$. This means that $B_{\epsilon}(x)\cap C=\cdots$, hence ...

Conversely, suppose that $C$ is a set that contains all its accumulation points. We need to show that $\mathbb{R}-C$ is open. Let $x\in\mathbb{R}-C$; then $x$ is not an accumulation point of $C$, so therefore (negating the definition), it follows that ....