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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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Your definition of "open set" is incorrect. The correct definition is that for all $x\in S$ there exists $\epsilon\gt 0$ such that $B_{\epsilon}\subseteq S$. Importantly, $x$ must be in $S$, and we only require the existence of some $\epsilon$. – Arturo Magidin Jan 24 '12 at 4:44
Since this is the definition in some contexts, in order for us to help you show that it is an "alternative definition" we have to know what your definition is. We could guess, but should not have to. Do you define a set to be closed if its complement is open (and define open according to Arturo's correction)? – Jonas Meyer Jan 24 '12 at 4:47
right -- $A\in\mathbb{R}$ is closed if $A^{c}$ is open – Emir Jan 24 '12 at 4:48
The following question is close to a duplicate:…. Although taken at face value it is a different question, "closed" there is defined in terms of your "alternative definition", and so it proves the equivalence from the other perspective. (It would be better to include your definition in your question, which you can edit.) – Jonas Meyer Jan 24 '12 at 4:50
@Emir: You mean $A\subseteq \mathbb R$ rather than $A\in \mathbb R$. – Jonas Meyer Jan 24 '12 at 4:56

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 ....

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