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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* $\epsilon$.2012-01-24
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    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)?2012-01-24
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    right -- $A\in\mathbb{R}$ is closed if $A^{c}$ is open2012-01-24
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    The following question is close to a duplicate: http://math.stackexchange.com/questions/30039/a-subset-g-of-mathbbrn-is-open-iff-the-complement-of-g-is-closed. 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.)2012-01-24
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    @Emir: You mean $A\subseteq \mathbb R$ rather than $A\in \mathbb R$.2012-01-24
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    It helps if you specify the topological structure of $(\mathbb{R},\cdot)$ in the question.2012-01-24
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    @HenryShearman: There is a standard topology on $\mathbb R$ that may generally be assumed unless another is specified, and with Arturo's correction we already have a definition of open. What is the "$\cdot$" in your comment?2012-01-24
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    Yes, the standard topology on $\mathbb{R}$ is what would be assumed. I was more implying that it should have been stated that it was a metric space. The balls make that obvious here though.2012-01-24
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    @HenryShearman: I don't understand what you mean by "$(\mathbb R,\cdot)$".2012-01-24
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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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