I need to prove that the intersection of all 1/n neighborhoods of a closed set gives me the set itself. However, i have no idea how to even start this proof (to me it seems like it's false) Does anyone have any tips/hints on how to start it?
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If a point does not belong to the set, is there a number $n$ such that the point does not belong to $1/n$ neighborhood of the set too? – Alex Ravsky Sep 30 '15 at 01:07
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What do you men by the 1/n neighborhood of the set? Aren't neighborhoods defined only for points? – Iliketoproveit Sep 30 '15 at 01:20
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@Alex: $B_\epsilon(S) = \bigcup_{x\in S} B_\epsilon(x) = { y \mid \exists x \in S : |x-y| < \epsilon }$. – user251257 Sep 30 '15 at 02:04
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Suppose x is not in the set, call it E. E is closed so x is not a limit point. So there is an $\epsilon$ such that B($\epsilon$, x) has no point of E. We can find an m such that 0 < $\frac{1}{m}$< $\epsilon$. So B($\frac{1}{m}$, x) has no point of E. So for all y in E, d(x, y) > $\frac{1}{m}$. So x is not in the $\frac{1}{m}$ set of E. So x is not in the intersection of $\frac{1}{n}$ neighborhoods. So the only x that are in the intersection must also be in E.
So the intersection is a subset of E.
And clearly E is a subset of any $\frac{1}{n}$ neighborhood so E is a subset of the intersections. So E = the intersection.
fleablood
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