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Let $(X,d)$ be a metric space and $S \subset X$. Show that $d_S(x):=\text{inf}\{d(x,s): s \in S\}=0 \Leftrightarrow x \in \overline S .$

Notes: $\overline S$ is the closure of S. Maybe you can use that a closed set is also closed for sequences in the set? I think the difficult part is when S is open, otherwise its trivial as the closure would be equal to S.

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    What is your definition of closure?, because that is the definition in some books2010-11-27
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    You should make it clear what you have tried. Have you managed to do either direction? If not, have you written out the definitions of the terms involved? Do you understand them (particularly the closure of a set)?2010-11-27
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    @user3123: please try again. 1) The right hand side of your equation does not contain $S$. 2) Please edit this into the question itself.2010-11-27
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    @Listing your definition for closure is not correct. $\overline{S}:\{x\in X: \forall~ e>0,B_e(x)\cap S\neq 0\}$.2015-10-20
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    @ShinningStar you are right, I deleted my incorrect comment as the definition given in your comment is the intended one.2015-11-06

2 Answers 2

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If $x\notin \overline{S}$, then there exists $r>0$ such that $B_r(x)\cap S=\phi$. It follows that $d(x,s)\ge r>0$ for all $s\in S$ and hence $d_S(x)>0$. And these steps can be reversed, i.e. the steps above are actually "if and only if". So you get the proof.

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    Thank you, very straight-forward proof2010-11-27
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    FYI: instead of `\phi` ($\phi$), you want `\emptyset` $(\emptyset)$. Unfortunately Math.SE doesn't have `\varnothing` which in my opinion looks nicer than `\emptyset`2010-11-27
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If $x \in \overline{S}$, then there is a sequence $(s_n)$ in $S$ converging to $x$ (e.g. take $s_n$ to be some element in $S \cap B(x;\tfrac{1}{n})$). Then $d_S(x) \leq \inf\lbrace d(x,s_n) \;\vert\; n \in \mathbb{N}\rbrace = 0$ (why?).

For the other directtion, we prove the contrapositive: If $x \notin \overline{S}$, then there is some ball around $x$ disjoint from $S$. This gives that the infimum in $d_S(x)$ must be greater than $0$ (why?).

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    Is it a problem that $\frac{1}{n}$ doesn't converge in every metric space?2010-11-27
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    No. $B(x;r)$ is the ball around $x$ with radius $r$. What did you think it were?2010-11-27