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.

  • $\begingroup$ What is your definition of closure?, because that is the definition in some books $\endgroup$ – Bryan Yocks Nov 27 '10 at 16:31
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    $\begingroup$ 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)? $\endgroup$ – Alex B. Nov 27 '10 at 16:33
  • $\begingroup$ @user3123: please try again. 1) The right hand side of your equation does not contain $S$. 2) Please edit this into the question itself. $\endgroup$ – Pete L. Clark Nov 27 '10 at 19:12
  • $\begingroup$ @Listing your definition for closure is not correct. $\overline{S}:\{x\in X: \forall~ e>0,B_e(x)\cap S\neq 0\}$. $\endgroup$ – Shinning Star Oct 20 '15 at 9:25
  • $\begingroup$ @ShinningStar you are right, I deleted my incorrect comment as the definition given in your comment is the intended one. $\endgroup$ – Listing Nov 6 '15 at 21:23

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.

  • $\begingroup$ Thank you, very straight-forward proof $\endgroup$ – Listing Nov 27 '10 at 16:47
  • $\begingroup$ FYI: instead of \phi ($\phi$), you want \emptyset $(\emptyset)$. Unfortunately Math.SE doesn't have \varnothing which in my opinion looks nicer than \emptyset $\endgroup$ – kahen Nov 27 '10 at 17:41

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?).

  • $\begingroup$ Is it a problem that $\frac{1}{n}$ doesn't converge in every metric space? $\endgroup$ – Listing Nov 27 '10 at 16:44
  • $\begingroup$ No. $B(x;r)$ is the ball around $x$ with radius $r$. What did you think it were? $\endgroup$ – kahen Nov 27 '10 at 17:40

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