How to show $d(x,A)=0$ iff $x$ is in the closure of $A$?

Question

This is a problem form Topology by Munkres:

Let $X$ be a metric space with metric $d$ and $A$ is a nonempty subset of $X$. Show that $d(x,A)=0$ if and only if $x$ is in the closure of $A$.

I think this problem is quite easy to understand emotionally but I don't know how to express the proof in standard math language. Thanks in advance!

See In a metric (X,d), prove that for each subset A, x $\in$ $\bar{A}$ if and only if d(x,A)=0 or If $x$ is not in $A$, a closed set in a metric space then $d(x,A)>0$ — Martin Sleziak, Jun 10 '13 at 14:27
@MartinSleziak thanks! I guess there are many questions asked before but How can I find the answer I want in the past problem? any skills in searching?thanks! — Brain Zhang, Jun 10 '13 at 14:53
I found this one by putting metric "d(x,A)=0" closure site:math.stackexchange.com into Google. You should also always check the list of related questions, which is automatically generated while you are writing a question. Some tips for searching can be found here and in other posts on meta tagged [meta-tag:search]. — Martin Sleziak, Jun 10 '13 at 15:08

score 17 · Accepted Answer · answered Jun 10 '13 at 14:24

17

We prove the result by equivalence: $x\in cl(A)\iff \forall \epsilon>0\ B(x,\epsilon)\cap A\not=\emptyset\iff\forall\epsilon>0\ \exists a\in A: d(x,a)<\epsilon\iff \inf_{a\in A}d(x,a)=0\iff d(x,A)=0$

answered Jun 10 '13 at 14:24

very succinct!! – Brain Zhang Jun 10 '13 at 14:31
Succinct, indeed! – amWhy Jun 06 '14 at 16:33

How to show $d(x,A)=0$ iff $x$ is in the closure of $A$?

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