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If $A$ is a closed in a metric space $(X,d)$ with $x\notin A$, I need to show that $d(x,A)>0$. Now assume $d(x,A)=0$ then $\exists x_n\in A $ s.t.$d(x_n,A)=0$ then there is a sequence in $A$ s.t. $x_n$ converges to $x$ in $A$ since $A$ is closed in an Metric space thus it is compact, therefore the limit of the sequence is inside the compact set $A$.

Ok here are my thoughts, but are fuzzy.. I need help. Thank you

Martin Sleziak
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Klara
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3 Answers3

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If $A$ is closed then $A^c$ is open. So if $x\not\in A$, there is some $\epsilon > 0 $ so that $B_\epsilon(x)\subseteq A^c$. Hence, $d(x, A) \ge \epsilon$.

ncmathsadist
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You can indeed do it with sequences if you want, but not the way you did.

If $d(x,A)=0$, by definition of an inf, we can find a sequence $a_n\in A$ such that $d(x,a_n)\leq 1/n$ for all $n$.

Then $a_n$ converges to $x$, so $x$ belongs to $\overline{A}$.

Since $A$ is closed, we have $x\in A$, which concludes the contrapositive.

Note that the same argument and its straighforward converse show more generally that $$ \overline{A}=\{x\in X\;;\; d(x,A)=0\}. $$

Julien
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It is known that $\overline{A}=\{x\in X:d(x,A)= 0\}$(See for instance Chapter 8 of Schaums Outline Series in General Topology). Because $A$ is closed, $\overline{A}=A$. If $x\notin A$, then $x\notin \overline{A}$ so that $d(x, A)>0$.

Juniven Acapulco
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  • I guess the OP wouldn't be asking about this if that was his characterization of the closure of a set. In fact, OP's question aims to prove that if $A$ is closed $$x\in A\iff d(x,A)=0$$ which is basically your claim. Note one direction is immediate, while the other is the OP's question. – Pedro Feb 28 '13 at 01:51
  • @Peter Tamaroff, Ah okay. Thanks for the comment.:) – Juniven Acapulco Feb 28 '13 at 01:55