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I am trying to prove this. I did not find in any book.

I was making some exercises where I had to prove that if $A$ is closed and $x\not\in A$ then $d(x,A) >0.$ Because of the condition "being closed" I asked myself: $d(x,A) = 0 \Leftrightarrow x \in \overline{A}?$

It is true? Any hints to prove? Any references?

Thanks!

L.F. Cavenaghi
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1 Answers1

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Sketch:

Suppose $d(x,A)=r>0$, then the open ball $B_x(r/2)$ does int intersect $A$, so $x$ is not in the closure of $A$.

Suppose $d(x,A)=0$. Then for all $n$, we can can find $a_n\in A$ such that $d(x,a_n)<\frac{1}{n}$. For any open set $U$ containing $x$, there is a ball $B_x(r)\subseteq U$. Let $n>\frac{1}{r}$, then $a_n\in U$, so $U\cap A\not=\emptyset$. Therefore, $x$ is in the closure of $A$.

Michael Burr
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