# $d(x,A)=0\iff$ every neighborhood of $X$ contains a point of $A$

Mendelson, Introduction to Topology, p.52

$(8)$. Let $A$ be a non-empty subset of a metric space $(X,d)$. Let $x\in X$. Prove that $d(x,A)=0$ if, and only if, every nieghborhood $V$ of $x$ contains a point of $A$.

DEFINITION Given a subset $A$ of a metric space $X$, and $x\in X$, the distance of $x$ to $A$ is defined as:

$$d(x,A)=\inf\{d(x,a):a\in A\}$$

COROLLARY 5.9 Let $(X,d)$ be a metric space, $a\in X$ and $A$ a non-empty subset of $X$. Then there is a sequence $\{a_n\}$ of points of $A$ such that $\lim \; d(a,a_n)=d(a,A)$

PROOF

$(\Rightarrow)$ Suppose every neighborhood of $x$ contains a point of $A$. We must prove that $\inf\{d(x,a):a\in A\}=0$ But since every neighborhood of $x$ contains a point of $A$, then there is a sequence of points $\{a_n\}$ of $A$ such that $\lim \;a_n=x$. It follows that $\lim \; d(x,a_n)=0$, and since $\{a_n\}\subset A$, $\inf\{d(x,a):a\in A\}=0$ since $d(x,a)\geq 0$ for any $a,x$.

$(\Leftarrow)$ Suppose $d(x,A)=0$. It follows by 5.9 that there is a sequence of points $\{a_n\}$ in $A$ such that $\lim \; d(x,a_n)=0$. But given a point $a\in X$, the function $f:X\to \Bbb R\;/\;f(x)=d(x,a)$ is continuous (just take $\epsilon =\delta$). Thus $\lim \; d(x,a_n)= d(x,\lim \;a_n)=0$. But $d(x,a)=0\iff x=a$, so $\lim \;a_n=x$. This means that for any neighborhood $V$ of $x$ there exists an $N$ such that $a_n \in V$ whenever $n>N$, so every neighborhood of $x$ contains some $a\in A$.

Is this alright? Is there any gap or circularity I'm missing?

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There is a problem with $\Longleftarrow$ part. The sequence $\{a_n\}$ doesn't have a limit in general. – Norbert Jul 16 '12 at 21:13
@Norbert What do you mean? – Pedro Tamaroff Jul 16 '12 at 21:17
You can't speak about $\lim a_n$ in equality $d(x,a_n)=d(x,\lim a_n)$. This limit may not exist – Norbert Jul 16 '12 at 21:19
It exists in your case since $\lim d(x,a_n)=0$ is equivalent to $\lim a_n=x$. But take $x=1/2$ and $(a_n)=(1,0,10,1)$. Under the usual metric, $\lim d(x,a_n)=1/2$, but $\lim a_n$ does not exist. – Michael Greinecker Jul 16 '12 at 21:22
@PeterTamaroff You don't know a priori that the limit exists. But you can argue directly that $\lim d(x,a_n)=0$ implies that $\lim a_n$ exists and its equal to $x$. – azarel Jul 16 '12 at 21:26

You can get a cleaner proof when you avoid sequences altogether:

Suppose that every neighborhood of $x$ contains a point of $A$. Obviously, $d(x,A)\geq 0$. Now let $\epsilon>0$. Then there exists by assumption $a\in A$ with $d(x,a)<\epsilon$. Since $\epsilon$ was arbitrary, $d(x,A)=0$.

Suppose that $d(x,A)=0$. Let $\epsilon>0$. By assumption, there is $a\in A$ with $d(x,a)<\epsilon$. Hence, every open $\epsilon$-ball around $x$ contain an element of $A$. Since every neighborhood of $x$ is a superset of such a ball, we are done.

Afficionados might note that one avoids having to make arbitrary choices in the sequence-free proof.

Edit: On your own proof. The original proof is correct except for your argument that $\lim d(x,a_n)=0$ implies $\lim a_n=x$. The result is true, but there is a subtle flaw in the argument. You seem to use the sequence characteriation of continuity there, so that $f$ is continuous at $y$ if $f(y_n)$ converges to $f(y)$ whenever $y_n$ converges to $y$. To apply that in your case, you need to assume that the sequence $(a_n)$ is convergent. It is, but that is something you have to prove first.

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How can it even be possible that $d(x,a)<0$? I understand the second part of your proof though. I'd like if you could correct my proof, too, which is in fact why I included it in the post. – Pedro Tamaroff Jul 16 '12 at 21:37
That was a typo and should have been an $\epsilon$. – Michael Greinecker Jul 16 '12 at 21:41
Oh, sure. Could you comment or expand a little on my proof? I'm mostly on my own on this, so understanding my own mistakes is something I could really use. – Pedro Tamaroff Jul 16 '12 at 21:43
It's a matter of taste, but I don't agree that this is necessarily cleaner than using sequences. Basically, it is the difference between working directly with neighborhoods, or in the language of sequences (or more generally nets). – wildildildlife Jul 16 '12 at 21:46
@wildildildlife: This might be an aesthetic tendency of those of us who prefer not to ivoke the axio of choice without necessity. – Michael Greinecker Jul 16 '12 at 21:52

Basically, you have proved $d(x,A)=0$ iff $x\in \overline{A}$ (i.e. every neighborhood of $x$ meets $A$), by using the intermediate equivalence between $x\in \overline{A}$ and the existence of $a_n\in A$ with $a_n\to x$.
As I mentioned in the comments, there is no need to introduce your function $f$, since $d(x,a_n)\to 0$ is equivalent to $\lim a_n=x$. Depending on your definition, this is either a tautology or a triviality. In any case I don't understand the objections in the comments about $\lim a_n$ which might not exist...of course it does, since $d(x,a_n)\to 0$...
And what should $\bar A$ be? – Pedro Tamaroff Jul 16 '12 at 21:41
@PeterTamaroff: I am sorry, $\overline{A}$ means the closure of $A$. – wildildildlife Jul 16 '12 at 21:43
@PeterTamaroff: don't worry, usually $\overline{A}$ is defined as $x\in \overline{A}$ iff every neighborhood of $x$ meets $A$, so this is just another way to say what Mendelson asked you to prove. – wildildildlife Jul 16 '12 at 21:47
So I can just say "...there is a sequence of points in $A$ such that $\lim\;d(x,a_n)=0$, which means $\lim\;a_n=x$." – Pedro Tamaroff Jul 16 '12 at 22:05