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 $\lim \; d(a,a_n)=d(a,A)$


$(\Rightarrow)$ 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?

  • 1
    $\begingroup$ There is a problem with $\Longleftarrow$ part. The sequence $\{a_n\}$ doesn't have a limit in general. $\endgroup$ – Norbert Jul 16 '12 at 21:13
  • $\begingroup$ @Norbert What do you mean? $\endgroup$ – Pedro Tamaroff Jul 16 '12 at 21:17
  • $\begingroup$ You can't speak about $\lim a_n$ in equality $d(x,a_n)=d(x,\lim a_n)$. This limit may not exist $\endgroup$ – Norbert Jul 16 '12 at 21:19
  • 2
    $\begingroup$ 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. $\endgroup$ – Michael Greinecker Jul 16 '12 at 21:22
  • 1
    $\begingroup$ @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 $x$. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Pedro Tamaroff Jul 16 '12 at 21:37
  • $\begingroup$ That was a typo and should have been an $\epsilon$. $\endgroup$ – Michael Greinecker Jul 16 '12 at 21:41
  • $\begingroup$ 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. $\endgroup$ – Pedro Tamaroff Jul 16 '12 at 21:43
  • $\begingroup$ 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). $\endgroup$ – wildildildlife Jul 16 '12 at 21:46
  • 1
    $\begingroup$ @wildildildlife: This might be an aesthetic tendency of those of us who prefer not to ivoke the axio of choice without necessity. $\endgroup$ – Michael Greinecker Jul 16 '12 at 21:52

Your proof is perfectly fine.

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 $d(x,a_n)\to 0$...

  • $\begingroup$ And what should $\bar A$ be? $\endgroup$ – Pedro Tamaroff Jul 16 '12 at 21:41
  • $\begingroup$ @PeterTamaroff: I am sorry, $\overline{A}$ means the closure of $A$. $\endgroup$ – wildildildlife Jul 16 '12 at 21:43
  • $\begingroup$ Oh. That is something I am not introduced to yet. =/ $\endgroup$ – Pedro Tamaroff Jul 16 '12 at 21:46
  • $\begingroup$ @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. $\endgroup$ – wildildildlife Jul 16 '12 at 21:47
  • $\begingroup$ So I can just say "...there is a sequence of points in $A$ such that $\lim\;d(x,a_n)=0$, which means $\lim\;a_n=x$." $\endgroup$ – Pedro Tamaroff Jul 16 '12 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.