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

Question

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?

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
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
@Norbert I'm writing $\lim ;d(x,a_n)= d(x,\lim ; a_n)$. Why wouldn't it exist? — Pedro, Jul 16 '12 at 21:20
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
@Norbert: I am confused. As mentioned, $d(x,A)=0$ implies the existence of $a_n\in A$ with $d(x,a_n)\to 0$. Bu definition the latter means $\lim a_n=x$. Therefore I don't understand Norbert's objection, and I don't understand why Peter introduces his function $f$. — wildildildlife, Jul 16 '12 at 21:24
@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
@MichaelGreinecker The sequence can't be finite. When I write $\lim $ I mean $\lim\limits_{n\to \infty}$. I'm sorry if that caused confusion. — Pedro, Jul 16 '12 at 21:26
@azarel Could you put that into an answer? (With the explanation) — Pedro, Jul 16 '12 at 21:26
@PeterTamaroff The equivalence between $\lim d(x,a_n)=0$ and $\lim a_n=x$ is tautological. So I don't know how to explain it. — azarel, Jul 16 '12 at 21:33
@azarel But that's what I'm saying (since the distance function is continuous). However Norbert seems to object that. — Pedro, Jul 16 '12 at 21:33
Your statement that there is a sequence of points of $A$ converging to $x$ in the $\Rightarrow$ direction is unsupported. To construct such a sequence, you must choose a sequence of shrinking neighborhoods of $x$, like $B(x,1), B(x,1/2), B(x,1/3), \dots$ and choose one point from $A$ for each such ball. — Carl, Jul 16 '12 at 21:35
@Peter: There are different definitions of convergence in metric saces, that are all equivalent. Whether $\lim a_n=x\iff \lim d(a_,x)=0$ is something that needs proof depends o whether you start with it as a definition. — Michael Greinecker, Jul 16 '12 at 21:39
@Carl Maybe, but I think it is clear. I just show the sequence exists, I don't need to construct it. — Pedro, Jul 16 '12 at 21:40
@Peter: You do not show that it exists - the corollary doesn't apply because you haven't shown that $d(x,A) = 0$ yet. — Carl, Jul 16 '12 at 21:46
@Carl I'm not using the corollary, I'm using the theorem that says that $\lim; a_n=a \iff$ for every nbhd $V$ of $a$,$a_n\notin V$ for only a finite number of $a_n$'s, or any equivalent form you like. — Pedro, Jul 16 '12 at 21:49
You may be interested in this related answer and the threads linked therein. — t.b., Jul 16 '12 at 21:51
@t.b. Thank you, but that is way out of my league for the moment. =) — Pedro, Jul 16 '12 at 21:54

Michael Greinecker · Accepted Answer · 2012-07-16T21:49:29.363

3

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.

edited Jul 16 '12 at 21:49

answered Jul 16 '12 at 21:34

Michael Greinecker

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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 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 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
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@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
@MichaelGreinecker About the edit: $f$ is continuous if and only if whenever $\lim ; a_n =a$ then $\lim; f(a_n)=f(a)$ so I think the justification is OK, isn't it? – Pedro Jul 16 '12 at 21:52
@Michael: You can always talk about nets instead! :-) – Asaf Karagila Jul 16 '12 at 21:53
@PeterTamaroff Yes, but $\lim f(a_n)$ can exist even when $\lim a_n$ does not. For eaxample, if $f$ is constant, every sequence converges after being transformed by $f$. – Michael Greinecker Jul 16 '12 at 21:55
@Asaf: I think to show that $x\in\text{cl}A$ implies the existence of a net in $A$ that converges to $x$ still requires AC. If not, I would be very surprised. – Michael Greinecker Jul 16 '12 at 21:57
@MichaelGreinecker Makoto Kato might be up for a proof of that. He seems particularily interested in proving stuff without AC. – Pedro Jul 16 '12 at 21:59
@Michael: Perhaps you are right, but in metrizable spaces I think we may be able to engineer some net using the metric. I will have to think about it, though. – Asaf Karagila Jul 16 '12 at 21:59
I must say your proof is very, very clear. Thanks. – Pedro Jul 16 '12 at 22:15
@Michael: Interestingly enough, I can't find much about this. Mostly because either AC is assumed and sequences are enough; or because people care about other properties broken without AC. I did glimpse and saw a situation where amorphous sets can be subsets of a metric space (non-trivially, I hope). I'll see if I can arrange such space, and how nets behave within amorphous sets. – Asaf Karagila Jul 16 '12 at 22:15
@Asaf: I think you can actually avoid AC. Let $x\in\text{cl}A$. Use the directed set of pairs $(U,u)$ with $U$ being a neighborhood of $x$ and $u\in A\cap U$, such that $(U,u)\geq (V,v)$ if $U\subseteq V$. Then the net $(U,u)\mapsto u$ should converge to $x$ and lie wholly in $A$ – Michael Greinecker Jul 16 '12 at 22:24
@Michael: Nice. I found a paper by Brunner but it is in German, so I'm not sure if it is relevant. http://dx.doi.org/10.1007/BF02007144 I will keep thinking about the question though. Interesting! Let me know if this paper has anything significant to add... – Asaf Karagila Jul 16 '12 at 22:32
@Asaf: i can't access the text, but it seems to be concerned with special cases of the Tychonov theorem. I found this relevant post by Gerald Edgar. Essentially, if you do not require nets to be indexed by posets, you can instead of making a choice make all choices simultaneously, which works without AC. Anyways, it's time to sleep for me... – Michael Greinecker Jul 16 '12 at 22:39
@MichaelGreinecker: Silly me. This is much simpler. There cannot be non-discrete amorphous metric spaces (fix $x$ and define the function $f(y)=d(x,y)$, the range must be finite, so almost all points have equal distance from $x$. This holds for all $x$, so every point is isolated.)... maybe if we go higher in the food-chain. Interesting... I'll have to think about it. – Asaf Karagila Jul 16 '12 at 23:11
(in reply to Michael): That's possible, I for one are happy to use AC at will in point-set topology. As for your proposed AC-less argument: how do you prove that your pre-ordered set of pairs $(U,u\in A\cap U)$ is directed? An upper bound for $(U,u)$ and $(V,v)$ is $(U\cap V,w)$ where $w\in U\cap V\cap A$ is chosen...does this choice (which must be done for all such $U,V$) not need AC? – wildildildlife Jul 17 '12 at 11:57
@wildildildlife: You don't have to construct a function that maps any two neighborhoods into a point with their intersection with $A$. That for any neighborhoods $U$ and $V$ there is some $x\in A\cap U\cap V$ follows from $U\cap V$ being a neighborhood and from $A\cap W\neq\emptyset$ for every neighborhood $W$ of $x$. The discussion here might be helpful. – Michael Greinecker Jul 17 '12 at 15:40
Right, you don't have to make a choice for all such $U,V$ simultaneously. It's somewhat remarkable (to me) that you can avoid AC by simply looking at a bigger directed set; I like your argument very much! – wildildildlife Jul 17 '12 at 16:11

score 1 · Answer 2 · answered Jul 16 '12 at 21:38

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

answered Jul 16 '12 at 21:38

wildildildlife

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And what should $\bar A$ be? – Pedro Jul 16 '12 at 21:41
@PeterTamaroff: I am sorry, $\overline{A}$ means the closure of $A$. – wildildildlife Jul 16 '12 at 21:43
Oh. That is something I am not introduced to yet. =/ – Pedro Jul 16 '12 at 21:46
@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 Jul 16 '12 at 22:05
@PeterTamaroff: Well, yes. As I said, the equivalence between $d(x,a_n)\to 0$ and $a_n\to x$ is either a definition or a triviality. Of course if it is not your definition, and you don't find it a triviality either, then you need to prove it for yourself. – wildildildlife Jul 17 '12 at 11:30
Sorry to tell you this so late, but indeed I have defined that for a sequence of points $a_n$ in a metric space, we write $\lim ; a_n=a$ if $d(a,a_n)\to 0$. – Pedro Jul 22 '12 at 00:44

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

2 Answers2