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A metric space $X$ is said to be discrete if every point is isolated.

A point $x ∈ A ⊂ X$ is an isolated point of $A$ if some open ball centred at $x$ contains no members of $A$ other than $x$ itself.

I am having troubles with proving the following statement:

Every infinite metric space $(X, d)$ contains an infinite subset $A$ such that $(A, d)$ is discrete.

I have spent some time on this problem. I am thinking that a constructive proof may be impossible. But even if I tried proof by contradiction, I still did not get much progress. Can someone help me? Thanks so much.

Alex M.
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Y.X.
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2 Answers2

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Suppose $X$ is not discrete (otherwise you're already done).

Then there is $a\in X$ which is not an isolated point; so, for every $n>0$, there is a point $x_n\in X$, $x_n\ne a$, such that $d(x_n,a)<1/n$.

(It is not difficult to build the sequence so that, for every $m$, $x_1,x_2,\dots,x_m$ are pairwise distinct, but it's not really required.)

Consider the set $A=\{x_n:n>0\}$. Then $A$ is infinite and has no limit point, because…

egreg
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  • To fill the "It is not difficult"-part with a rigorous argument, you should pick the $x_n$ inductively with $d(a,x_0)<1$ and $d(x_{n+1},a)<\frac{1}{2}d(x_n,a)$. This also eases the argument that the set ${x_n}$ is discrete. – MooS Mar 22 '17 at 11:53
  • @MooS Yes, of course. That was left to the OP. ;-) – egreg Mar 22 '17 at 12:09
  • @MooS Anyway, the set is discrete even if the terms are chosen with no particular care (a convergent sequence has a single limit point). – egreg Mar 22 '17 at 12:34
  • @Moos Is it true that $A$ is infinite is because $X$ is infinite (and by axiom of choice)? – Y.X. Mar 22 '17 at 13:02
  • @Y.X. You need some form of choice, that's right. The fact that $A$ is infinite follows from $a$ not being isolated. – egreg Mar 22 '17 at 13:05
  • @egreg Here is a little proposition which I think can explain this : if $x$ is a limit point of $A \subset X$, then every open ball centred at $x$ contains an infinite number of points from $A$. We just take $A=X$ in this case. Am I right? – Y.X. Mar 22 '17 at 13:07
  • @Y.X. No, you can't. Consider the standard $\mathbb{R}$ example. – egreg Mar 22 '17 at 13:07
  • @egreg This makes me a little worried. In your answer, you talked about a limit point of $X$. I think it only makes sense to talk about a limit point of a subset of $X$. Does it make sense to talk about a limit point of the whole metric space? – Y.X. Mar 22 '17 at 13:10
  • @Y.X. I took a non isolated point of $X$, then defined a suitable subset $A$. – egreg Mar 22 '17 at 13:11
  • @egreg Are you saying that "a non-isolated point" is different from "a limit point"? I think a point is either a limit point or an isolated point in a metric space? – Y.X. Mar 22 '17 at 13:12
  • @Y.X. If you prefer calling it a limit point, fine. And so? What's the point? You obviously can't choose $A=X$, because $X$ is assumed not discrete to begin with. – egreg Mar 22 '17 at 13:15
  • @egreg Sorry, I guess my problem would be in the second line of your proof, you wrote "there is a point which is not an isolated point (presumably of what)"? My question is (which I don't know the answer): does it make sense to talk about a limit point (or a non-isolated point) of the whole metric space? (because it seems that in all textbooks, we talk about a limit point of a certain subset of a metric space?) – Y.X. Mar 22 '17 at 13:19
  • @Y.X. A space is a subset of itself. – egreg Mar 22 '17 at 13:24
  • @egreg Ok I see what the problem is. In our previous discussion, I presented the proposition :" if x is a limit point of $A \subset X$,..." In this general proposition,$A$ does not mean the $A$ in the question. It just means a certain subset of a metric space.I am sorry. – Y.X. Mar 22 '17 at 13:29
  • @egreg I am definitely not an expert on questions like "Do we need choice for that?". But honestly, I do not see how we need AC here? The existence of such points $x_n$ is just a consequence of the definition of a non-isolated point. And to pick one element of a non-empty set, we do not need to invoke choice, I think. – MooS Mar 22 '17 at 13:50
  • @MooS You need to pick infinitely many points and put them together in a set. That's AC, to me. A weaker form suffices, but some form is needed. – egreg Mar 22 '17 at 13:53
  • This means, in general, when we construct an infinite set inductively, we always need AC? To me, it looks like this: Given ${x_0, x_1, \dotsc, x_n }$, we can pick $x_{n+1}$ without AC. Thus we can - for any $n \geq 0$ - construct the set $A_n ={x_0, x_1, \dotsc, x_n }$ without AC. Then let $A$ be their union. – MooS Mar 22 '17 at 14:04
  • @MooS You're wrong, I'm afraid. This is “countable choice”. – egreg Mar 22 '17 at 14:08
  • At which point exactly do I fall short without (countable) choice? Im just curious. For the working mathematician, this is of course not relevant... – MooS Mar 22 '17 at 14:10
  • @MooS Comments are not for discussion. Ask a new question if searching the site gives no suitable topic. – egreg Mar 22 '17 at 14:13
  • @egreg I started to think the proof is actually not correct. For example, for $x_1$, no matter how small you choose the radius, the open ball with that radius will contain something other than $x_1$. This is because of the following proposition : "if $a$ is a limit point of any subset $B$ of a metric space $X$, then every open ball centred at $a$ contains an infinite number of points from $B$". In your proof, $B=X$. – Y.X. Mar 22 '17 at 14:14
  • @egreg To be more precise, say for $x_1$, you choose $r$ as radius, then I can choose $r'$ as radius with $a$ being the centre such that the ball centred at $x_1$ and the ball centred at $a$ intersects. In this case, since $a$ is a limit point, by definition, the ball I constructed (which centred at $a$ and has $r'$ as radius will contain something other than $a$. How can you make sure that extra point will not be in your ball for $x_1$? – Y.X. Mar 22 '17 at 14:22
  • @MooS "their union" is the problem. The union of which family of sets? Given $A_n$, there are usually many possibly ways to extend it to $A_{n+1}$, but no distinguished way. So you need to choose $A_{n+1}$ (or equivalently $x_{n+1}$) given $A_n$ to get an infinite family of sets whose union you can take. Done in this way, you even need dependent choice, countable choice is not enough. Countable choice suffices if you choose $x_n \in B_{1/n}(a)$ for all $n\in \mathbb{N}\setminus {0}$, but that allows for multiple $x_k$ to be the same point (we still get an infinite discrete subset, though). – Daniel Fischer Mar 22 '17 at 14:30
  • @DanielFischer Do you mind if I ask you to take a look at my argument to see whether that is a disproof of the answer? – Y.X. Mar 22 '17 at 14:35
  • The answer is perfectly valid. Our discussion was just a nitpicking of whether one needs AC or not. And it seems one needs. But this is redundant for the sake of just proving the original statement. – MooS Mar 22 '17 at 14:41
  • @Y.X. The ball around $x_1$ may contain other points of $X$ for every positive radius, but if the radius is small enough it doesn't contain another point of $A$. Since $d(x_1,a) > 0$, there is an $m$ such that $d(x_1,a) > 1/m$. Thus the ball around $x_1$ with radius $\frac{1}{m(m+1)}$ contains no $x_n$ for $n > m$. $B = { x_k : 2 \leqslant k \leqslant m} \setminus {x_1}$ is a finite set, so $\rho = \min:\bigl( { d(x_1,b) : b \in B} \cup \bigl{\frac{1}{m(m+1)}\bigr}\bigr) > 0$. Then $A \cap B_{\rho}(x_1) = {x_1}$. – Daniel Fischer Mar 22 '17 at 14:42
  • @DanielFischer OK. I got it. My understanding was wrong. For $x_1 \in A \subset X$ to be isolated, what I need to show is that there is some open ball centred at $x_1$ such that this open ball contains no element (other than $x_1$) from $A$ not from $X$! Did I state this correctly this time? – Y.X. Mar 22 '17 at 14:50
  • @Y.X. Yes. The subspace $(A,d)$ is discrete if every $x\in A$ is an isolated point of $A$. Consider $A = \mathbb{Z}\subset \mathbb{R} = X$ for a situation where $X$ has no isolated points at all. – Daniel Fischer Mar 22 '17 at 14:55
  • @DanielFischer Sorry. I actually need to show that the set $B$ is infinite. May I please ask why the set $\lbrace x_k \rbrace \setminus \lbrace x_1 \rbrace$ becomes a finite set? – Y.X. Mar 23 '17 at 18:01
  • @Y.X. The set $B$ in my comment is finite. It consists of those points in $A$ different from $x_1$ that might be closer to $x_1$ than $\frac{1}{m(m+1)}$, and that can only happen for $1 < k \leqslant m$. This is used to show that $x_1$ is isolated in $A$. It's the set $A$ that needs to be (and is) infinite. – Daniel Fischer Mar 23 '17 at 18:37
  • @DanielFischer OK. Suppose that $d(x_1,a)=5/12$. Then obviously, there are infinitely many choices of $m$ I can make. Note that $1/3<5/12<1/2$. So I will just take $m=3$ (presumabally, I can take $m$ to be anything less than three, right?). Then by your argument in the set $B$, there are only two points $p,q$ where $d(p,a)<1/2, d(q,a)<1/3.$ Is this what you mean? – Y.X. Mar 24 '17 at 02:03
  • @Y.X. I suppose it's a typo, but "take $m$ anything larger than three", not less than. We need $\frac{1}{m} < d(x_1,a)$. If we choose $m = 3$, then $B$ contains at most two points, $x_2,x_3$. It is possible that $x_2 = x_1$ or $x_2 = x_3$ in this situation, so $B$ may be a singleton set. But the generic situation is that $B$ contains the two points $x_2$ and $x_3$. – Daniel Fischer Mar 24 '17 at 13:41
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Split $X$ into countably many nonempty disjoint sets $X_{1}, X_{2}, \dots$. By the axiom of choice, the set $\{ \xi \mid\ \forall n \in \mathbb{N}, \exists !\ \xi' \in X_{n}\ \text{s.t.}\ \xi = \xi' \}$ exists. Let $a_{1} \in X_{1} $ such that $d(a_{1}, X_{2}) > 0$; let $a_{n} \in X_{n}$ such that $d(a_{n}, X_{n-1}), d(a_{n}, X_{n+1}) > 0$ for all integers $n \geq 2$. Then the set $\{ a_{n} \mid n \in \mathbb{N} \}$ is $\subset X$, infinite, and discrete.

Yes
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