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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.

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  • $\begingroup$ Look at the set $\{1,\frac{1}{2},\frac{1}{3}, \dotsc \}$. This set is infinite and discrete. The idea is that if $X$ is not discrete itself, it always has a subset, which somehow looks like the above. $\endgroup$ – MooS Mar 22 '17 at 11:46
  • $\begingroup$ It's true already for infinite Hausdorff spaces. No metric is needed $\endgroup$ – Henno Brandsma Mar 22 '17 at 20:24
  • $\begingroup$ See math.stackexchange.com/a/601210 $\endgroup$ – Henno Brandsma Mar 22 '17 at 20:52
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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…

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  • $\begingroup$ 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. $\endgroup$ – MooS Mar 22 '17 at 11:53
  • $\begingroup$ @MooS Yes, of course. That was left to the OP. ;-) $\endgroup$ – egreg Mar 22 '17 at 12:09
  • $\begingroup$ @MooS Anyway, the set is discrete even if the terms are chosen with no particular care (a convergent sequence has a single limit point). $\endgroup$ – egreg Mar 22 '17 at 12:34
  • $\begingroup$ @Moos Is it true that $A$ is infinite is because $X$ is infinite (and by axiom of choice)? $\endgroup$ – PropositionX Mar 22 '17 at 13:02
  • $\begingroup$ @Y.X. You need some form of choice, that's right. The fact that $A$ is infinite follows from $a$ not being isolated. $\endgroup$ – egreg Mar 22 '17 at 13:05
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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.

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