Suppose $X$ is a separable metric space, and $C$ is an uncountable subset of $X$. Prove that there is a point $x \in C$ such that for each $\epsilon>0$, $B(x;\epsilon)\cap C$ is uncountable.

I missed first couple of classes so I don't really know what is the general strategy to prove something is uncountable (or countable)? The only thing that comes to my mind is that, since $X$ is separable, then for any open cover, it has a countable sub cover (may not be finite though), and I was hoping I could get a contradiction out of here?

I tried contradiction, but that would means for every $x\in C$, there is a corresponding $\epsilon(x)$ such that $B(x;\epsilon(x))\cap C$ is countable. But I cannot deduce that $C$ is then countable because that intersection may be empty.

Any idea?


Even more is true, and not much harder to prove.

HINT: Let $\mathscr{B}$ be a countable base for $X$. Let $$\mathscr{B}_0=\{B\in\mathscr{B}:B\cap C\text{ is countable}\}\;.$$

Let $C_0=C\setminus\bigcup\mathscr{B}_0$. Show that $C_0$ is actually an uncountable set such that $B\cap C_0$ is uncountable for every open nbhd $B$ of every $x\in C_0$.

  • $\begingroup$ OK, so I'll have to look up what a countable base is, too... $\endgroup$ – 3x89g2 Apr 21 '15 at 5:14
  • $\begingroup$ @Misakov: As long as you know what a base for a topology is, you're okay: it's just a base (or, as some prefer, a basis) that contains only countably many sets. For example, the family of open intervals with rational endpoints is a countable base for the usual topology on $\Bbb R$. $\endgroup$ – Brian M. Scott Apr 21 '15 at 5:23
  • $\begingroup$ I don't think we ever talked about what a base is, but I'll look it up. Thanks! $\endgroup$ – 3x89g2 Apr 21 '15 at 6:40
  • $\begingroup$ @Misakov: Short definition: if $\tau$ is a topology on a set $X$ (i.e., the collection of all open sets), then a family $\mathscr{B}$ of subsets of $X$ is a base for $\tau$ if $\mathscr{B}\subseteq\tau$, and for each $U\in\tau$ there is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$. One of the basic(!) theorems of metric spaces is that every separable metric space has a countable base: if $D$ is a countable dense subset of $X$, $\{B(x,r):x\in D\text{ and }r\in\Bbb Q^+\}$ is a countable base for the the topology on $X$. $\endgroup$ – Brian M. Scott Apr 21 '15 at 6:55

Initially I was proving existence of $x\in X$ since I misread the problem this way. Now I made the small changes to prove existence of such $x\in C$.

Let $\{x_n\}\subset C$ be dense in $C$. Assume no such $x\in C$ exists. Then there are $r_n>0$, largest, such that $B(x_n,r_n)\cap C$ is countable because, in particular, none of the $x_n$ is a candidate for $x$.

But $\bigcup_n B(x_n,r_n)\supset C$. This implies that $C$ is countable.

Assume $y\in C\setminus\bigcup_n B(x_n,r_n)$ (in particular $y\neq x_n$ for all $n$). Let $r>0$ be such that $B(y,r)\cap C$ is countable, which by the assumption exists. Let $x_n$ be inside $B(y,r/4)$. Then $r_n\leq d(y,x_n)<r/4$. But $B(x_n,3r/4)\subset B(y,r)$ and therefore $B(x_n,3r/4)\cap C$ is countable contradicting the choice of $r_n$ (since $3r/4>r/4>r_n$).

Therefore $\bigcup_n B(x_n,r_n)\supset C$ as claimed.

Existence of the largest $r$:

Under the assumptions that $C$ is uncountable, and that $x$ is such that there is $\epsilon>0$ such that $B(x,\epsilon)\cap C$ is countable. Let $r:=\sup\{s\in\mathbb{R}^+:\ B(x,s)\cap C\text{ is countable}\}$. Since $B(x,r)=\bigcup_{n\in\mathbb{N}}B(x,r-1/n)$ we have that $B(x,r)\cap C=\bigcup_{n\in\mathbb{N}}\left(B(x,r-1/n)\cap C\right)$ is countable.

  • $\begingroup$ Okay, you're now taking advantage of the fact that separability is hereditary in metric spaces (which is not true for arbitrary topological spaces). $\endgroup$ – user642796 Apr 21 '15 at 6:09
  • $\begingroup$ @ArthurFischer ... and which is a pretty trivial fact to prove. $\endgroup$ – Alamos Apr 21 '15 at 6:11

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