Prove that if $S\subset \mathbb{R}^n$ is not countable, then there exists $x \in S$ such that $x$ is a condensation point.

Question

Let $S \subset \mathbb{R}^n$ with the usual metric. A point $x \in \mathbb{R}^n$ is said to be a condensation point of $S$ if for all $r>0$, $B(x,r)\cap S$ is not countable. Show that if $S$ is not countable then there exists a point $x\in S $ that is a condensation point of $S$.

I am not bein able to show this, and I don't know how to approach it. Any hint is more than welcome!

Answer 1

Hint: There is a countable base for the topology of $\mathbb R^n$, e.g., the set of all open balls with rational center and radius. Let $\mathcal B$ be a countable base.

Assume that $S\subseteq\mathbb R$ is uncountable. Define $$\mathcal B'=\{B\in\mathcal B:S\cap B\text{ is countable}\}.$$ Then $S\cap\bigcup\mathcal B'$ is countable. Choose a point $x\in S\setminus\bigcup\mathcal B'$ and show that $x$ is a condensation point of $S$.