Let $(X,d)$ be a metric space. Let $A \subset X$. We say $x \in X$ is a condensation point of $A$ if every neighborhood of $x$ contains an uncountable number of elements of $A$. Prove every uncountable subset of $\mathbb{R}$ has a condensation point.
My current thought is proving by contradiction. Suppose no point of $A$ is a condensation point, every uncountable subset of $\mathbb{R}$ has countable number of elements around a point $x$. However, I don't know how to proceed from here.
Best Answer
Yes, that's a good starting point. For each $n\in\mathbb Z$, let $A_n=A\cap[n,n+1]$. Then $A=\bigcup_{n\in\mathbb Z}A_n$ and therefore some $A_n$ is uncountable too. Suppoes that $A_n$ has no accumulation point. For each $x\in[n,n+1]$, let $U_x$ be an open set such that $x\in U_x$ and that $A_x$ contains only countably many points of $A$. Since $[n,n+1]$ is compact, there is a finite set $\{x_1,x_2,\ldots x_N\}\subset[n,n+1]$ such than$$A_n\subset[n,n+1]\subset\bigcup_{j=1}^NU_{x_j}.$$This is impossible, because each $U_{x_j}$ only has countably many elements of $A$, whereas $A_n$ is uncountable.