Is it true that an uncountable subset of any compact metric space condenses at uncountably many points of itself? How about of any separable metric space? I know that the first fact is true when you take the metric space $\mathbb{R}$, but is it true for any compact metric space?
Thanks.
To check that we're using the same definitions: $x$ is a condensation point of $E$ if every open neighborhood of $x$ contains uncountably many points of $E$. Your question is whether an uncountable set must contain uncountably many of its own condensation points.
The answer is yes, in any separable metric space, or more generally in any second-countable topological space.
Lemma. If $X$ is second countable, every uncountable $E \subset X$ contains at least one of its own condensation points.
Proof. Let $\{U_n\}$ be a countable base for $X$. Suppose to the contrary that every point $x \in E$ is not a condensation point of $E$. Then for every $x \in E$ there is an open set $V_x$ containing $x$ and such that $U \cap E$ is (at most) countable. Since $\{U_n\}$ is a base, we can find $U_{n_x}$ with $x \in U_{n_x} \subset V$; then $U_{n_x} \cap E$ is also countable. We now have $E = \bigcup_x (U_{n_x} \cap E)$, where the right side is a countable union of countable sets. This is absurd since $E$ is uncountable.
Now suppose $E$ is an uncountable set. Let $C$ be the set of points in $E$ that are also condensation points of $E$. If $C$ is countable then $E \setminus C$ is an uncountable set which contains none of its own condensation points, contradicting our lemma.
