Subsets of the Cantor set A copy of the Cantor set is a space homeomorphic to $2^{\omega}$.
Suppose that $X$ is a Hausdorff space that contains a copy $C^{\prime}$ of the Cantor set. Let $U$ be a nonempty subset open in $C^{\prime}$, also let $D$ be a countable set dense in $X$ such that $D\cap U$ is dense in $U$. Does anyone have any idea how to prove that the set of accumulation points of $D\cap U$ is infinite?
 A: To answer the question: every point of $U$ is an accumulation point of $D\cap U$, hence there are continuum many accumulation points. The ambient space $X$ plays no role here; everything takes place in the Cantor set.
A: If $A\subseteq X$, denote by $A^{\mbox{d}}$ the set of accumulation points of $A$. Thanks to YCor, Saúl Rodríguez Martín and KP Hart. Putting together the previous ideas and being more explicit, we have that:
As $U$ is a nonempty subset open in $C^{\prime}$, then $U$ is infinite, in fact $|U|=2^{\aleph_0}$. We claim that $U\subseteq (D\cap U)^{\mbox{d}}$, for this, let $x\in U$, and $V$ a nonempty open set with $x\in V$. We will show that $V\cap (D\cap U)\setminus\{x\}\not=\emptyset$. In fact, $x\in V\cap U$. As $C^{\prime}$ is perfect, and $V\cap U$ is an open neighborhood in $C^{\prime}$, then $(V\cap U)\setminus\{x\}\not=\emptyset$. Also $(V\cap U)\setminus\{x\}=(V\setminus\{x\})\cap U$ is a nonempty set open in $U$, then $\emptyset\not=(D\cap U)\cap (V\setminus\{x\})\cap U=V\cap (D\cap U)\setminus\{x\}$.
