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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?

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    $\begingroup$ If $U$ is empty the assertion is false. So it should be assumed that $U$ is nonempty. $\endgroup$
    – YCor
    Commented Mar 5, 2022 at 20:59
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    $\begingroup$ If $U$ is assumed non-empty, just being a nonempty open subset of a Cantor set, it is homeomorphic to Cantor minus one point (or equivalently to Cantor $\times$ discrete countable). So the point is to check that every dense subset $D$ of $U$ has infinitely many accumulation points inside $D$. Indeed every element of $D$ is an accumulation point of $D$ (this just follows from the fact that $U$ has no isolated point). $\endgroup$
    – YCor
    Commented Mar 5, 2022 at 21:01
  • $\begingroup$ @YCor just wanted to point out that $U$ can also be homeomorphic to the Cantor set. Also, the same reasoning can be applied without knowing how open subsets of the Cantor set are up to homeomorphism: it is enough to notice that $U$ is uncountable if it is non empty (because every nonempty open set of the Cantor set contains a small copy of the Cantor set), and every point of $U$ is an accumulation point of $D\cap U$. $\endgroup$
    – Saúl RM
    Commented Mar 5, 2022 at 21:11
  • $\begingroup$ @SaúlRodríguezMartín oops, my mistake, yes of course: if closed (=clopen) nonempty it's homeomorphic to Cantor, and if not closed it's homeomorphic to Cantor minus singleton. Anyway this leaves the reasoning unchanged: it works in any Hausdorff $U'$ with no isolated point. $\endgroup$
    – YCor
    Commented Mar 5, 2022 at 21:17

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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.

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  • $\begingroup$ I guess OP means accumulation points of $D\cap U$ within itself (so there are infinitely many, but not continuum). $\endgroup$
    – YCor
    Commented Mar 7, 2022 at 11:30
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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 $\emptyset\not=(D\cap U)\cap (V\setminus\{x\})\cap U=V\cap (D\cap U)\setminus\{x\}$.

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