Suppose, in a separable (so it has a countable base) metric space $X$, a subset $E$ is uncountable. Now, $P$ be the set of all points $p$ of $X$ such that every neighbourhood of $p$ contains uncountably many points from $E$. I want to prove that any point in $P$ is a limit point of $P$. If we take any neighbourhood of a point $p$ in $P$, then why should it contain another point in $P$. Suppose all the other points in the neighbourhood of this point has at least one neighbourhood which has only countable number points from $E$. Then what is the contradiction ? Can I construct a neighbourhood of $p$ which contains only countable number of points from $E$. Stuck here


It’s a little easier to think about if you let $C$ be the set of $x\in X$ such that some nbhd of $x$ contains only countably many points of $E$; clearly $C=X\setminus P$. Let $\mathscr{B}$ be a countable base for $X$, and for each $x\in C$ let $B_x\in\mathscr{B}$ be an open nbhd of $x$ such that $B_x\cap E$ is countable. Let $\mathscr{B}_0=\{B_x:x\in C\}$; $\mathscr{B}$ is countable, so $\mathscr{B}_0$ is countable as well. Let $U=\bigcup\mathscr{B}_0$.

  • Use the countability of $\mathscr{B}_0$ to show that $U\cap E$ is countable.
  • Then show that $E\setminus C\subseteq P$.
  • Conclude that if $p\in P$, and $V$ is any open nbhd of $p$, then $V\cap P$ is uncountable.

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