I now that there are many questions about this theorm on here already (such as Theorem 2.43 in Baby Rudin: How to understand the proof?) but after reading them, I noticed that none of them answer my question.


Let $P$ be a non-empty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.

Let $(X,d)$ be a metric space, and let $P \subset X$. Then $P$ is perfect if it is closed (i.e. it contains all of its limit points) and every point of $P$ is also a limit point of $P$.

Since $P$ has limit points, $P$ must be infinite. Suppose $P$ is countable, and denote the points of $P$ by $x_1, x_2, x_3, \ldots$. We shall construct a sequence $\{V_n\}$ of neighborhoods as follows:

Let $V_1$ be any neighborhood of $x_1$. If $V_1$ consists of all $y \in \mathbb{R}^k$ such that $\vert y - x_1 \vert < r$, the closure $\overline{V_1}$ of $V_1$ is the set of all $y \in \mathbb{R}^k$ such that $\vert y - x_1 \vert \leq r$.

Supose $V_n$ has been constructed, so that $V_n \cap P$ is not empty. Since every point of $P$ is a limit point of $P$, there is a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_n$, (ii) $x_n \not\in \overline{V_{n+1}}$, (iii) $V_{n+1} \cap P$ is not empty. By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed.

Put $K_n = \overline{V_n} \cap P$. Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact. Since $x_n \not\in K_{n+1}$, no point of $P$ lies in $\cap_1^\infty K_n$. Since $K_n \subset P$, this implies that $\cap_1^\infty K_n$ is empty. But each $K_n$ is non-empty, by (iii), and $K_n \supset K_{n+1}$, by (i); this contradicts the Corollary to Theorem 2.36.

I can follow the proof and understand it but I do not get why this implies that $\cap_1^\infty K_n$ is empty. I know that none of the $x_n$ will belong to the intersection but we have not checked the points contained in their neighborhoods. Why is it enough to know the intersection is empty just by knowing that none of $x_n$ belong?

Can someone please tell me what I am missing?

  • $\begingroup$ You seem to have missed the words "Since $K_n\subset P$..." $\endgroup$ Commented Jun 28, 2019 at 18:39
  • $\begingroup$ At some point we must use the fact that the metric on $\Bbb R$ is complete. E.g. if $X=\Bbb Q=P$ with the usual metric then $P$ is closed in $X$ and each point in $P$ is a limit point of $P$ but $P$ is countable. $\endgroup$ Commented Jun 28, 2019 at 20:32

1 Answer 1


Suppose that there is some $x\in\bigcap_{n\in\mathbb N}K_n$. Then $x\in P$, since each $K_n$ is a subset of $P$. So, $x=x_n$, for some $n\in\mathbb N$. This is impossible, since $x=x_n\notin K_{n+1}$.

  • $\begingroup$ Thank you, I was confused at first but totally get it now $\endgroup$ Commented Jun 28, 2019 at 18:17
  • $\begingroup$ I'm glad I could help. $\endgroup$ Commented Jun 28, 2019 at 18:18

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