Theorem 2.43 Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.

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

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

Suppose $V_n$ 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) $x_n\notin \overline{V_{n+1}}$, (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 $\mathbf{x_{n}}\notin K_{n+1}$, 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 nonempty, by (iii), and $K_n\supset K_{n+1}$, by (i); this contradicts the Corollary to Theorem 2.36.

I have difficulty in understanding the third paragraph. So I have been reading answers on the following related links on MSE (see [a] and [b]) to strengthen my understanding, but I am unable to get the motivation behind this construction. Also, I would be grateful if someone could explain this idea to me in English.

  • $\begingroup$ I am self-studying analysis, will be grateful if anyone could help. $\endgroup$
    – Student
    Commented Jan 26, 2017 at 11:30

2 Answers 2


To better understand the third paragraph of Rudin's proof, it may be useful to see some of the smaller induction cases worked out in detail.

To start off, pick $\mathbf{x'}\in V_1\cap P$ where $\mathbf{x'}\ne \mathbf{x_1}$. Such a point exists because $P$ is perfect.

Let $r'=|\mathbf{ x'}−\mathbf{x_1}|$. Pick $r_2$ such that $0<r_2<\min \left( {r',r - r'} \right)$ and set $V_2 = N_{r_2}(\mathbf{x'})$, where I’m using Rudin’s notation for a neighborhood. Then clearly $\overline{V_2}\subset V_1$, $\mathbf{x_1}\notin V_2$, and $V_2\cap P$ is not empty (the latter, since $V_2$ is a neighborhood of $\mathbf{x'}\in P$).

Next consider what to do for $V_3$. If $\mathbf{x_2}\notin V_2$, then we can just shrink the neighborhood about $\mathbf{x'}$ to obtain $V_3$: that is, pick any $r_3$ such that $0<r_3<r_2$. If we then set $V_3=N_{r_3 } \left(\mathbf{x'} \right)$, we have that $\overline{V_3}\subset V_2$, $\mathbf{x_2}\notin V_3$, and $V_3\cap P$ is not empty.

If $\mathbf{x_2}\in V_2$ and $\mathbf{x_2}\ne \mathbf{x'}$, we can still have $V_3$ be a neighborhood of radius $r_3$ centered about $\mathbf{x'}$ if we simply ensure that $r_3$ is set to some positive value less than $\left| {\mathbf{x_2}-\mathbf{x'}} \right|$.

If, on the other hand, we actually have that $\mathbf{x_2}=\mathbf{x'}$, then we can construct $V_3$ similarly to how we constructed $V_2$. That is, set $V_3 = N_{r_3} \left( \mathbf{x''} \right)$ where $\mathbf{x''}$ is some point in $V_2\cap P$ and $r_3$ is chosen so that $V_3$ is strictly contained in $V_2$ but excludes $\mathbf{x_2}$. For example, set $r'' =|\mathbf{ x''}−\mathbf{x_2}|$ and pick $r_3$ such that $0<r_3<\min(r'', r- r'')$. Again, we have $\overline{V_3}\subset V_2$, $\mathbf{x_2}\notin V_3$, and $V_3\cap P$ is nonempty.

We can continue this construction indefinitely.


Paragraph 3 is what trips most people up. I try to show how you can construct such a $V_{n+1}$ here: Proof of Baby Rudin Theorem 2.43


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