I will include the proof here and highlight the parts that are giving me trouble.

Theorem $\hspace{5 pt}$ Let $P$ be a nonempty perfect set in $\mathbb{R}^k$. Then $P$ is uncountable.

Proof $\hspace{5 pt}$ 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 $x_1$.

1) ^ Are we using the Axiom of Choice here? How can we have an arbitrary set in a construction?

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

2) ^ It makes sense intuitively, but how do we prove this last statement?

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.

3) ^ I really don't get this whole paragraph much at all. Could someone explain it in a more step-by-step way?

Put $K_n = \overline{V_n} \cap P$. Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact.

4) ^ "closed" comes from it being a closure and "bounded" comes from the definition of neighborhood, correct?

Since $x_n \notin K_{n+1}$, no point of $P$ lies in $\bigcap_1^\infty K_n$. Since $K_n \subset P$, this implies that $\bigcap_1^\infty K_n$ 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.


2 Answers 2


$1):$ I don't think AC is used here. We just look at all possible neighborhoods of $x_1$ and pick one. After that, the construction can proceed.

$2):$ If $(y_n)$ is a sequence in $V_1$ and $y_n\rightarrow y$, then for all $\epsilon>0$ there is an $y_n$ such that: $|x_1-y|\le |x_1-y_n|+|y_n-y|\le r+\epsilon$. Then we must also have $|x_1-y|\le r$.

$3):$ Since $x_n$ is a limit point of $P$ and $V_n$ a neighborhood of $x_n$, there is a $p\in V_n\cap P$ such that $p\neq x_n$. Since $V_n$ is open, there is an $\epsilon$ such that $B_{\epsilon}(p)\subset V_n$. Put $\delta:= \min \left\{\frac{\epsilon}{2},\frac{|p-x_n|}{2}\right\}$. Then $x_n\not\in \overline{B_{\delta}(p)}$ and $\overline{B_{\delta}(p)} \subset V_n$. So we define $V_{n+1}:=B_{\delta}(p)$. Since $p\in V_{n+1}$, $p$ is a limit point of $P$ and $V_{n+1}$ is open, we have $V_{n+1}\cap P\neq \emptyset$.

$4):$ Yes, the sets are closed and bounded since they are closed balls with radius $r$.

  • $\begingroup$ Thank you for your answer. I don't understand all of your answer to part 3), but I will give it some time and come back to it later. $\endgroup$ Commented Apr 13, 2014 at 10:42
  • $\begingroup$ What does $B_\epsilon(p)$ denote? Is it a closed ball? $\endgroup$
    – Student
    Commented Jan 26, 2017 at 9:51
  • $\begingroup$ @ShreyAryan No, it is the open ball with radius $\epsilon$. $\endgroup$ Commented Jan 26, 2017 at 11:07
  • $\begingroup$ Could you justify your choice of $\delta$? $\endgroup$
    – Student
    Commented Jan 26, 2017 at 11:13
  • 1
    $\begingroup$ @Math_QED I've thought about it and concluded that Rudin's argument is incomplete. Property (i) would imply that $x_{n+1}\in V_n$, but I do not see any reason for this to be true. What I think actually happens is that he constructs a subsequence $(x_{n_k})$ and neighborhoods $V_{n_k}$ such that (i) and (iii) hold (after replacing $n$ by $n_k$), but with the condition (ii') that $x_n \not\in \overline{V}_{n_{k+1}}$ for all $n<n_{k+1}$. Now the induction hypothesis gives you that $p = x_m$ with $m>n_k$ and we can assume that $m$ is minimal with this property. You put $n_{k+1} = m$ and the rest $\endgroup$ Commented Jul 16, 2018 at 9:24

My proof here includes a step-by-step construction of $V_{n+1}$ for step 3:

Proof of Baby Rudin Theorem 2.43

You might find it helpful. Step 3 is what most people (myself included) found puzzling about 2.43 when they first encountered it.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .