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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.

AmadeusDrZaius
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2 Answers2

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$1):$ 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 $r$.

Vincent Boelens
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  • 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. – AmadeusDrZaius Apr 13 '14 at 10:42
  • What does $B_\epsilon(p)$ denote? Is it a closed ball? – Student Jan 26 '17 at 09:51
  • @ShreyAryan No, it is the open ball with radius $\epsilon$. – Vincent Boelens Jan 26 '17 at 11:07
  • Could you justify your choice of $\delta$? – Student Jan 26 '17 at 11:13
  • I imagine two balls, one contained in the other. – Student Jan 26 '17 at 11:13
  • We need $\delta \le \frac{\epsilon}{2}$ to ensure $\overline{B_{\delta}(p)} \subset V_n$ and $\delta \le \frac{|x_n-p|}{2}$ to ensure $x_n\not \in V_n$. – Vincent Boelens Jan 26 '17 at 11:25
  • @ShreyAryan See comment above, I forgot to ping you. – Vincent Boelens Jan 26 '17 at 12:38
  • @VincentBoelens Could you give some motivation for this argument in English, please.. Like what is the main idea involved here? – Student Jan 26 '17 at 12:50
  • @ShreyAryan In the comment above it should be $x_n \not \in V_{n+1}$. I'm afraid I can't say much more, since it was a long time ago. But probably I just tried something until it worked. – Vincent Boelens Jan 26 '17 at 13:11
  • This answer doesn't seem quite correct. Why can you pick the balls such that the ball $V_n$ has center $x_n$? This is not adressed in the induction hypothesis and I doubt it is possible. –  Jul 12 '18 at 16:00
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    @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 – Vincent Boelens Jul 16 '18 at 09:24
  • goes through as above. I can edit the answer if you want more details. – Vincent Boelens Jul 16 '18 at 09:27
  • No, it's fine for me. Rudin's argument is correct but there are missing some details. I just wanted to point out that your answer missed this point. –  Jul 16 '18 at 09:44
  • @VincentBoelens, I don't think that (i) implies that $x_{n+1} \in V_n$. You can easily come up with cases where it is not. – James Shapiro Mar 03 '19 at 16:34
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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.

James Shapiro
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