# Problems in Theorem 2.43 of baby Rudin

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 $$|y−x_1|, the closure $$\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 not been able to understand third paragraph of Walter's proof. I would like to understand why the neighborhood $$V_{n+1}$$ exists with properties (i),(ii) and (iii), using only the previous definitions and theorems of the Book.

• If X is a non-empty completely metrizable space with no isolated points then X has a subspace homeomorphic to the Cantor set. Oct 19, 2015 at 5:42

For the 3rd paragraph:

$V_n = N_{\epsilon_n}(p_n)$ some $p_n \in P, \epsilon_n > 0$. $\exists$ infinitely many elements of $P$ in $V_n$, so pick one $p_{n+1}$ that is not equal to $x_n$ or $p_n$. Let $\epsilon=d(p_n, p_{n+1})$, $\epsilon'=d(p_{n+1}, x_n)$, $\epsilon''=\epsilon_n-\epsilon>0$. Choose $\epsilon_{n+1} < \min\left\{\epsilon, \epsilon', \epsilon''\right\}$, then:

• $d(x_n, p_{n+1}) > \epsilon_{n+1}$ so certainly $x_n \not\in \overline{V}_{n+1}$
• if $d(e, p_{n+1}) \leq \epsilon_{n+1}$ then \begin{align}d(e,p_n) &\leq d(e,p_{n+1}) + d(p_n, p_{n+1}) \\ &\leq \epsilon_{n+1} + \epsilon \\ &< (\epsilon_n - \epsilon) + \epsilon = \epsilon_n\end{align} so $\overline{V}_{n+1} \subset V_n$.
• $p_{n+1} \in V_{n+1} \cap P$ so intersection is non-empty.
• I don't think that's the case. After all, as Rudin writes: "Let $V_1$ be any neighborhood of $x_1$". What if the neighborhood was $N_\epsilon(x_1)$ with $\epsilon < d(x_1, x_2)$? Then you couldn't construct a neighborhood $N$ of $x_2$ with $N \subset V_1$. (in response to suggestion that $V_n$ must contain $x_n$. Comment was deleted which is a shame as it was a natural question to raise) Oct 18, 2015 at 14:20
• Sorry, you're right. I delete it because Rudin just say "Let $V_1$ be any naighborhood of $\mathbf{x_1}$" and this doesn't imply that $V_n$ must be a neighborhood of $\mathbf{x_n}$. Oct 18, 2015 at 14:31
• After all, the point is to construct telescoping subsets of $P$ which exclude successive $x_n$ and yet are still compact, so that their intersection is non-empty but exclude all $x_n$. Oct 18, 2015 at 14:31
• @Fonseca it's a good question to ask in understanding the prrof Oct 18, 2015 at 14:32

I explain why there is a neighborhood $V_n$ satisfying (i), (ii) and (iii). We know that $P$ is perfect. That is, every point in $P$ is a limit point of $P$. Especially, $x_n$ is a limit point of $P$ so every neighborhood $V$ of $x_n$ contains a point in $P$ differ from $x_n$. Let it say $y$.

You can take a neighborhood $W = B(y, \rho)$ of $x_{n+1}$ which is not contain $x_n$ and contained in $V_n$. If we take $V_{n+1} = B(y,\rho/2)$, then $\overline{V_{n+1}} \subseteq W\subseteq V_n$. Since $x_n\notin W$, $\overline{V_{n+1}}$ does not contain $x_n$. Moreover, since $y\in P$ and $y\in V_{n+1}$, $V_{n+1}\cap P$ is not empty.

Let me know if you need more details or have another question.

• Sorry, what is the meaning of "Let $V_1$ be any neighborhood of $\mathbf{x_1}$"? I thoughtthat its meaning is that $V_1 = B(x_1,r)$, where $r$ is any real positive number I also thought that each $V_n$ must be of the form $B (x_n, r_n)$. Oct 18, 2015 at 13:21
• @Fonseca Any open set $V_1$ containing $x_1$ is called a neighborhood of $x_1$. Open balls $B(x_1,r)$ are neighborhoods of $x_1$, of course. However there are lots of neighborhoods of $x_1$ that is not an open ball. Oct 18, 2015 at 13:27
• Even that fact, in many cases, we need not consider another open neighborhoods, since every open nbhds of $x_1$ contains some open ball $B(x_1,r)$. The philosophy of nbhds is "smallness", so in many cases considering certain type of nbhds is sufficient. Oct 18, 2015 at 13:27

$$V_1 \cap P \ni x_1$$. So, $$V_1 \cap P \neq \emptyset$$.

Suppose that $$V_n \cap P \neq \emptyset$$.

(1) If $$x_n \in V_n$$, there exists a neighborhood $$N(x_n) \subset V_n$$.
Because $$x_n \in P$$ and $$P$$ is a perfect set, $$x_n$$ is a limit point of $$P$$.
So, $$N(x_n)$$ contains $$x_k$$ which is not equal to $$x_n$$.
Let $$V_{n+1}$$ be a neighborhood whose center is $$x_k$$ and whose radius is less than "the radius of $$N(x_n)$$" minus $$|x_k - x_n|$$ and whose radius is less than $$|x_k - x_n|$$.
Then, $$V_{n+1}$$ satisfies (i) and (ii) and (iii).

(2) the case when $$x_n \notin V_n$$.
Becase $$V_n \cap P \neq \emptyset$$, there exists $$x_k \in V_n$$.
Let $$V_{n+1}$$ be a neighborhood whose center is $$x_k$$ and whose radius is less than "the radius of $$V_n$$" minus "the distance between the center of $$V_n$$ and $$x_k$$".
Then, $$V_{n+1}$$ satisfies (i) and (ii) and (iii).

I show how such a neighborhood can be constructed in great detail here: Proof of Baby Rudin Theorem 2.43