# Proof of Baby Rudin Theorem 2.43

December 1st, 2020

## Theorem 2.43

Note: I originally submitted this proof on StackOverflow. Reposting here to test out $\LaTeX$.

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

**Proof**: Since $P$ is non-empty, and every point of $P$ is a limit point, $P$ contains at least one limit point. Hence, $P$ is infinite (by 2.20).

Suppose $P$ is countable and denote the points of $P$ by $x_1, x_2, x_3, \dots$

We can construct a sequence $\{V_n\}$ of neighborhoods as follows:

Let $V_1$ be **any** neighborhood of $x_1$. Then $V_1 \cap P$ is non-empty, because $x_1$ is a limit point of $P$.

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 $\{y | |y-x_1| \le r\}$ (note: I will not prove this rigorously, but see Lemma 1 below for a sketch of the proof).

Suppose $V_n$ has been constructed so that $V_n \cap P$ is not empty and $V_n = N_r(p)$ for some $p \in P$. 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, and $V_{n+1} = N_r(p)$ for some point in $P$.

By (iii), $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed.

We now show **how** we can always construct such a neighborhood $V_{n+1}$. Let's say that $V_n = N_{r_n}(p_n)$ for some $p_n \in P$.

Just a quick note: it's a common point of confusion that people believe that $x_n$ must be in $V_n$. While $x_1 \in V_1$, we cannot assume that this holds true for any other $x_n$, and at no point does the proof assume this (or need to assume this).

Let $p_{n+1}$ be some point in $V_n \cap P$, such that (a.) $p_{n+1} \not = p_{n}$, (b.) $p_{n+1} \not = x_n$, and (c.) $p_{n+1} \not = x_{n+1}$. It follows from Lemmas 2 and 3 that such a point exists.

**Lemma 2:** For $n>1$, there exists some point $q \in V_n \cap P,$ such that $q \not = p_n$, and such that $d(p_n, q) < d(p_n, x_n)$ and $d(p_n, q) < d(p_n, x_{n+1})$. Suppose not. Note that for $n > 1$, we have that $p_n \not = x_n$ and $p_n \not = x_{n+1}$ (by our choice of $p_n$). Therefore, $d(p_n, x_n) > 0$ and $d(p_n, x_{n+1}) > 0$. So there is some neighborhood of $p_n$ (namely, the neighborhood with radius $r = d(p_n, x)$, where $x \in \{x_n, x_{n+1}\}$) such that the only point of that neighborhood in $P$ is the point $p_n$, which contradicts our assumption that $p_n$ is a limit point of $P$.

**Lemma 3:** For $n=1$, there exists some point $q \in V_1 \cap P,$ such that $q \not = p_1$, $d(p_1, q) < d(p_1, x_2)$. Suppose not. Note that $p_1 = x_1$. Therefore, $d(p_1, x_2) = d(x_1, x_2) > 0$. So there is some neighborhood of $p_n$ (namely, the neighborhood with radius $r = d(p_1, x_2))$, such that the only point of that neighborhood in $P$ is the point $p_1$, which contradicts our assumption that $p_1$ is a limit point of $P$.

Let $V_{n+1} = N_{r_{n+1}}(p_{n+1})$ with $r_{n+1}$ chosen subject to the following conditions.

(1.) $r_{n+1} \le d(p_{n+1}, x_n)$, and

(2.) $r_{n+1} < r_n - d(p_n, p_{n+1})$.

By our choice of $p_{n+1}, r_{n+1}$, and $V_{n+1} = N_{r_{n+1}}(p_{n+1})$ we have the following:

(I) **$V_{n+1}$ satisfies (i)**:

If $y \in \overline{V_{n+1}}$, then

$d(p_n, y)$

$\le d(p_n, p_{n+1}) + d(p_{n+1}, y)$ [by the properties of a metric space]

$\le d(p_n, p_{n+1}) + r_{n+1}$

$< d(p_n, p_{n+1}) + r_n - d(p_n, p_{n+1})$ [by our choice of $r_{n+1}$]

$= r_n$.

Thus, $y \in V_n$. Hence, $V_{n+1}$ satisfies (i).

(II) **$V_{n+1}$ satisfies (ii)**:

If $y \in V_{n+1}$, then $d(p_n, y) < r_{n+1} \le d(p_{n+1}, x_n)$. Thus, $x_n \not \in V_{n+1}$. Hence, $V_{n+1}$ satisfies (ii).

(III) **$V_{n+1}$ satisfies (iii)**:

Because $p_{n+1}$ was chosen to be in $P$, we have that $N_r(p_{n+1}) \cap P$ is non-empty for all neighborhoods of $p_{n+1}$. Thus, $V_{n+1}$ satisfies (iii).

Let $K_n = \overline{V_n} \cap P$. Since $\overline{V_n}$ is closed and bounded in $\mathbb{R^k}$, $\overline{V_n}$ is compact (by 2.41). $P$ is closed (because $P$ is perfect). Thus, $\overline{V_n} \cap P$ is closed (by 2.24(b)). Hence $K_n$ is compact (by 2.35, because $K_n$ is a closed subset of a compact set).

Since $x_n \not \in K_{n+1}$, no point of $P$ lies in $\cap_{1}^{\infty} K_n$ (this is implied by the fact that $P$ is countable, hence for every $x_i \in P$, there is a $K_{i+1}$ that excludes $x_i$ from $\cap_{1}^{\infty} K_n$). But $K_n \subset P$, so 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)). But this contradicts the corollary to 2.36. The theorem follows. $\blacksquare$

**Lemma 1: $y$ is a limit point of $V_1$ if and only if $|y - x_1| = r$ or $y \in V_1$. Proof Sketch:** suppose $|y - x_1| > r$. Then $|y - x_1| = r + \epsilon$ for some $\epsilon > 0$. So $N_{\epsilon/2}(y) \cap V_1 = \emptyset$ and $y$ is not a limit point of $V_1$. Now suppose $|y - x_1| = r$. Then every neighborhood of $y$ contains a point in $V_1$, so $y$ is a limit point of $V_1$).