# Proof of Baby Rudin Theorem 2.43

I came up with the following fleshed out proof of Theorem 2.43 from Principles of Mathematical Analysis.

Although this is a duplicate of several other questions, I found many of the other proofs of 2.43 on StackExchange verbose or unclear (and I didn't see this proof covered in the Harvey Mudd or Scripps YouTube lectures), so I hope that this version of the proof can serve as a more helpful reference to anyone else who gets stuck.

At the same time, I'm not confident that my reasoning is sound, so I hope that if there are any major flaws or oversights in my proof that a critical reader can point them out, or confirm that I'm correct about property (iii).

Theorem 2.43

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.

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 $$V_1$$).

• Do you have a question? Mar 3, 2019 at 17:40
• Yes, I'm unclear about property (iii) and am hoping someone can either confirm it or point out why it is wrong. Mar 3, 2019 at 17:52
• It looks fine to me. One way you can clarify your construction is to be more explicit about the induction step, regarding what objects are relevant to the induction. In particular: you are inductively constructing a sequence of point $p_n \in P$, a sequence of real numbers $r_n > 0$, and a sequence of open intervals $V_n = N_{r_n}(p_n)$. You were rather vague at first about the nature of the neighborhoods $V_n$, making your proof hard to follow. Mar 3, 2019 at 18:11
• Thank you for taking the time to read this over. Mar 4, 2019 at 8:37
• I have not studied your work... By a method resembling the construction of the Cantor set $C$ we can show that if $P$ is a non-empty complete metric space with no isolated points then $P$ has a subspace $E$ homeomorphic to $C$. (Although when we reach the part in the proof where we see that the constructed set $E$ has the cardinal of $\Bbb R$ we can, for the purposes of your Q, stop without showing that $E$ is homeomorphic to $C$.) Jun 24, 2019 at 8:39

Let $$(X,d)$$ be a complete metric space and let $$P$$ be a non-empty closed subspace of $$X$$ with no isolated points. Then P is uncountable.

Notation: $$B_d(x,r)=\{y:d(x,y)

By contradiction suppose $$P=\{x_n:n\in \Bbb N\}.$$

Let $$f(1)=r(1)=1$$ and let $$S(1)=B_d(p_{f(1)},r(1)\,)=B_d(x_1,1).$$

For $$n\in \Bbb N$$ inductively presume that $$f(n)\in \Bbb N,$$ that $$r(n)>0$$, and that $$S(n)=B_d(x_{f(n)},r(n)\,).$$ Let $$f(n+1)=\min \{m>f(n): x_{f(n)}\ne x_m\in S(n)\}.$$ Note that $$f(n+1)$$ exists because $$S(n)\cap P$$ is an infinite set.

Then let $$r(n+1)=\frac {1}{2}\min (\,d(x_{f(n)},x_{f(n+1)}),\, r(n)-d(x_{f(n)},x_{f(n+1)})\,)$$

and let $$S(n+1)=B_d(x_{f(n+1)},r(n+1)\,).$$

Observe that $$f:\Bbb N\to \Bbb N$$ is strictly decreasing and that $$x_{f(n)}\not \in\overline {S(n+1)}\subset S(n)$$ for each $$n\in \Bbb N.$$

Now $$d(x_{f(n)},x_{f(n+1)})<2^{-(n-1)}$$ so $$(x_{f(n)})_{n\in \Bbb N}$$ is a $$d$$-Cauchy sequence of members of $$P$$. Since $$P$$ is closed and $$d$$ is complete there exists $$k\in \Bbb N$$ such that this sequence converges to $$x_k.$$ For each $$n\in \Bbb N$$ we have $$\{x_{f(m)}:m>n\}$$ $$\subset$$ $$\overline {S(n+1)}$$ so $$x_k\in \overline {S(n+1)}.$$ But $$x_{f(n)}\not \in \overline {S(n+1)}.$$

So $$x_k\ne x_{f(n)}$$ and $$k\ne f(n)$$ for each $$n.$$

Now $$k\ne f(1)=1$$ so $$k>f(1)$$, and $$f$$ is strictly increasing. So let $$n_0=\max \{n\in \Bbb N: k>f(n)\}.$$

We have $$f(n_0)

And we have $$x_k\in \overline {S(n_0+1)}\subset S(n_0)$$ and $$x_k\ne x_{f(n_0)}$$

so $$k\in \{m>f(n_0): x_{f(n_0)}\ne x_m\in S(n_0)\}$$

so $$k\ge \min \{m>f(n_0):x_{f(n_0)}\ne x_m\in S(n_0)\}=f(n_0+1)$$ by def'n of $$f(n_0+1).$$

So $$k\ge f(n_0+1),$$ contradictory to $$f(n_0)

• This method was used by Cantor for a particular case. That is, to prove $\Bbb R$ is uncountable Jun 24, 2019 at 17:11

I was also kind of curious, as an exercise, to see why this proof would not apply in the same way to $$P = \mathbb{Q} \cap [0,1]$$ and (wrongly) suggest that $$P$$ in that case would be uncountable. The rational numbers are dense among the real numbers, so for any point in $$\mathbb{Q} \cap [0,1]$$, and for any radius, we can find an infinite number of other points in $$\mathbb{Q} \cap [0,1]$$ to select as $$p_{n+1}$$, and we have a lot of freedom in our choice of $$r_{n+1}$$.

I believe that the place where the proof breaks in this case is here:

In the original proof we say,

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).

However, in the case where $$P = \mathbb{Q} \cap [0,1]$$, it is no longer accurate to say that $$P$$ is closed. In fact, there are many limit points of $$\mathbb{Q} \cap [0,1]$$ that are not in $$\mathbb{Q} \cap [0,1]$$ (for example, $$\sqrt{2}/2$$, or any irrational number in $$[0,1]$$ for that matter). Thus, we cannot assume that $$K_n$$ is compact, and the contradiction does not follow.