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\LaTeX.

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

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

Suppose PP is countable and denote the points of PP by x1,x2,x3,x_1, x_2, x_3, \dots

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

Let V1V_1 be any neighborhood of x1x_1. Then V1PV_1 \cap P is non-empty, because x1x_1 is a limit point of PP.

If V1V_1 consists of all yRky \in \mathbb{R}^k such that yx1<r|y-x_1| < r, the closure V1\overline{V_1} of V1V_1 is {yyx1r}\{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 VnV_n has been constructed so that VnPV_n \cap P is not empty and Vn=Nr(p)V_n = N_r(p) for some pPp \in P. Since every point of PP is a limit point of PP, there is a neighborhood Vn+1V_{n+1} such that

(i) Vn+1Vn\overline{V_{n+1}} \subset V_n,

(ii) xn∉Vn+1x_n \not \in \overline{V_{n+1}},

(iii) Vn+1PV_{n+1} \cap P is not empty, and Vn+1=Nr(p)V_{n+1} = N_r(p) for some point in PP.

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

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

Just a quick note: it's a common point of confusion that people believe that xnx_n must be in VnV_n. While x1V1x_1 \in V_1, we cannot assume that this holds true for any other xnx_n, and at no point does the proof assume this (or need to assume this).

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

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

Lemma 3: For n=1n=1, there exists some point qV1P,q \in V_1 \cap P, such that qp1q \not = p_1, d(p1,q)<d(p1,x2)d(p_1, q) < d(p_1, x_2). Suppose not. Note that p1=x1p_1 = x_1. Therefore, d(p1,x2)=d(x1,x2)>0d(p_1, x_2) = d(x_1, x_2) > 0. So there is some neighborhood of pnp_n (namely, the neighborhood with radius r=d(p1,x2))r = d(p_1, x_2)), such that the only point of that neighborhood in PP is the point p1p_1, which contradicts our assumption that p1p_1 is a limit point of PP.

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

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

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

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

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

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

d(pn,y)d(p_n, y)

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

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

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

=rn= r_n.

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

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

If yVn+1y \in V_{n+1}, then d(pn,y)<rn+1d(pn+1,xn)d(p_n, y) < r_{n+1} \le d(p_{n+1}, x_n). Thus, xn∉Vn+1x_n \not \in V_{n+1}. Hence, Vn+1V_{n+1} satisfies (ii).

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

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

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

Since xn∉Kn+1x_n \not \in K_{n+1}, no point of PP lies in 1Kn\cap_{1}^{\infty} K_n (this is implied by the fact that PP is countable, hence for every xiPx_i \in P, there is a Ki+1K_{i+1} that excludes xix_i from 1Kn\cap_{1}^{\infty} K_n). But KnPK_n \subset P, so this implies that 1Kn\cap_{1}^{\infty} K_n is empty.

But each KnK_n is nonempty (by (iii), and KnKn+1K_n \supset K_{n+1} (by (i)). But this contradicts the corollary to 2.36. The theorem follows. \blacksquare

Lemma 1: yy is a limit point of V1V_1 if and only if yx1=r|y - x_1| = r or yV1y \in V_1. Proof Sketch: suppose yx1>r|y - x_1| > r. Then yx1=r+ϵ|y - x_1| = r + \epsilon for some ϵ>0\epsilon > 0. So Nϵ/2(y)V1=N_{\epsilon/2}(y) \cap V_1 = \emptyset and yy is not a limit point of V1V_1. Now suppose yx1=r|y - x_1| = r. Then every neighborhood of yy contains a point in V1V_1, so yy is a limit point of V1V_1).

James Shapiro, 2021