Problem 1

Exercise 1.1 (Perfect sets are uncountable)   Let $A\subseteq \mathbb{R}$ be closed and perfect, and nonempty. Then $A$ is uncountable.

Proof. It is first easy to show that $A$ is at least countable. Let $x\in A$ and select $x_1\in B_1(x)$. For each $n=1,2,\dots$, select

$$x_{n+1} \in (A\cap B_{\epsilon_n}(x))\setminus\{x\} \quad \epsilon_n = d(x_n,x)$$ (1.1)

This definition implies $d(x_n,x_m)>0$ for $n\neq m$. Suppose that $d(x_n,x_m)=0$, then

$$\epsilon_n = d(x_n, x) \leq d(x_n,x_m) + d(x_m,x) = \epsilon_m$$ (1.2)

$$\epsilon_m = d(x_m, x) \leq d(x_m,x_n) + d(x_n,x) = \epsilon_n$$ (1.3)

Therefore, $\epsilon_n=\epsilon_m$, indicating $n=m$, since $\epsilon$ strictly decreases. Therefore, $A$ is at least countable.

As a closed subset of $\mathbb{R}$, we know $A$ is a complete metric space, so an application of the Baire Category Theorem reveals that

$$A \neq \bigcup_{n=1}^\infty \{y_n\}$$ (1.4)

for any sequence $y_n$, so that $A$ is uncountable. $\qedsymbol$