Reading this, I wanted to do the classic demonstration again by myself but there are points that bother me.
Let $$C_0=[0,1], C_1=[0,\frac13]\cup[\frac23,1]...$$We have the classical definition of the Cantor set $$\mathbb K_3:=\bigcap_{n=0}^{+\infty }C_n$$ and we know that $$\mathbb K_3=\{x\in [0,1]:x=\sum_{n=1}^{+\infty}\frac{a_n}{3^n} (\forall i , a_i\in \{0,2\})\}$$ I want to prove that $\mathbb K_3$ is the attractor $\mathcal A$ of $\{f_1,f_2\}$, where $$f_1:[0,1]\to[0,1], x\mapsto \frac13 x \text{ and }f_2:x\mapsto \frac13(x+2)$$
I know that
- $(\mathbb R,d)$ is a metric space, with $d(x,y):=|x-y|$. $\mathbb R$ is complete by construction;
- $\forall x,y \in \mathbb R, |f_2(x)-f_2(y)|\leq \frac13|x-y|$(the same for $f_1$)
- $\mathcal A$ is the unique compact of $\mathbb R$ such that $\mathcal A=f_1(\mathcal A)\cup f_2(\mathcal A) \color{red}{(*)}$;
- $\color{blue}{\text{I don't remember how to prove that K3 is compact}} $;
- Using $\color{red}{(*)}$, it is then easy to prove that $$\boxed{\mathbb K_3=\mathcal A(\{f_1,f_2\})}.\square$$ I'd love to hear from you about this demonstration and help me complete it. Thanks in advance.
Yes, the Cantor set is compact by the Heine–Borel theorem, as it's a closed and bounded subset of $[0, 1] \subset \mathbb{R}$ under the standard topology. Boundedness is obvious, and the fact that it's closed is most easily seen by observing that its complement is a union of open intervals (the "middle thirds").
You can read more about its topological properties here.