Let $A,B\in\mathbb{R}$ be two sets and $\lambda \in\mathbb{R}$ a scalar. We define the following operations: $$A+B\triangleq \{a+b \mid a\in A, b\in B\}$$ $$\lambda A \triangleq \{\lambda a \mid a\in A \}$$ We define a sets sequence $C_0 , C_1 , ... , C_n ,...$: $$C_0=[0,1]\\C_n=\frac{1}{3}C_{n-1}\cup\bigg(\frac{2}{3}+\frac{1}{3}C_{n-1} \bigg)$$ And the cantor set is defined: $C=\cap_{i=0}^{\infty}C_i$.
Show that the Cantor set is self-similar in the following sense: $$C=\frac{1}{3}C \cup \bigg(\frac{2}{3}+\frac{1}{3}C\bigg)$$
My approach is to show that $C \subseteq \frac{1}{3}C \cup \bigg(\frac{2}{3}+\frac{1}{3}C\bigg)$ and $\frac{1}{3}C \cup \bigg(\frac{2}{3}+\frac{1}{3}C\bigg) \subseteq C$, thus proving the identity.
I was able to show that $C \subseteq \frac{1}{3}C \cup \bigg(\frac{2}{3}+\frac{1}{3}C\bigg)$, but not the other way. I get stuck right from the start:
Let $c \in\frac{1}{3} C \cup \bigg(\frac{2}{3}+\frac{1}{3}C\bigg)$. Meaning, $c \in \frac{1}{3}C$ or $c \in \bigg(\frac{2}{3}+\frac{1}{3}C\bigg)$.
Case 1: $c \in\frac{1}{3}C$. So there exists $c'\in C$ s.t. $c=\frac{1}{3}c'$. Why does $c\in C$?
Any kind of help would be much appreciated.
Let $c \in \dfrac13C$. Then $3c \in C$.
Now, we need to prove that $c \in C$, which is for every $k$, $c \in C_k$.
We know that for every $m$, $3c \in C_m$. Now let $m=k-1$: $3c \in C_{k-1}$.
$C_k = \dfrac13C_{k-1} \cup \left(\dfrac23 + \dfrac13C_{k-1}\right) \supseteq \dfrac13C_{k-1} \ni \dfrac13(3c) = c$.
Therefore, $c \in C$ is proved.
A similar method can be used to prove case $2$.