Let C be the Cantor set. Let 3C = {3x : x ∈ C}, and prove that 3C ∩ [0, 1] = C. I understand that theres a bijection between the Cantor set and the interval [0,1], but I'm stuck on how to interpret 3C. 3C does contain C, but how do i prove that C is the only intersection?
 A: The Cantor set is the union of the following sets.
$$\{0,1\}, \{0,1/3,2/3,1\},\{0, 1/9,2/9,3/9,6/9,7/9,8/9,1\},...$$
When you multiply each set by $3$ and intersect it with $[0,1]$, it is the previous one.
Thus $$3C\cap [0,1] \subseteq C$$
A: Define$$C_0:=(0,\,1),\,C_{n+1}:=\tfrac13C_n\cup\left(\tfrac23+\tfrac13C_n\right)$$so$$C_n=3C_{n+1}\cap[0,\,1],\,C=\bigcap_{n\ge0}C_n=3\bigcap_{n\ge0}C_n\cap[0,\,1]=3C\cap[0,\,1].$$
A: The Cantor set $C$ is equal to $\bigcap_{n\in\mathbb Z^+}\bigcup_{k=1}^{2^n}I_{n,k}$, where:


*

*$I_{0,1}=[0,1]$;

*$I_{1,1}=\left[0,\frac13\right]$ and $I_{1,2}=\left[\frac23,1\right]$;

*$I_{2,1}=\left[0,\frac19\right]$, $I_{2,2}=\left[\frac29,\frac13\right]$, $I_{2,3}=\left[\frac23,\frac79\right]$, and $I_{2,4}=\left[\frac89,1\right]$
and so on.
But:


*

*$3I_{0,1}\cap[0,1]=[0,1]$

*$3\left(I_{1,1}\cup I_{1,2}\right)\cap[0,1]=[0,1]$;

*$3\left(I_{2,1}\cup I_{2,2}\cup I_{2,3}\cup I_{2,4}\right)\cap[0,1]=I_{1,1}\cup I_{1,2}$
and so on.
Can you take it from here?
A: Since $C$ is only defined on the interval $[0, 1]$, you've so far shown that $C \subseteq 3C \cap [0, 1]$. Then all you have to do is take some element $x \in 3C \cap [0, 1]$ (what might it look like?), and show that it's in $C$!
A: A quick and dirty proof is that if $0 \le x \le 1$ has a ternary expansion containing no "1"s, then it's in $C$ (and vice-versa). Multiplying such a number by $3$ moves its ternary expansion to the left by one place. The result is either a number bigger than $1$ (in which case you chuck it out) or another number whose ternary expansion contains no $1$s. That proves that $3C \cap [0, 1] \subset C$. 
Now suppose that $x \in C$; it has a ternary expansion containing no ones. Shift that ternary expansion one place to the right (by prepending a $0$, so that, for instance, 
.20220002 ->
.020220002

The resulting number $y$ (which is just $x/3$) clearly also has a ternary expansion containing no $1$s, hence is in $C$. That shows that the map $x \mapsto 3x$ has the entire cantor set in its image, i.e, that $C \subset 3C$. So we're done. 
(Of course, all this relies on a bunch of small theorems about convergence of infinite series, and one the claim that the cantor set consists of numbers of the form I've described, so it's hardly a self-contained proof.)
