Cantor set question Is it true that for any $x, 0<x<1$ there are two points $a,b$ inside the Cantor set such that $a-b=x$?
I suppose that I should use ternary representation, but I can't find a way to do a proper proof.
Can anyone help ?
 A: As an intersection of closed sets $\bigcap_i C_i$ (with $C_0 = [0,1]$, $C_1 = [0,1] \setminus (1/3,2/3)$, etc), the Cantor set $C$ is closed.
Hence if a sequence $(u_n)$ satisfying $u_n \in C_n$ for all $n$ converges, its limit $u$ lies in $C$.
Indeed, the sequence of sets $(C_n)$ is decreasing for the inclusion, so if $u \not\in C$ there would exist a $N$ such that $u \notin C_N$. So there would be an open neighborhood $U$ of $u$ such that $U \cap C_N = \emptyset$, thus $U \cap C_M = \emptyset$ so $u_M \not\in U$ for all $M\geq N$ which contradicts $u_n \rightarrow u$.
So you just have to show by induction that given $x \in [0,1]$, there exists for all $n$ a pair of elements $a_n, b_n \in C_n$ verifying $a_n - b_n = x$. 
Indeed, then you can just use compactness to extract two converging sequences $(a'_n)$ and $(b'_n)$ from $(a_n)$ and $(b_n)$ respectively. You would then have :


*

*$\forall n, \quad a'_n - b'_n = x$

*$\forall n, \quad a'_n, b'_n \in C_n$

*$a'_n \rightarrow a, \quad b'_n \rightarrow b$


Thus, $a-b=x$, and by the argument above you have $a,b\in C$.
