A property of the Cantor set. Let $C$ be the Cantor set. Prove $C-C=[0,1]$ where
$A-B$ is defined to be $\{x-y\,:\, x\in A,\,y\in B\}$.
We can't use the ternary expansion.
 A: You can use self similarity of the Cantor set:
$$C = (C/3) \cup (C/3 + 2/3)$$
this leads to
$$C-C = (C/3) \cup (C/3+2/3) - (C/3) \cup (C/3+2/3) = C/3 \cup (C/3+2/3) \cup (C/3-2/3)$$
Which means that $C-C$ is self similar and especially it has the property that $a,b\in C-C$ implies that the equidistant points $(a+b)/3, (a+2b)/3\in C-C$ too (this follows from $x\in C-C$ implies that $x/3\in C-C$ and $x/3\pm2/3\in C-C$). If you repeat this it follows that $a + j(b-a)3^{-k}\in C-C$ for $k>0$ and $0\le j\le3^k$. 
This means that $C-C$ is dense in $[a,b]$, and since $-1,1\in C-C$ we have that $C-C$ is dense in $[-1, 1]$ and since $C$ is closed then so is $C-C$ and therefore $C-C = [-1,1]$.

You can also use the construction of the Cantor set where you have sets $C_j$ that are constructed by removing middle section and $C=\bigcap C_j$ and that $C_j-C_j = [-1,1]$.
First of all it's obvious that $C-C \subseteq [-1,1]$ (since $C\subseteq[0,1]$). So we have to prove that $[-1,1] \subseteq C$.
Suppose now we have an element $c$ in $[-1,1]$, now since $C_j=[-1,1]$ we have $a_j,b_j\in C_j$ such that $a_j-b_j = c$. Now since $[-1,1]$ is compact and $C_j$ is a chain of subsets we can chose the sequences $a_j$ and $b_j$ to be convergent and set $a=\lim_{j\to\infty}a_j$ and $b=\lim_{j\to\infty}b_j$.
Now we know that $a-b=c$ and the only thing remaining is to show that $a,b\in C$. 
To do that we assume that it is not, this means that there for some $C_j$ happens to be that $a\notin C_j$ (or that $b$ is not). But since $C_j$ is closed you would have an neighborhood of $a$ not intersecting $C_j$ and
$a_k\in C_j$ for $k>j$ which contradicts $\lim_{j\to\infty}a_j=a$.
