Sum of Cantor Sets Let C denote the usual cantor set(which is obtained from the interval [0,1]).Then what is the sum of three such cantor sets i.e. what is C+C+C=?
 A: Well, $C+C=[0,2],$ so it follows readily that $C+C+C=[0,3],$ by noting that $$[0,3]=[0,2]+\{0,1\}\subseteq[0,2]+C=C+C+C\subseteq[0,3].$$
As a hint for how to prove the first result, you should first prove that $\frac12C+\frac12C=[0,1],$ using the fact that the elements of $C$ are those that can be expressed as sums $$\sum_{k=1}^\infty\frac{t_k}{3^k},$$ where each $t_k\in\{0,2\}.$ On the other hand, every number in $[0,1]$ can be expressed as such a sum where $t_k\in\{0,1,2\}$.

Added: To prove the first result, we make use of a few properties of set addition and scalar multiplication, which we define for $\alpha\in\Bbb R$ and $X,Y\subseteq\Bbb R$ by
$$X+Y:=\{x+y:x\in X,y\in Y\}$$ and $$\alpha X:=\{\alpha x:x\in X\}.$$
In particular, we have the following:

*

*For any $D,E\subseteq\Bbb R$ and any $\alpha\in\Bbb R,$ we have $\alpha D+\alpha E=\alpha(D+E).$

*For any $B,F\subseteq R$ and any non-zero $\alpha\in\Bbb R,$ we have $\alpha B=\alpha F$ if and only if $B=F.$

*For any $D,E,F\subseteq\Bbb R$ and any non-zero $\alpha\in\Bbb R,$ we have $\alpha D+\alpha E=\alpha F$ if and only if $D+E=F.$
The third property follows directly from the first two. (Do you see how?) Let me prove the first one. (I leave the proof of the second to you.)

On the one hand, suppose that $x\in\alpha D+\alpha E,$ meaning that $x=y+z$ for some $y\in\alpha D$ and some $z\in\alpha E.$ But then $y=\alpha s$ for some $s\in D$ and $z=\alpha t$ for some $t\in E.$ Then $s+t\in D+E$ and $x=y+z=\alpha s+\alpha t=\alpha(s+t),$ so $x\in\alpha(D+E).$ Since this holds for all $x\in\alpha D+\alpha E,$ then $\alpha D+\alpha E\subseteq\alpha(D+E).$
On the other hand, suppose that $x\in\alpha(D+E).$ Then $x=\alpha y$ for some $y\in D+E,$ so $y=s+t$ for some $s\in D$ and some $t\in E,$ so $\alpha s\in\alpha D$ and $\alpha t\in\alpha E,$ and $x=\alpha y=\alpha(s+t)=\alpha s+\alpha t,$ so $x\in\alpha D+\alpha E.$ Since this holds for all $x\in\alpha(D+E),$ then $\alpha D+\alpha E\supseteq\alpha(D+E).$

Hence, if we can show that $\frac12C+\frac12C=[0,1],$ then since $[0,1]=\frac12[0,2],$ it follows by the third property that $C+C=[0,2],$ as desired.
