I'm trying to understand the following lemma(See here on googlebooks) on page $37$, Nonmeasurable Sets and Functions, Alexander Kharazishvili(2004):
$\qquad$**Lemma $\bf 1.$** Let $C$ denote the Cantor discontinuum on the unit segment $[0,1]$. Then the set $$C+C=\{x+y:x\in C,y\in C\}$$ coincides with the segment $[0,2]$.
It seems to me the Cantor discontinuum is the Cantor Ternary Set in $\Bbb R$, not a perfect and nowhere dense subset of of $\Bbb R$ in general. But I'm not sure.
In addition, I have no clue why "from the geometric viewpoint, it is almost evident that $$C \times C = \bigcap\{Z_n : n < \omega\}$$ $\ldots$"
Here's the proof:
Proof. We can use the standard geometric argument presented, for example, in $[155]$. Namely, let us introduce a mapping $$\phi:\bf\mathbf R\times\mathbf R\to\mathbf R$$ defined by $$\phi(x,y)=x+y\qquad(x\in\mathbf R,\:y\in\mathbf R).$$ This mapping can be described in another way. Denoting $$l=\{(x,y)\in\mathbf R\times\mathbf R:x+y=0\},$$ we see that $\phi$ is identical with the projection $$pr_l:\mathbf R\times\mathbf R\to\mathbf R\times\{0\}$$ whose direction is determined by the straight line $l$. Now, from the geometric viewpoint it is almost evident that $$C\times C=\cap\{Z_n:n\lt\omega\},$$ where $\{Z_n:n\lt\omega\}$ is some decreasing (with respect to inclusion) sequence of compact subsets of the unit square $[0,1]\times[0,1]$ and, in addition, the equality $$pr_l(Z_n)=[0,2]$$ holds for any natural number $n$. This fact readily implies, by virtue of the compactness of all sets $Z_n$, that the relation $$pr_l(C\times C)=[0,2]$$ holds true, too, which is equivalent to the relation $C+C=[0,2]$. The lemma has thus been proved.
