Proof related to the Cantor set I was wondering if anyone knew a proof for the fact that the endpoints of each of the divisions in the Cantor set is everywhere dense in the Cantor set? I understand it intuitively, but the rigorous proof escapes me.
And how does one prove that x + y, where x and y are elements of the Cantor set, fill the whole interval [0,2]?
Thanks in advance for any help!
 A: The first question:
A subset $A$ is dense in $X$ if every point in $X$ either belongs to $A$ or is a limit point of $A$.  
Thus the set of $E$ Cantor set endpoints is dense in the Cantor set $C$ if every 
point in the cantor set is either an endpoint or a limit point of $E$.
Expressing this in the language of ternary expansions of real numbers (decimals, for people who have worked in a sawmill and lost seven fingers), $E$ is $0$, $1$ union-ed with the set of terminating decimals which have no $1$s before the last digit.  And $C$ is the set of those plus non-terminating decimals containing no $1$s.
Now consider any point $t$ in $C$ with ternary expansion $0.t_1t_2t_3t_4\cdots$, and this sequence of points in $E$:
$$0.t_1, \,\,0.t_1t_2,\,\, 0.t_1t_2t_3,\,\, 0.t_1t_2t_3t_4,\,\,\cdots$$
This sequence demonstrates that $t$ is a limit point of $E$.
Special case arguments are needed for the endpoints $1$ and $0$ but these are also easy.
Your second question has a constructive proof, choosing the two numbers $x,y$ one ternary digit at a time among $0$ and $2$, in such a way that the ternary digits of the desired real number are attained.  You have to be a bit careful not to devise a proof that relies on the axiom of countable choice.
