I'm not saying I understand what this is saying, but it might point you in the right direction.
Thanks, I think I figured out how to prove it. It turns out that I should have mentioned something earlier. This something is the actual definition of the cantor set. Here it is. Let T=[0,1]. Then, remove the midle 3rd open segment to form the set
= [0,1/3]U[2/3,1]. Then, remove the middle 3rd open segement in each of those intervals whose union forms to get . Continue this k times to get . Then, define the cantor set as .
You have to proceed by induction. Writing x=1/3 as 0.022222222222222222222222222222222..., to handle the endpoints. You suppose that x is in the cantor set, and that the nth digit of x is either 0 or 2. Then, x is in . Then, the kth dight of x sort of tells you which 3rd of a segment from x is in. It dosen't matter which one, as long as it's either the first or the 2nd. On the k+1th iteration, the k+1th digit must be either 0 or 2 since it's either in the first segment or the last segment of whatever segment it was in before. Hence, x only has 0s or 2s in its base 3 expansion. Then, you do something similar to prove the converse