Notice that like Lavinia said,the terms in the (Standard) Cantor set C have no 1's in
their base-3 expansion. Now try to show,given x in C --so that there are no
1's in the decimal expansion of x -- that, no matter how close you go about x in
(x-e,x +e ) , you will hit a number y in (x-e,x+e) ,whose decimal expansion _does_
have a 1 in it . Hint: you can cut-off the decimal expansion of x at any point,
as far back as you want.