I've been watching a lecture and didn't understand why there wouldn't be any point whose neighbourhood is completely surrounded by the cantor set. Oh I think I start to grasp it now, since every interval's "middle" is removed and that process goes on forever, every point's neighbourhood becomes somewhat "incomplete". Am I wrong?
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.
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Thank to all of you. But 1/3 is included in the cantor set and in trinary it's 0.1 isn't it? I see there is no finite trinary number in the set but why is 0.1 included?
edit: Now I think I'm okay. In the trinary number system, 1/3 can be written not only as 0.1 but also as 0.222222...