- Thread starter
- #1

I believe this has to do with always taking a 3rd away.

- Thread starter dwsmith
- Start date

- Thread starter
- #1

I believe this has to do with always taking a 3rd away.

- Moderator
- #2

- Feb 7, 2012

- 2,753

At the first stage of the construction, you remove the open interval $(1/3,2/3)$, or in ternary notation $(0.1,0.2)$. That removes all the numbers that have a 1 as the first digit in their ternary expansion. The left-hand endpoint of the interval, the point 0.1, is not removed. At first sight, it looks as though this fails to remove a point with a 1 as the first digit in its ternary expansion. However, the point 0.1 can also be represented as 0.022222... (recurring). So this point can in fact be represented without any 1s in its ternary expansion, and we have only removed those points in the unit interval which

I believe this has to do with always taking a 3rd away.

Similarly, the second stage of the Cantor construction removes all those points in the unit interval which

Thus the Cantor set consists of all those points in the unit interval which can be represented without a 1 anywhere in their ternary expansion.