The Cantor set comprises all numbers between 0 and 1 that can be expressed in ternary expansion without the digit 1. The construction process involves sequentially removing open intervals, specifically starting with the interval (1/3, 2/3) in the first stage, which eliminates numbers with a 1 as the first digit in their ternary representation. Subsequent stages continue this pattern, removing points based on their ternary digit positions. Ultimately, the Cantor set includes only those numbers that can be represented solely with the digits 0 and 2 in their ternary expansion.
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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 must have a 1 as the first digit in their ternary expansion.dwsmith said:Explain why the Cantor set consists precisely of all the numbers between 0 and 1 (including 0 and 1) which can be represented by a ternary expansion in which the digit 1 does not appear anywhere in the expansion.
I believe this has to do with always taking a 3rd away.