MHB Cantor Set Numbers: Explaining the Unique Ternary Expansion

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The Cantor set includes all numbers between 0 and 1 that can be expressed in ternary form without the digit 1. The construction process begins by removing the open interval (1/3, 2/3), eliminating numbers with 1 as the first digit in their ternary expansion. Subsequent stages continue this pattern, removing points with 1 as the second, third, and further digits. Notably, numbers like 0.1 can still be represented without 1s, such as 0.022222..., which remains in the set. Consequently, the Cantor set is defined by its unique ternary expansions devoid of the digit 1.
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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.
 
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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.
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.

Similarly, the second stage of the Cantor construction removes all those points in the unit interval which must have a 1 as the second digit in their ternary expansion, and so on.

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.
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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