I Understanding the Paradox of the Cantor Set: A Closer Look at Its Derivation

[rmberwin]

I am puzzled by the derivation of the Cantor set. If the iteration of removing the middle-thirds leaves an uncountable set of points, it seems the iteration had to be performed an uncountably infinite number of times. Is this the case? If so, that seems paradoxical to me.

 

[fresh_42]

Why? Remove $\{\frac{1}{2}\}$ from $[0,1]$ and there are uncountably many points left. Remove all $\{\frac{1}{n}\,\vert \,n \in \mathbb{N}\}$ from $[0,1]$ and there are still uncountable many points left. The iteration for the Cantor set goes with $n \in \mathbb{N}$ ergo by countably many steps.

 

[rmberwin]

Why? Remove $\{\frac{1}{2}\}$ from $[0,1]$ and there are uncountably many points left. Remove all $\{\frac{1}{n}\,\vert \,n \in \mathbb{N}\}$ from $[0,1]$ and there are still uncountable many points left. The iteration for the Cantor set goes with $n \in \mathbb{N}$ ergo by countably many steps.

Ah, I see your point. But the Cantor set also has zero measure, which (I assume) means that all the points are disconnected. So I don't see how taking the limit at countable infinity would get to the final result. Probably a failure of imagination on my part.

 

[Mark44]

I am puzzled by the derivation of the Cantor set. If the iteration of removing the middle-thirds leaves an uncountable set of points, it seems the iteration had to be performed an uncountably infinite number of times. Is this the case? If so, that seems paradoxical to me.

As @fresh_42 said, the removals happen a countably infinite number of times. If you follow through what happens, you are removing 1/3, then 2(1/9), then 4(1/27), and so on. You are removing a set of intervals whose combined length is $\frac 1 3 + \frac 2 9 + \frac 4 {27} + \frac 8 {81} + \dots$. In closed form, this is $\sum_{n = 0}^\infty \frac {2^n}{3^{n + 1}} = \frac 1 3 \sum_{n = 0}^\infty \frac {2^n}{3^n}$, a convergent geometric series that converges to 1. In essence, you are removing a set of measure 1 from an interval of the same length.

 

[WWGD]

Ah, I see your point. But the Cantor set also has zero measure, which (I assume) means that all the points are disconnected. So I don't see how taking the limit at countable infinity would get to the final result. Probably a failure of imagination on my part.

Notice you are removing uncountably many points in each step. EDIT, yes, the set is totally-disconnected,
meaning singletons are the components. Assume your set was connected. Connected EDIT (plus open ), in the Reals implies path-connected. This means there is a path joining two points in the set. This path is a sub(interval) , say (a,b) with measure m(a,b)=b-a >0. So measure zero, by contraposition, implies totally-disconnected. EDIT2: You can also argue, using ternary representation , that, given any point c in the Cantor set, that points will be removed about any open set containing c, and no remaining 'hood will be open.