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.fresh_42 said: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.
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.rmberwin said: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.
rmberwin said: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.