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

In summary, the Cantor set is derived by removing the middle-thirds of an interval an infinite number of times, resulting in an uncountably infinite number of points being removed. Despite this, the set is still totally-disconnected and has measure zero, as each step only removes a set of measure 1.
  • #1
rmberwin
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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.
 
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  • #2
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.
 
  • #3
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.
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.
 
  • #4
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.
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.
 
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  • #5
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.

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.
 
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1. What is the Cantor set?

The Cantor set is a set of real numbers that is created by repeatedly removing the middle third of a line segment. It was first introduced by German mathematician Georg Cantor in the late 1800s and has since become an important example in fractal geometry and topology.

2. How is the Cantor set derived?

The Cantor set is derived by starting with a line segment and removing the middle third of it. Then, the same process is repeated for the remaining two line segments, and this process continues infinitely. The resulting set is the Cantor set.

3. What are the properties of the Cantor set?

The Cantor set is a self-similar set, which means that it has the same structure at every scale. It is also a perfect set, meaning that it is closed, has no isolated points, and has a Lebesgue measure of zero. Additionally, the Cantor set is uncountable, meaning that it contains an infinite number of points.

4. What are some applications of the Cantor set?

The Cantor set has many applications in mathematics and other fields. It is an important example in fractal geometry, and its self-similar structure can be seen in various natural phenomena. It also has applications in computer science, such as in data compression and generating pseudorandom numbers.

5. Are there any variations of the Cantor set?

Yes, there are variations of the Cantor set that involve removing different portions of the line segment, such as removing the middle fourth or middle fifth. These variations can also be extended to higher dimensions by removing portions of a cube or other shapes. Additionally, there are generalized Cantor sets that involve removing more than one portion of the line segment in each iteration.

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