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Note I am not asking about a proof that Cantor set is uncountable. I want to get some "natural" answer on my question (see below).

A Construction of the Cantor set

Let us change a bit the famous construction of Cantor set. Instead of removing intervals we will add points at each iteration. So we build Cantor set by induction:

  1. At the first iteration $n=0$ there is a set $A_0 = \{0,1\}$
  2. After $A_n$ has been constructed $A_{n+1}$ is obtained as the folllowing: $$A_{n+1} = A_n\cup \{a_i+\frac{(-1)^{3^{n-1} a_i}}{3^n}\,|\, a_i \in A_n\}$$

By the above costruction Cantor set is the union of all $A_n$'s: $$\mathcal{C} = \bigcup_{n=0}^{\infty}A_n$$

Question

From this construction the following statement arises $$\xi\in\mathcal{C}\Rightarrow \xi \in \mathbb{Q}$$ So if each element in $\mathcal{C}$ is rational how it (set) is uncountable?

P.S. I am not sure that my construction of Cantor set is absolutely correct. I will appreciate any ideas, advices, corrections etc.

Community
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LRDPRDX
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2 Answers2

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You seem to think the only elements in the Cantor set are the endpoints of the intervals you have deleted. This is not true.

Ted
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  • Thank you for your reference. Now it is clear for me. – LRDPRDX Oct 29 '17 at 06:32
  • Another question: Is the following true: each element of Cantor set either rational or have periodic ternary expansion. – LRDPRDX Oct 29 '17 at 06:38
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    Your question doesn't make sense. If a number has periodic ternary expansion, then it is rational. In any case, I suggest you read the linked question and further links from there if necessary. Everything you could want to know about the Cantor set is there :) – Ted Oct 29 '17 at 06:44
  • Reading this after all these years makes me smile :) – LRDPRDX Jan 07 '24 at 04:36
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I just upvoted this old question because I think it has much more merit than it seems at first glance.

I know we are not supposed to use this space to voice an opinion, and I'm not even a mathematician, but please let me elaborate. This fits better as an answer than as a comment. Besides, the OP did ask for "ideas, advice, corrections".

Contrary to some comments, I believe the proposed construction may very well be a valid alternative construction of the Cantor set. As such, whether it is true or not, it is an interesting idea. IMO, the proposition cannot be easily dismissed without proof. It is known that unexpected things can happen when the limit is applied to a seemingly simple and predictable process.

As the formula is stated, there are minor formal errors in the indexing, and the final union should not be needed as the union is already embedded in each iteration step. So you could say simply $$C_{alt}=\lim A_n=A_\infty.$$

Answering the question:

The error in your question is "each element in $C$ (or $C_{alt}$) is rational". It isn't, for the same weird reasons as in the normal construction. Each iteration $A_n$ involves a finite amount of rational numbers, yet the application of the limit process brings out an uncountable number of irrational numbers.

I think one can look at it in this way: the infinite iteration of $A_n$ bifurcates into an infinite number of simultaneous infinite sequences of rational numbers. It is known that the limit of any one such sequence can be irrational.

Finally here is a slightly more formal argument that the construction might actually be correct, and should in any case be uncountable: Each iteration doubles the size (cardinality) of the set $A_n$. Therefore $C_{alt}=A_\infty$ has cardinality $(\lim 2^n)$ which is indeed uncountable. And being uncountable, $C_{alt}$ cannot contain only rational numbers.

Anders H
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  • The endpoints are all rational numbers because they have a finite number of digits when expressed in base 3 notation. Sure, the $n$th level has twice as many points as the previous level because the added ternary digit is 0 or 2, but the number of points at each level is still finite, so this construction produces a countable number of points. Your argument about $\lim 2^n$ is bogus as you could same the same thing about $|\mathbb{N}|$ being $\lim 10^n$ by writing the numbers in decimal with each added digit giving ten times as many numbers. It is still countable because each level is finite. – Jaap Scherphuis Jan 05 '24 at 11:02
  • Jaap, I disagree with your last argument. We are looking at cardinality here, and $2^{\aleph_0}$ is uncountable. At the infinity level you would have an infinite number of strings of infinite number of digits. You might as well write your numbers after the decimal point: at each step you get only ten times as many rational numbers as before, but in the end you get all the real numbers (edit: between 0 and 1). Same thing. – Anders H Jan 07 '24 at 00:09
  • No. Let $A_n$ be the set of numbers with $n$ digits after the decimal point. Each of these sets contains only rationals and is countable. The union of these sets also contains only rationals and is also countable. While it contains for example $0.3$, $0.31$, $0.314$, $0.3141$, …, which are rational approximations to $\pi/10$, it does not contain $\pi/10$. It has cardinality $\aleph_0$, not $2^{\aleph_0}$. – Jaap Scherphuis Jan 07 '24 at 08:42
  • Jaap, I'm mot a mathematician and I should back down, but your argument is confusing me. If you are right, then please explain to me why the "standard" construction of the Cantor set is working. To me, it is a set iteration of exactly the same type. At each step some intervals are removed, leaving a finite number of intervals bounded by rat nums. In the limit, this math-magically becomes the Cantor set. ...Could some third person please comment on this? – Anders H Jan 07 '24 at 18:35
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    In the ordinary Cantor set construction, at each iteration you have a finite number of intervals. Those intervals are bounded by rational endpoints, but each interval is of course uncountable, and contains many irrational numbers. Even in the limit, there are many numbers (both rational and irrational) that are not ever removed and are also not endpoints of any interval. For example, the ternary number $0.02020202..._3$ (which is actually $\frac14$) is one. Those extra points make it uncountable. The rational interval endpoints are a countable set which is dense in the uncountable Cantor set. – Jaap Scherphuis Jan 07 '24 at 20:41
  • Thank you for taking the time to explain, Jaap, and I think I technically understand the difference now. I am still reluctant to accept the implications of it, but that could be me in need of maturing with these concepts. I suggest we may end this discussion here. – Anders H Jan 08 '24 at 00:15