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: $$\mathcal{C} = \bigcup_{n=0}^{\infty}A_n$$


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.

  • 2
    $\begingroup$ Not every element of the Cantor set is rational. $\endgroup$ – Lord Shark the Unknown Oct 29 '17 at 6:22
  • $\begingroup$ Said differently, your set is not a Cantor set. $\endgroup$ – quasi Oct 29 '17 at 6:23
  • $\begingroup$ Your construction is definitely not the usual one. $\endgroup$ – user99914 Oct 29 '17 at 6:23
  • $\begingroup$ Why do you call the set $C$ you have constructed the "Cantor set", and why do you suppose that it is uncountable? $\endgroup$ – Lord Shark the Unknown Oct 29 '17 at 6:24
  • 2
    $\begingroup$ Nothing is wrong with it. It's just not the Cantor set. Your set is definitely a subset of the set of rational numbers. $\endgroup$ – quasi Oct 29 '17 at 6:25

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

  • $\begingroup$ Thank you for your reference. Now it is clear for me. $\endgroup$ – LRDPRDX Oct 29 '17 at 6:32
  • $\begingroup$ Another question: Is the following true: each element of Cantor set either rational or have periodic ternary expansion. $\endgroup$ – LRDPRDX Oct 29 '17 at 6:38
  • 1
    $\begingroup$ 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 :) $\endgroup$ – Ted Oct 29 '17 at 6:44

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