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Let $$C_0 = \left\{ 0.0 \right\}$$

Define, for $$n∈ℕ : n> 0$$ subsequently:

$$C_n = \left(C_{n-1} + \frac{2}{3^n} \right) \bigcup C_{n-1}$$

Let $C$ be the Cantor Set, then

$$C = \bigcup_{n=1}^\infty C_n $$

My concern is that members like $0.1$ are never defined but $0.0222\dots$ might be, which is kind of the same thing.

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    $\begingroup$ The set you obtain this way is countable. $\endgroup$
    – Andrés E. Caicedo
    Apr 18, 2017 at 14:11
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    $\begingroup$ You are correct that your set will not contain all of the numbers. $0.1=0.222\dots$ (ternary, of course) is in the Cantor set, but not in any of the $C_n$. More generally, any number which has infinitely many $2$s in the ternary expansion will be in $C$ but not in $C_n$. $\endgroup$
    – Wojowu
    Apr 18, 2017 at 14:12

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As Wojowu pointed out in the comments, your $C$ is missing all of the points with infinitely many $2$'s in their decimal expansion. However, your $C$ is certainly dense in the Cantor set - that is, given any open interval $I$ that intersects the Cantor set, $I \cap C$ is nonempty. So you could define the Cantor set to be the closure of $\bigcup_{n = 1}^{\infty}C_n$. If you're not familiar with topological notions like "closure", think of it as adding all of the points that are almost in the set; so, for example, $1$ is not in your set, but points arbitrarily close to it are, so $1$ would be in the closure.

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