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$C$ = Cantor set. $C$ is closed, so $C^c$ is the countable union of disjoint open intervals. So let $$C^c = \bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} A_i $$

where $A_i$ are disjoint open intervals.

For each $i \in \mathbb{N},\ A_i\ $ differs from a closed set by just two points, namely the endpoints of the interval $A_i.$ For each $i \in \mathbb{N},\ $ let

$$ B_i = \{x_{i1}, x_{i2} \} = \{ \text{ the two endpoints of } A_i \}. $$

Then $\left|\bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} B_i\right| \leq 2 \left|\mathbb{N}\right| $ is countable. However, $\bigcup\limits_{\substack{i~\in~\mathbb{N} \\}} B_i$ is surely the Cantor set, which is uncountable.

Something is incorrect in all of this, but what?

David C. Ullrich
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Adam Rubinson
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1 Answers1

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There are lots of points in the Cantor set other than the end points of the intervals $A_i$. Infinite sums of the type $\sum \frac {a_i} {3^{i}}$ ($a_i \in \{0,2\}$) are in $C$ and they are not end points unless the sum reduces to a finite sum.

Kavi Rama Murthy
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  • Oh I didn't know that. – Adam Rubinson Jan 28 '21 at 12:50
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    Ah wikipedia says ( https://en.wikipedia.org/wiki/Cantor_set#Composition ): "It may appear that only the endpoints of the construction segments are left, but that is not the case either. The number $\frac{1}{4}$, for example, has the unique ternary form $0.020202... $. It is in the bottom third, and the top third of that third, and the bottom third of that top third, and so on. Since it is never in one of the middle segments, it is never removed. Yet it is also not an endpoint of any middle segment, because it is not a multiple of any power of $\frac{1}{3}.$" – Adam Rubinson Jan 28 '21 at 12:57