I saw a post somewhere that stated that "each point in the Cantor set is the endpoint of some closed interval of the form $\big[\frac{n}{3^k}, \frac{n+1}{3^k}\big]$, with $n, k \in \mathbb{N}$, so each point in the Cantor set is rational." Is that true?
Is every point in the Cantor set rational?
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real-analysis
rational-numbers
cantor-set
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1We have some fine answers already posted. A related question: Is there an algebraic irrational in the Cantor set? Answer: of course not, but no one knows how to prove it. – 2017-02-19
2 Answers
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No: the middle-thirds Cantor set is uncountable, while the set of rational numbers is countable. Therefore not every point in the Cantor set is rational.
It turns out that the elements of the Cantor set are precisely the real numbers in $[0,1]$ which have a ternary expansion consisting only of $0$s and $2$s. Note that this includes many points besides the endpoints of the intervals removed.
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Note that $$\frac{1}{4}\in C$$ is a rational but not an endpoint.
(plus those uncountably many irrationals hiding in the corners...)