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I am stumped on a problem in a text book. This is not homework. I'm a physicist doing some self study on Lebesgue integrals and Fourier theory. I'm starting with the basics, and reading up on measure theory.

The problem is to show that $\frac{1}{4}$ is an element of the Cantor set. My first thought would be t0 find a ternary expansion consisting of only 0's and 2's.

However, what I'm having trouble with is imagining that anything remains following the infinite intersection creating the Cantor set other than the interval endpoints. I imagine that if I pick a real number not lying on some interval endpoint I could find an $N$ large enough that the portion of the real line the number belongs to would be deleted. I'd like to see why this argument breaks down.

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    Thank you for this question. I was wondering exactly the same thing. This post and the subsequent answer helped me a lot.2013-01-07

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Your first thought is a good one. $(\frac 14)_{10}=0.\overline{02}_3=\sum \frac 2{9^i}=\frac {\frac 29}{1-\frac 19}$. Since each stage deletes the numbers left that have a $1$ at that point in the expansion, it never gets deleted. It isn't the endpoint of an interval, but it is a limit of endpoints of intervals.

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    @ncRubert: I would say it is a consequence of the Cantor set being closed. We know it is because we have removed a union of open intervals, which is therefore open.2012-10-12