Consider the Cantor set $C$, and negative integer powers $2^{-k}$.

Clearly, for $k=1$, $2^{-1} \notin C$ since $1/2 \in (1/3, 2/3)$, the first deleted open interval.

It is known that $1/4 = 2^{-2} \in C$, since it alternates between the lower and upper thirds of each non-deleted interval in every iteration, i.e., it lies in $L_1 = [0, 1/3], U_2 = [2/9, 1/3] ...$ and so on.

Since $2^{-3} \in (1/9, 2/9)$, a deleted interval, $2^{-3} \notin C$.

A slightly more tedious procedure shows that since $2^{-4} \in (1/27, 2/27)$, $2^{-4} \notin C$.

My question is as follows:

Which negative powers of 2 belong to the Cantor set? Does it contain any powers other than $1/4$?

It seems easy enough to determine this question for special cases (as shown above), but I wonder about the general case.

PS: The alternating interval proof for $1/4$ is from the text by Kolmogorov and Fomin.

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    $\begingroup$ My (unhelpful) intuition is that this is a hard problem. Well asked, though! $\endgroup$ – Greg Martin Aug 11 '14 at 15:57
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    $\begingroup$ This isn't descriptive set theory, so I removed that tag. The question looks like number theory to me, but I'm reluctant to add that tag without understanding the problem better. $\endgroup$ – Andreas Blass Aug 11 '14 at 16:06

In this paper by Charles R. Wall, it is shown that there are only fourteen numbers with terminating decimal expansion in the Cantor set (sixteen if you include $0$ and $1$), they are:

$$\frac{1}{4}, \frac{3}{4}, \frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}, \frac{1}{40}, \frac{3}{40}, \frac{9}{40}, \frac{13}{40}, \frac{27}{40}, \frac{31}{40}, \frac{37}{40}, \frac{39}{40}.$$

As every number of the form $2^{-k}$ has finite decimal expansion, any such number in the Cantor set must appear in the above list. As such, the only number of the form $2^{-k}$ which is an element of the Cantor set is $\frac{1}{4}$.

  • $\begingroup$ This is very interesting - thanks! $\endgroup$ – user_of_math Aug 11 '14 at 17:24

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