7
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ My (unhelpful) intuition is that this is a hard problem. Well asked, though! $\endgroup$ – Greg Martin Aug 11 '14 at 15:57
  • 1
    $\begingroup$ 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
6
$\begingroup$

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 $\frac{1}{4}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.