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I know that the Cantor Set contains no segment of the form $$\left(\frac{3k+1}{3^m}, \frac{3k+2}{3^m}\right)$$ for any integers $k$ and $m$. If we can prove that every real segment contains a segment of that form, then certainly the Cantor Set contains no segment.

How do I show that for every $a, b \in \mathbb{R}$, there exist $k, m \in \mathbb{Z}$ such that $$\left(\frac{3k+1}{3^m}, \frac{3k+2}{3^m}\right) \subset (a, b)?$$

Presumably we can use the Archimedean Property of the real numbers, but I'm hazy on the details...

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3 Answers 3

up vote 6 down vote accepted

Or a slightly different approach: for any integer $m \geq 0$ the Cantor set is a subset of a disjoint union of $2^m$ closed intervals of length $3^{-m}$. Therefore it cannot contain an interval of length $> 3^{-m}$. Since this holds for any $m$, the Cantor set does not contain any interval of positive length.

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This is a great answer. But what about segments? Just because the Cantor Set cannot contain an interval of length $> 3^{-m}$, doesn't mean it can't contain a segment of that length... –  jamaicanworm Feb 29 '12 at 1:23
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@jamaicanworm: you can choose any $m$ so the only possible segments would be $\[a,a\]$ with length zero, if you call that a segment at all. –  WimC Feb 29 '12 at 5:39

Your way of expressing the question suggests this way of looking at it. The numbers $(3k+1)/3^m$ and $(3k+2)/3^m$ have terminating ternary expansions. After some point, both expansions consist only of $0$s. Then ask whether somewhere between those two numbers there is a number that has a $1$ in its ternary expansion. Hint: Try to locate the $1$ after the $m$th place in the expansion!

The Archimedean property is involved in proving that ternary expansions, or binary or decimal, etc., exhaust the whole set of real numbers. If there were infinitesimals, then two numbers differing by an infinitesimal would have identical ternary expansions, since every place in the ternary expansion is in a finite position.

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I prefer Wim's answer, but here is an explicit answer based on the way you look at the question.

Let $m\in\mathbb{N}$ such that $3^m>4/(b-a)$. Then $$ \frac{3^mb-2}3-\frac{3^ma-1}3=\frac{3^m(b-a)-1}3>\frac{4-1}3=1. $$ So there exists $k\in\mathbb{N}$ with $$ \frac{3^ma-1}3<k<\frac{3^mb-2}3. $$ These two inequalities are $$ a<\frac{3k+1}{3^m},\ \ \ \frac{3k+2}{3^m}<b. $$ So $$ \left(\frac{3k+1}{3^m},\frac{3k+2}{3^m}\right)\subset (a,b). $$

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