# a very small question in baby rudin

I know this is kind of minor. But I want to know if I understand this correctly. In "baby rudin", on page 42, is the condition

$$3^{-m}\lt \frac{\beta-\alpha}{6}$$

tight? I thought $$3^{-m}\lt \frac{\beta-\alpha}{4}$$ is sufficient, am I correct?

Thanks.

Edit: Okay, the background is: the Cantor set (i.e. the union of all intervals after repeatedly removing all the middle third interval) does not contain any point lying in the segment

$(\frac{3k+1}{3^m},\frac{3k+2}{3^m})$

And he was saying every segment $(\alpha, \beta)$ contains a segment of the form above, if

$$3^{-m}\lt \frac{\beta-\alpha}{6}$$

But I thought this condition is too loose.

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It might help if you gave some background. –  picakhu Jan 24 '11 at 4:48
You could post more information. What are $m, \alpha$ and $\beta$ and how are they related? Without that, if $\beta=\alpha$ both claims are false. But it could well be that the first is tight and the second sufficient, as the second is looser and we do not know how you are using it. –  Ross Millikan Jan 24 '11 at 4:49

I would agree that if $3^{-m}\lt \frac{\beta-\alpha}{4}$ there will be three intervals in $(\alpha, \beta)$ of length $3^{-m}$, one of which will be of the form $(\frac{3k+1}{3^m},\frac{3k+2}{3^m})$. But you have tight vs. loose backwards. Rudin's condition, with 6 in the denominator, may force a larger $m$ than your condition with the 4 (or they may be the same). So your condition is looser (it may allow more values of $m$) than Rudin's.
Yes you're right. The condition $\frac{\beta - \alpha}{6}$ is too strong and $\frac{\beta - \alpha}{4}$ is sufficient.
The length of the interval $I_{k} = (\frac{3k+1}{3^m},\frac{3k+2}{3^m})$ is precisely $\frac{1}{3^{m}}$. Of four consecutive intervals at least one is of this form and entirely contained in an interval $(\alpha,\beta)$ of length $> \frac{4}{3^{m}}$.
Suppose $\alpha = \frac{1}{3^{m}} + \varepsilon$ with $\varepsilon$ very small. Therfore $(\alpha,\beta)$ does not contain $I_{0}$ The next interval of the form $I_{k}$ is $I_{1} = (\frac{4}{3^{m}},\frac{5}{3^{m}})$. It follows that $\beta$ must be at least $\frac{5}{3^{m}}$, so $\beta - \alpha$ should be at least $\frac{4}{3^{m}}$. And this gives your bound $3^{-m} < \frac{\beta - \alpha}{4}$. All other cases are similar but easier.