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.
 A: 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.
A: 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.
