Cantor set does not contain any interval Cantor set does not contain any interval of the form $(\frac{3k+1}{3^m}, \frac{3k+2}{3^m})$ where $k$ and $m$ are positive integers. Now suppose it contains an interval of the form $(\alpha, \beta)$ then it will contain a segment of the above form if $$3^{-m} < \frac{\beta-\alpha}{6}$$. I did not understand how this expression came. Should not the value of $k$ also be specified ?
 A: For given $m$, we can split up the interval $[0..1]$ in segments of the form:
$$\left(\frac{3k}{3^m}..\frac{3k+3}{3^m}\right)$$
simply by letting $k$ run from $0$ to $3^{m-1}-1$ (dealing with the points $\frac{3k}{3^m}$ is a detail I will not go into -- it's not hard). Each of these intervals contains a unique interval of the form:
$$\left(\frac{3k+1}{3^m}..\frac{3k+2}{3^m}\right)$$
The condition that is specified ensures that $(\alpha..\beta)$ is more than $2\cdot 3^{1-m}$ wide. Therefore, $\dfrac{\alpha+\beta}2$ is more than $3^{1-m}$ away from $\alpha$ and $\beta$.
Now, (a number arbitrarily close to) $\dfrac{\alpha+\beta}2$ needs to be in one of the large intervals $(\frac{3k}{3^m}..\frac{3k+3}{3^m})$.
As the length of this interval is $3^{1-m}$, it must be contained in $(\alpha..\beta)$. 

Remark We see that the bound is naively tight: $3^{-m} < \dfrac{\beta-\alpha}6$ is necessary to be sure that the larger interval fits. However, since $\left(\frac{3k+1}{3^m}..\frac{3k+2}{3^m}\right)$ is the middle third of $(\frac{3k}{3^m}..\frac{3k+3}{3^m})$, we can get away with one third smaller, i.e. $3^{-m} < \dfrac{\beta-\alpha}9$, at the cost of more complicated reasoning.

Transitivity of subsets now establishes that:
$$\left(\frac{3k+1}{3^m}..\frac{3k+2}{3^m}\right) \subseteq (\alpha ..\beta)$$
for some suitable $k$.
In principle, we could express $k$ in terms of $\alpha$ and $\beta$, but this is fiddly, and not important for the conclusion that is to be attained.
