Answer
a) $\frac{1}{2}\Sigma_{i=1}^{\infty}(\frac{2}{3})^i = \ 1$.
Two numbers that is in Cantor's set: $\frac{1}{3}$ and $\frac{2}{3}$.
b) $\frac{1}{8}\Sigma_{i=1}^{\infty}(\frac{8}{9})^i =1$
Work Step by Step
a) Here we start by summing the sizes of the ranges, as the set is defined recursively, we need to understand the pattern, which is $\frac{1}{3} + 2 \frac{1}{3^2} + 2^2 \frac{1}{3^3} + ...$, because every time we take 1/3 of an interval we are dividing it into two parts.
This formula is equivalent to $\frac{1}{2}\Sigma_{i=1}^{\infty}(\frac{2}{3})^i =\frac{1}{2} \frac{2/3}{1-2/3} \text{(by infinite geometric sum formulae)}
=1$.
b) Almost the same of part a), we only need to understand the pattern, which is $(\frac{1}{3})^2 + 8 (\frac{1}{3})^2 + 8^2 (\frac{1}{3})^2+ ... = \frac{1}{8}\Sigma_{i=1}^{\infty}(\frac{8}{9})^i = \frac{1}{8} 8 = 1$
