Hint:
- There is a theorem that if $C_n$ is a descending sequence of measurable sets,
$C = \bigcap C_n$ then $\lim_{n \rightarrow \infty}m(C_n)= m(C)$.
Here, you know $C_n$ is a descending sequence, $m(C_n) $= (2/3)^n, and you want to know m(C)
- to prove $C$ is uncountable:
Express a number $x$ between $[0,1]$ in base $3$:
$x =0.x_1x_2x_3...$
In the 1st step, we remove the middle third from $[0,1]$
We express 0=.0, $\frac{1}{3} = .1, \frac{2}{3} = .2$, 1= .222222...
We have 3 intervals: [.0 , .1] , (.1, .2), [.2 , .22222...] and we remove the middle interval. The removed interval (.1 , .2) consists of all numbers with $x_1$ = 1, except the endpoint $.1$ of [.0 , .1 ], however, we can express .1 as .02222.....
So we can use the rule: whenever we have a number of form 0.x1(x is a sequence consisting of 0,1,2) as the end point of an interval, we express as 0.x0222....So in this step, we remove all numbers with $x_1$ = 1. The remaining intervals are [.0, .0222...] and [.2, .222...]
Similarly, we can prove that in the n-th step, we keep only those numbers with $x_n$ = 0 or 2:
So the Cantor set contains of all numbers of the form $.x_1x_2..$ with $x_i =0 $ or 2.
There exists a bijection between $E$ and $[0,1]$:
If you consider the new set E, with each member of E is a member of the Cantor set with every digit is divided by 2. E consists of all sequence with $x_i$ is 0 or 1. $|E| = |C|$
There exists an injective map from $[0,1]$ to this new set E. So you can prove it's uncountable.
A number with a $1$ in the ternary representation of a number corresponds to a a "middle third" of some interval...
Look at how the ternary expansion works: $0.1a_1\dots a_n\dots$ is necessarily in the middle third.
$0.01a_2\dots$ in the middle third of the first third.
$0.21a_2\dots$ in the middle third of the third third, etc...
This is because, in ternary, $0.a_1\dots a_n\dots=\sum_{i=1}^\infty \frac{a_i}{3^i}$, where $a_i=0,1$ or $2\,\forall i$.
Best Answer
Notation $$ \frac{[0,1]}{3} = \left\{\frac{t}{3}\;:\;t \in [0,1]\right\} , \\ \frac{2}{3} + \frac{[0,1]}{3} = \left\{\frac{2}{3}+\frac{t}{3}\;:\;t \in [0,1]\right\} . $$ So $$ \frac{[0,1]}{3} = \left[0,\frac{1}{3}\right] , \\ \frac{2}{3} + \frac{[0,1]}{3} = \left[\frac{2}{3},1\right] , \\ C_1 = \frac{[0,1]}{3} \cup \left(\frac{2}{3} + \frac{[0,1]}{3}\right)= \left[0,\frac{1}{3}\right]\cup \left[\frac{2}{3},1\right] . $$