I have to prove two things.
First is that the Cantor set has a lebesgue measure of 0. If we regard the supersets $C_n$, where $C_0 = [0,1]$, $C_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$ and so on. Each containig interals of length $3^{-n}$ and by construction there are $2^n$ such intervals. The lebesgue measure of each such interval is $\lambda ( [x, x + 3^{-n}]) = 3^{-n}$ therefore the measure of $C_n$ is $\frac{2^n}{3^n} = e^{(\ln(2)-\ln(3)) n }$ which goes to zero with $n \rightarrow \infty$. But does this prove it?
The other thing I have to prove is that the Cantor set is uncountable. I found that I should contruct a surjectiv function to $[0,1]$. But im totaly puzzeld how to do this.
Thanks for help
Best Answer
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)
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.