Proof Cantor set is uncountable I'm trying to prove the cantor set is uncountable. This is what I ended up with:
We know that $[0,1]$ is uncountable. Then, after taking out the middle third, we end up with $[0, \frac{1}{3}] \cup [\frac{2}{3},1]$. Let's focus on $[0, \frac{1}{3}]$. $[0, \frac{1}{3}]$ is also uncountable. (By defining $f: [0, \frac{1}{3}] \rightarrow [0,1]$ such that for $b \in [0, \frac{1}{3}], f(b) = 3b$)
Now, if we repeat this process and take out the middle third, we end up with $[0, \frac{1}{9}] \cup [\frac{2}{9}, \frac{1}{3}]$. Let's now focus on $[0, \frac{1}{9}]$, that is also uncountable. We can keep repeating this process indefinitely, and also for $[\frac{2}{3},1]$. This way, if we then take the cantor set $C$, $C$ is the union of these sets, which are uncountable. Thus, $ C $ is uncountable.
I feel something is not entirely right with this proof, but I can't figure out what. Any feedback is appreciated.
 A: Show there exists a bijection between the Cantor set and the set of sequences with entries 0 or 1. Then you can show this latter set is uncountable by doing something closely related to Cantor’s diagonal argument which proves (0,1) is uncountable.
A: Here is a way to associate to each element of the cantor set $C$ an infinite sequence of $0$'s and $1$'s and vice versa. I leave the details to the OP.

*

*Notice that at each step of the construction of the Cantor set, say step $n$, a middle interval is removed from each of the $2^{n-1}$ subintervals obtained in the $n-1$-th step.

*Thus, if $I_{k,n-1}$, $k=1,\ldots, 2^{n-1}$, counted from left to right, is one subinterval obtained in the $n-1$-th step, in the $n$-step one removes the middle interval from $I_{k,n-1}$ leaving  a subinterval $I_{k',n}$ interval to the left of the middle and another $I_{k'',n}$ to the right of the middle.

*Let is denote with $0$ and $1$ left and right of middle respectively.

*Each subintervals obtained in the $n$-th step can be codified uniquely with a string of $0$ and $1$'s of length $n$, and conversely each string of length $n$ of $0$'s and $1$'s correspond to one (and only one) of the subintervals left on the $n$-th step of the construction.

*For example $00$ corresponds to first choosing the left subinterval at step 1 of the construction, and then from this subinterval one choses the left subinterval in step 2, that is, the subinterval $[0,1/9]$. Similarly, $01$ stands for $[3/9,1/3]$.

*Denote by  $I^n_{\iota}$, $\iota\in\{0,1\}^n$ the subinterval obtained in the $n$-step which is codified by the string $\iota$. $C=\bigcap_n\bigcup_{\iota\in\{0,1\}^n}I^n_{\iota}$.

*Then, each $x\in C$ can be associated with one (and only one) sequence $\ell=(x_1,x_2,\ldots)$ of $0$'s and $1$'s which describes for each $n$, the precise subinterval  $I^n_\iota$ ($\iota=(x_1,\ldots,x_n)$) in the $n$-th step $x$ belongs.

*This association is a bijection $C\rightarrow\{0,1\}^{\mathbb{N}}$. This implies that $C$ is not countable.

