Cantor Sets/nonempty/cardinality Let $S_0=[0,1]$ and define every $S_k$ for $k\geq 1$ 
\begin{align*}
S_1&=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3}, 1\right],\\
S_2&=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9}, \frac{3}{9}\right]\cup\left[\frac{6}{9}, \frac{7}{9}\right]\cup\left[\frac{8}{9},1\right],\\
S_3&=\left[0, \frac{1}{27}\right]\cup\left[ \frac{2}{27}, \frac{3}{27}\right]\cup\left[ \frac{6}{27}, \frac{7}{27}\right]\cup\left[ \frac{8}{27}, \frac{9}{27}\right]\cup\left[ \frac{18}{27}, \frac{19}{27}\right]\cup\left[ \frac{20}{27}, \frac{21}{27}\right]\cup\left[ \frac{24}{25}, \frac{25}{27}\right]\cup\left[ \frac{26}{27},  1\right]\\
\vdots
\end{align*}
Now let $C=\bigcap_{k=0}^\infty S_k$. This is known as a Cantor set. 
I have to prove that C is non-empty, by showing that 1/4 is in C. I also have find the cardinal of C and prove my answer. 
I have some ideas how to prove non-emptiness, but I am stuck as to how to specifically prove that 1/4 is in C. Also I think that the cardinal is equal to the cardinal of the reals $2^{\aleph_0}$. I probably need to find a bijection to prove this right?
 A: One way to think of the cantor set is in terms of ternary expansions. We typically write a real number $r \in [0, 1]$ in decimal notation as
$$
r = 0.a_1a_2a_3a_4\cdots
$$
where $a_i \in \{0, \ldots, 9\}$. But we can pick other bases such as base 2 ($a_i \in \{0, 1\}$) or for our purposes, base 3 ($a_i \in \{0, 1, 2\}$).
Now if you think of numbers in terms of their ternary expansions, you should note that the Cantor set simply removes any number whose ternary expansion includes the digit 1 (that is, essentially, what throwing out the middle interval does). So to show that $1/4$ is in the Cantor set, it would suffice to show that the ternary expansion of $1/4$ does not include a 1.
Furthermore, if you think about this a little more, you should be able to adapt this to showing the uncountability of the cantor set. Remember that every number in the unit interval can be written in a binary expansion, which only involves 0 and 1...
A: Let's try the coding approach. We will take $0$ for left and $2$ for right. Then
$$C_0:=[0,1/3] \text{  and  } C_2:=[2/3,1]$$
and $S_1 = C_0\cup C_2$. Similarly the left and right subintervals of $C_0$ are coded $C_{00}$ and $C_{02}$ and for $C_2$ we have $C_{20}$ and $C_{22}$; for example $C_{02}:=[2/9,3/9]$. This gives us 
$$S_2=C_{00}\cup C_{02}\cup C_{20}\cup C_{22}$$
The intervals in $S_3$ are specified for string of length three. In general we can use induction and shows that the intervals of $S_n$ are specified for string of $n$ elements each of which is a chain of $0$ and $2$'s. Now if we have an infinite string  of $0$´s and $2$'s, i.e., $a_1a_2a_3a_4\ldots$we have the intervals 
$$C_{a_1}\supset C_{a_1a_2}\supset C_{a_1a_2a_3}\supset \ldots C_{a_1\ldots a_n}\supset \ldots$$
The intersection of which is a point in the Cantor set. So any infinite address string defines a point in $C$. Conversely any point $p$ in $C$ defines an infinite address, where its first $n$ symbols are specified by its location in $S_n$. Notice that this addressing is in fact injective, any other point $q$ in the Cantor set  has different address, because at some $n$, $p$ and $q$ are in different $C_{\alpha}$ and $C_{\beta}$ in $S_n$. Thus each point in $C$ is in a 1-1 correspondence with the collection of all address.
If each $2$ is replaced by $1$, we have a base two expansion of some $x\in [0,1]$. So as there are uncountable many points in $[0,1]$, uncountable many base two expansion and uncountable many addresses so uncountable many points in $C$. 
Now you can interpret the addresses as a base three expansion of points $x\in [0,1]$ as we have shown there is a 1-1 correspondence between the Cantor set and the address and the base three expansion of $1/4_3= .02020202\ldots_3$, so is in a point in $C$.
