$C=\{\sum_{k=1}^\infty a_k 3^{-k} : a_k \in \{0,2\} \forall k\}$
The Cantor set is generated in the following way. $C_0 = [0,1], C_1 =[0,1/3] \cup [2/3,1], C_2=[0,1/9] \cup [2/9,1/3] \cup [2/3,7/9] \cup [8/9,1]$ etc.
First it is shown easily (by induction) that the left end points (of the disjoint intervals) of $C_k$ are of the form $\{\sum_{n=1}^\infty a_n 3^{-n} : a_n \in \{0,2\}~\forall 1 \leq n \leq k, a_n = 0, k < n\}$ and conversely.
For $k=0$, this is indeed true. Assume this to be true for $k$ and let $[c,d]$ be one of the disjoint intervals in $C_{k}$. Now, this interval gives rise to the intervals $[c,e]$, $[f,d]$ in $C_{k+1}$. In the first case, $c$ already has the required form by induction hypothesis. In the second case, $f=c+2(e-c)/3=c+2/3^{k+1}$ as $e-c = 1/3^k$ by construction. Therefore, $f$ has the required form as well.
Conversely, suppose a sequence $\{a_1,a_2,\cdots,\}$ where $a_n \in \{0,2\}, 1 \leq n \leq k, a_n = 0, k < n$ is given. This sequence represents the left end point chosen in the following way:- If $a_1 = 0$ then choose the left end point of the left partition of $[0,1]$ and if $a_1=2$ then choose the left end point of the right partition of $[0,1]$. If $a_2=0$ then choose the left end point of the left partition of the partition chosen in the previous step and so on. For example, $\{0,2,2,0,0,\cdots\}$ represents choosing the left end point of the left partition of $[0,1]$ (which is $[0,1/3]$) followed by the choosing the left end point of right partition of $[0,1/3]$ (which is $[2/3,1]$) followed by choosing the left end point of the right partition of $[2/3,1]$ (which is $[8/27,1/3]$). Therefore $\{0,2,2,0,0,\cdots\}$ represents the end point $8/27$ (in $C_3$).
The Cantor set contains other points (which are uncountably many) too. Infact all the end points are countable, so the other points are what make up the majority of the cantor set. Take $x \in C$ then $x \in C_k~\forall k$ and hence there exists $[s_k,t_k]$ such that $x \in [s_k,t_k] \subset C_k$. By construction of the Cantor set, $|x-s_k| < \frac{1}{3^k}$ and hence $s_k \to x$. But $s_k = \sum_{n=1}^k a_n3 ^{-n}$ and $\lim_{k \to \infty} s_k = \lim_{k \to \infty} \sum_{n=1}^k a_n3 ^{-n}$ where $a_n \in \{0,2\}~\forall n$.
For $k=0$, this is indeed true. Assume this to be true for $k$ and let $[c,d]$ be one of the disjoint intervals in $C_{k}$. Now, this interval gives rise to the intervals $[c,e]$, $[f,d]$ in $C_{k+1}$. In the first case, $c$ already has the required form by induction hypothesis. In the second case, $f=c+2(e-c)/3=c+2/3^{k+1}$ as $e-c = 1/3^k$ by construction. Therefore, $f$ has the required form as well.
Conversely, suppose a sequence $\{a_1,a_2,\cdots,\}$ where $a_n \in \{0,2\}, 1 \leq n \leq k, a_n = 0, k < n$ is given. This sequence represents the left end point chosen in the following way:- If $a_1 = 0$ then choose the left end point of the left partition of $[0,1]$ and if $a_1=2$ then choose the left end point of the right partition of $[0,1]$. If $a_2=0$ then choose the left end point of the left partition of the partition chosen in the previous step and so on. For example, $\{0,2,2,0,0,\cdots\}$ represents choosing the left end point of the left partition of $[0,1]$ (which is $[0,1/3]$) followed by the choosing the left end point of right partition of $[0,1/3]$ (which is $[2/3,1]$) followed by choosing the left end point of the right partition of $[2/3,1]$ (which is $[8/27,1/3]$). Therefore $\{0,2,2,0,0,\cdots\}$ represents the end point $8/27$ (in $C_3$).
The Cantor set contains other points (which are uncountably many) too. Infact all the end points are countable, so the other points are what make up the majority of the cantor set. Take $x \in C$ then $x \in C_k~\forall k$ and hence there exists $[s_k,t_k]$ such that $x \in [s_k,t_k] \subset C_k$. By construction of the Cantor set, $|x-s_k| < \frac{1}{3^k}$ and hence $s_k \to x$. But $s_k = \sum_{n=1}^k a_n3 ^{-n}$ and $\lim_{k \to \infty} s_k = \lim_{k \to \infty} \sum_{n=1}^k a_n3 ^{-n}$ where $a_n \in \{0,2\}~\forall n$.
$\{0,1\}^\infty$ is a compact metric space
Note that the metric on $\{0,1\}^\infty$ is given by
$d(x,y) = \sum_{i=1}^\infty \frac{1}{2^i}\frac{d_i(x_i,y_i)}{1+d_i(x_i,y_i)}$ where $d_i(x_i,y_i) = |x_i-y_i|$.
Let $f : \{0,1\}^\infty \to C$ be given by $f(\{x_1,x_2,\cdots,\} = \sum_{k=1}^\infty 2x_k 3^{-k}$. Now it is shown that $f$ is a homeomorphism ($C$ is given the
subspace topology induced from $\mathbb{R}$)
- $f$ is injective. Assume that $a = (a_0,a_1,\cdots) \neq b = (b_0,b_1,\cdots)$. Let $k$ be the smallest integer for which $a_k \neq b_k $. WLOG assume that $a_k=0,b_k=1$. Then, $\sum_{n=k+1}^\infty 2a_n3^{-n} \leq \sum_{n=k+1}^\infty 2*3^{-n} \leq 1/3^k < 2*3^k + \sum_{n=k+1}^\infty 2b_n3^{-n}$. Therefore, $f(a) < f(b)$.
- $f$ is onto. If $y \in C$, then $y=\sum_{k=1}^\infty a_k 3^{-k}$ for some $a_k \in \{0,2\} \forall k$. Choose $x_k = a_k/2$. Then $f(x) = y$.
- $f$ is continuous. $|f(x)-f(y)| \leq |\sum_{k=1}^\infty \frac{2}{3^k} (x_k-y_k)| \leq \sum_{k=1}^\infty \frac{2}{3^k} |x_k-y_k| \leq 2d^\prime(x,y)$ where $d^\prime(x,y) = \sum_{i=1}^\infty \frac{1}{2^i} d_i(x_i,y_i)$. As $d$ and $d^\prime$ are topologically equivalent , it follows that $f$ is uniformly continuous.
- $f^{-1}$ is continuous: Suppose $d(f(x),f(y)) = |\sum_{k=1}^\infty \frac{2}{3^k} (x_k-y_k)| < 3^{-N}$. Let $M$ be the first mismatch between $x_k,y_k$ i.e. $x_M \neq y_M$ and $x_k = y_k, 1 \leq k \leq M-1$. Then $|\sum_{k=1}^\infty \frac{2}{3^k} (x_k-y_k)| = |\sum_{k=M}^\infty \frac{2}{3^k} (x_k-y_k)|$. Assume now that $1 \leq M \leq N$. This gives \begin{array} $|\sum_{k=M}^\infty \frac{2}{3^k} (x_k-y_k)| &= |\pm \frac{2}{3^M} + \sum_{k=M+1}^\infty \frac{2}{3^k} (x_k-y_k)| \\ &\geq |\frac{2}{3^M} | - \sum_{k=M+1}^\infty \frac{2}{3^k} (x_k-y_k)| | \\ & \geq \frac{2}{3^M} - \sum_{k=M+1}^\infty |\frac{2}{3^k}| \\ &\geq \frac{1}{3^M} \\ &\geq \frac{1}{3^N}. \end{array} Hence the first mismatch can only occur at $ M > N$. This gives that $d^\prime(x,y) \leq 2^{-N}$.
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