Let $X$ be the Cantor space and let $Y$ be the Cantor middle thirds set. Define the function $f:X\rightarrow Y$ such that:
$$
f(x) = f((x_0, x_1, …)) = 0.y_0y_1…\quad\text{where }\begin{cases}
y_i = x_i&\text{if }x_i = 0 \\
y_i = 2&\text{if }x_i = 1
\end{cases}
$$
where the output is a decimal expansion in ternary base. Clearly this is a bijective function since the Cantor set does not contain any numbers with decimal expansion in ternary base containing something other than 0 or 2. Let $U$ be open in $Y$, then under the product topology all but finitely many coordinates of any $x\in U$ are the same for each $x$. So there exists a $n$ such that for all $x\in U$, and for all $N\geq n$ $x_N$'s are equal. Observe that each element in $f^{-1}(U)$ is then of the form $0.y_0y_1…y_ny_\alpha y_\beta…$, where $y_0,…,y_n$ are fixed but $y_\alpha, y_\beta,…$ can be either 0 or 2. So $f^{-1}(U)$ is some $\epsilon/\left(3^n\right)$-ball intersected with the Cantor set, generating an open set in the subspace topology inherited from $\left[0,1\right]$. Hence $f$ is continuous. Let $V$ be open in $X$, then $V$ is some $\epsilon'$-ball intersecting with the Cantor set. Note then that $V = \epsilon/\left(3^n\right)$, where $n = \min\left\lbrace i \right\rbrace$ such that $\epsilon/\left(3^i\right)\leq\epsilon'$. Therefore elements in $V$ are of the form $0.y_0y_1…y_ny_\alpha y_\beta…$, where $y_0,…,y_n$ are fixed but $y_\alpha, y_\beta,…$ can be either 0 or 2. So $f(V) = (f^{-1})^{-1}$ consists of elements $x$ such that for all $N\geq n$ $x_N$'s are equal. Hence $f(V)$ is open in the product topology on the Cantor space. Hence $f^{-1}$ is continuous. This shows that $f$ is a homeomorphism.
I would like some advice on how to clean up the proof or if I forgot anything in the proof because it looks somewhat messy to me.
Best Answer
Unfortunately your proof is not correct.
As reuns observed in his comment, in your arguments you have to exchange $X$ and $Y$ ("Let $U$ be open in $Y$, then under the product topology ..." does not make sense). But that is easily repaired.
A more serious mistake is this:
Let $U$ be open in $X$, then under the product topology all but finitely many coordinates of any $x \in U$ are the same for each $x$. So there exists an $n$ such that for all $x \in U$ and for all $N \ge n$ the $x_N$'s are equal.
This is not true. In fact, if $x \in U$, there exists a basic open $V = \prod_{i=1}^\infty V_i$ (i.e. $V_i \subset \{0,1\}$ open such that $V_i = \{0,1\}$ for almost all $i$) with $x \in V$. This means that for $x \in U$ the coordinates $x_N$ are arbitrary for $N \ge n$.
On the other hand, it seems that in the rest of your proof you do not use that for all $N \ge n$ the $x_N$'s are equal, but that they are arbitrary!
To correct your proof, I recommend to write for $x = (x_1,x_2,x_3\ldots)$ [note that I start the sequence with index $1$] $$f(x) = \sum_{i=1}^\infty \frac{2x_i}{3^i} .$$ Let us moreover define $$f_N(x) = \sum_{i=N}^\infty \frac{2x_i}{3^i} .$$ Then for all $x, y$ we have $$\lvert f_N(x) - f_N(y) \rvert = \left\lvert \sum_{i=N}^\infty \frac{2x_i - 2y_i}{3^i} \right\lvert \le \sum_{i=N}^\infty \frac{2}{3^i} = \frac{1}{3^{N-1}} .$$ Given $x$, let $V_i = \{ x_i\}$ for $i < N$ and $V_i = \{ 0,1\}$ for $i \ge N$. Then $V_N(x) = \prod_{i=0}^\infty V_i$ is an open neigborhood of $x$ such that for $y \in V_N(x)$ we have $y_i = x_i$ for $i < N$.
You can easily see that $f$ is continuous because for $y \in V_N(x)$ $$\lvert f(x) - f(y) \rvert = \lvert f_N(x) - f_N(y) \rvert \le \frac{1}{3^{N-1}} .$$ To see that $f^{-1}$ is continuous it suffices to show that for each $x$ and each $N$ there exists $\epsilon > 0$ such that $\lvert f(x) - f(y) \rvert < \epsilon$ implies $y \in V_N(x)$.
So let $\epsilon = \frac{1}{3^{N-1}}$. We have $$\left\lvert\frac{2x_1 - 2y_1}{3} \right\lvert = \left\lvert f(x) - f_2(x) - f(y) + f_2(y) \right\rvert \le \left\lvert f(x) - f(y) \right\rvert + \left\lvert f_2(x) - f_2(y) \right\rvert < \frac{1}{3} + \frac{1}{3} = \frac{2}{3}.$$ This implies $x_1 = y_1$ because otherwise $\left\lvert\frac{2x_1 - 2y_1}{3} \right\lvert = \frac{2}{3}$ which is impossible.
Proceed inductively to show that $\left\lvert\frac{2x_i - 2y_i}{3} \right\lvert < \frac{2}{3^i}$ for $i < N$ (using that $x_j = y_j$ for $j=1,\ldots,i-1$). Thus $y_i = x_i$ for $i < N$. This shows $y \in V _N(x)$.