For each $n\in \mathbb N$, let $X_n=\{0,2\}$ and let $\mathscr T_n $ be the discrete topology on $X_n$. Let $X=\prod_{n=1}^\infty X_n,$ and $\mathscr T$ be the product topology on $X$. Prove that the Cantor set is homeomorphic to $(X,\mathscr T)$.
Proof. Let $C$ be the Cantor set, Let $(C,\mathscr U_C)$ be the subspace topological space of usual topology on $\mathbb R$. Let the map $\psi:(C,\mathscr U_A)\to (X,\mathscr T)$ be defined as $\psi(x)=(x_1,x_2,...,x_n,....),$ where $x\in C$ and $x=\sum_{i=1}^\infty \frac{x_i}{3^i},x_i=0$or $2$. $\psi$ is well defined if we take the ternary expansion of $1/3,1/9$,...etc with $0$ and $2$ s. We can easily prove that $\psi$ is injective.
How do I prove that $\psi$ is surjective? What is the guarentee that, If $y\in X$, $\exists x\in C$:$f(x)=y$?
For continuity of $\psi$, Let $\mathscr B$ be the basis of the product topology on $X$. It is enough to show that $\psi^{-1}(B)\in \mathscr U_C$ for every $B\in \mathscr B$. Let $B \in \mathscr B$, $B=\prod_{i=1}^n B_i$, $B_i \in \mathscr T_i$, $B_i=X_i$ for all but finitely many $i$. I want to prove $\exists$ some open interval ($I$(say) in $\mathbb R$) such that $\psi^{-1}(B)=C\cap I$. How do I prove the existence of $I$?
For completing the proof of homeomorphism, I need to prove that $\psi $ is open. Let $U\in \mathscr U_C$, then $U=C\cap J$ for some interval $J$ in $\mathbb R$. then $\psi(U)=\psi(C\cap J)=\psi(C)\cap\psi(J)=X\cap \psi(J).$ How do I complete the proof. Please help me.