Define an injection from $2^{\Bbb N}$ to $\Bbb R$
Does this proof look fine or contain logical gaps and flaws? Thank you for your help!
We define a mapping $F:2^{\Bbb N} \to \Bbb R$ by $$\forall f\in2^{\Bbb N}: F(f)=\sum_{n=0}^{\infty}\dfrac{f(n)}{3^{n+1}}$$
It's clear that for all $f\in 2^{\Bbb N}$: $$F(f)=\lim_{k\to\infty}\sum_{n=0}^{n=k}\dfrac{f(n)}{3^{n+1}}$$ and $$F(f)=\sum_{n=0}^{\infty}\dfrac{f(n)}{3^{n+1}}\le \sum_{n=0}^{\infty}\dfrac{1}{3^{n+1}} = \dfrac{1}{2}$$
Thus $F(f)$ is the limit of a strictly increasing and bounded above sequence of rational numbers. Hence $F(f)$ exists for all $f\in 2^{\Bbb N}$.
For $f_1,f_2\in 2^{\Bbb N}$ and $f_1\neq f_2$, let $N=\min\{n\in\Bbb N\mid f_1(n)\neq f_2(n)\}$.
$|F(f_1)-F(f_2)|=\left |\sum_{n=0}^{\infty}\dfrac{f_1(n)}{3^{n+1}}-\sum_{n=0}^{\infty}\dfrac{f_2(n)}{3^{n+1}} \right |=\left |\sum_{n=0}^{\infty}\dfrac{f_1(n)-f_2(n)}{3^{n+1}} \right |= \left |\sum_{n=N}^{\infty}\dfrac{f_1(n)-f_2(n)}{3^{n+1}} \right |= \left |\dfrac{f_1(N)-f_2(N)}{3^{n+1}}+\sum_{n=N+1}^{\infty}\dfrac{f_1(n)-f_2(n)}{3^{n+1}} \right | \ge \left |\dfrac{f_1(N)-f_2(N)}{3^{N+1}}\right |- \left |\sum_{n=N+1}^{\infty}\dfrac{f_1(n)-f_2(n)}{3^{n+1}} \right |=\dfrac{1}{3^{N+1}}- \left |\sum_{n=N+1}^{\infty}\dfrac{f_1(n)-f_2(n)}{3^{n+1}} \right |\ge \dfrac{1}{3^{N+1}}- \sum_{n=N+1}^{\infty}\left |\dfrac{f_1(n)-f_2(n)}{3^{n+1}} \right | \ge \dfrac{1}{3^{N+1}}- \sum_{n=N+1}^{\infty}\dfrac{1}{3^{n+1}} =\dfrac{1}{3^{N+1}}- \dfrac{1}{2.3^{N+1}}=\dfrac{1}{2.3^{N+1}}>0.$
Hence $f_1\neq f_2\implies F(f_1)\neq F(f_2)$. Thus $F$ is injective.