Consider the binary function $f: \{0,2\}^{\mathbb{N}}\to [0,1]$ by
$$
f(\{a_n\})=\sum_{n=1}^\infty\frac{a_n}{3^n}
$$
for $a_n\in \{0,2\}$.
I am working on the proof that $f$ is injective…
The proof will show $x \neq y$ implies $f(x) \neq f(y)$ which is the contrapositive of the definition of injectivity.
Since for $\{a_n\}\neq \{b_n\}$, there exists $N>0$ so that $a_n=b_n$ for $1\le n\le N-1$ but $a_N=b_N$, then
$$
f(\{a_n\})=\lim_{N\to \infty} \sum_{k=1}^N\frac{a_k}{3^k}=:\lim_{N\to \infty} S_N
$$
and
$$
f(\{b_n\})=\lim_{N\to \infty} \sum_{k=1}^N\frac{b_k}{3^k}=:\lim_{N\to \infty} L_N
$$
But what I am stuck is that the partial sum is
$$
S_N-L_N=\frac{a_N}{3^N}-\frac{b_N}{3^N}\to 0…
$$
This result is quite strange, as it suggests that when $N$ approaches infinity, the limits of the two are the same?
How to show that $f(\{a_n\})\neq f(\{b_n\})$?
Best Answer
Let $N$ be the first number such that $a_N \neq b_N$. WLOG assume that $b_N - a_N = 2$. Now observe
$$ \frac{a_N}{3^N} + \sum_{n = N+1} \frac{a_n}{3^n} \leq \frac{a_N}{3^N} + \sum_{n = N+1} \frac{2}{3^n} \leq \frac{a_N+1}{3^N} < \frac{b_N}{3^N}. $$
You can take it from here.