# Finding a bijection between set of infinite sequences of $0,1,2$ and the open interval $(0,1)$

I've been asked to find bijection between $$\{0,1,2\}^\mathbb{N}$$, the set of infinite sequences of $$0,1$$ and $$2$$, and the open interval $$(0,1)$$, and am told that it follows a similar logic to the proof creating a bijection between $$10^\mathbb{N}=\{0,1,2,3,4,5,6,7,8,9\}$$ and the open interval $$(0,1)$$.

I can find the bijection between $$10^\mathbb{N}=\{0,1,2,3,4,5,6,7,8,9\}$$ and $$(0,1)$$ by picking some $$a \in 10^\mathbb{N}$$ and showing that it converges to some number $$b \in (0,1)$$. Noticing that it is possible for two sequences to converge to the same number $$b \in (0,1)$$, such as $$0.5$$ and $$0.4\bar{9}$$, we create $$A=$$ "set of sequences with two representations" and $$B=$$ "numbers with 2 representations". I can then show that these two sets have a bijection between them without much difficulty.

What I do not understand is how to transfer this framework once we no longer have the ability to create an infinite sequence corresponding to any number in $$(0,1)$$ and I am struggling to find a function that allows me to do so. Any tips would be appreciated.

• Hint. Think about expressing the numbers in $(0,1)$ using base $3$ decimals instead of base $10$ decimals. – Ethan Bolker Nov 6 '18 at 18:47
• I was trying to think of something like $0=00$, $1=01$, $2=02$, $3=10$, ... $8=22$, But then I don't have a unique 2 digit sequence to represent a 9. Could I simply do a unique 3 digit sequence for each number? – Corran Horn Nov 6 '18 at 18:51
• There are $27$ three digit sequences. If you use $10$ of them to encode the $10$ decimal digits your map won't use all the infinite sequences you care about. You are overthinking this. Just use base $3$. – Ethan Bolker Nov 6 '18 at 18:56

Hint: The bijection for the $$10^{\mathbb N}$$ looks probably like this: \begin{align*} (x_k)_{k \in \mathbb N} \mapsto \sum_{k = 1}^\infty \frac{x_k}{10^k} \end{align*} (afterwards fixing the problem with $$0. ....9999....$$ that you mentioned). Try something similar.