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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.

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    $\begingroup$ Hint. Think about expressing the numbers in $(0,1)$ using base $3$ decimals instead of base $10$ decimals. $\endgroup$ – Ethan Bolker Nov 6 '18 at 18:47
  • $\begingroup$ 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? $\endgroup$ – Corran Horn Nov 6 '18 at 18:51
  • $\begingroup$ 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 $3$. $\endgroup$ – Ethan Bolker Nov 6 '18 at 18:56
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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.

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