Finding injective mapping $f : \{0,1,2\}^\mathbb{N} \rightarrow \{0,1\}^\mathbb{N}$ I know that 
$g:\mathbb N\rightarrow\{0,1,2\}$ is a sequence. For example $a=(a_0,a_1,a_2,...)$ where every $a_i$ belongs to $\{0,1,2\}$ and I have to change it in to another sequence $b \in \{0,1\}^\mathbb{N}$ so I found out that I can double for zero and one, (I mean that $0 \rightarrow (0,0) ; 1\rightarrow (1,1))$ and do this $2 \rightarrow (0,1)$ for two.
But I'm not sure if it comes to the notification and I also don't know how to prove that this will be injective.
So please help with proof and notification.
 A: Here’s how to do what you want:
Start with a sequence $g\colon\mathbb Z\to\{0,1,2\}$ That is, for each $n$, you have $0\le g(n)\le2$. Now the new sequence, I’ll call it $\hat g$, is to be defined this way, following your suggestion: if $g(n)=0$, then $\hat g(2n)=\hat g(2n+1)=0$; if $g(n)=1$, then $\hat g(2n)=\hat g(2n+1)=1$; and if $g(n)=2$, then $\hat g(2n)=0$ and $\hat g(2n+1)=1$. Now you need to check that if $\hat g=\hat h$, it must be that the original sequences $g$ and $h$ were the same. But you should be able to do that, just look at $\hat g(2n)$ versus $\hat h(2n)$ and at $\hat g(2n+1)$ versus $\hat h(2n+1)$.
EDIT: In response to your request to show that the correspondence $g\mapsto\hat g$ is one-to-one, let me go into some detail. Suppose that $g$ and $h$ are sequences such that $\,\hat g=\hat h$. This means in particular that for every even-odd pair $(2n,2n+1)$, $\,\hat g$ and $\,\hat h$ agree are the same there, that is $\,\hat g(2n)=\hat h(2n)$ and $\,\hat g(2n+1)=\hat h(2n+1)$. According to your construction, the only way that this can happen is for the original sequences to have been equal at $n$, that is, $g(n)=h(n)$. This argument is valid for every integer $n$, so that $g$ and $h$ are the same sequence, which is what needed to be shown.
A: What do you mean by notification?
If two seqeunces $a$ and $b$ differ on position $i$ then (if we use your proposed $g$) the seuences $g(a)$ and $g(b)$ should differ on at least one of the positions $2i+1$ and $2i+2$.
A: There are many set theoretic tricks which are hard to write formally but easy to explain.
One of the classical tricks like that is to use $0$ as delimiters and $1$ to encode the data. By this I mean, given a sequence in $\{0,1,2\}^\Bbb N$ (and this trick works even for $\Bbb{N^N}$) we write each value in base $1$ (i.e. a sequence of $1$'s whose sum is the coordinate value), add a $0$ to the end, and concatenate it to the previous strings we have.
Slightly more formally we consider for $0,1,2$ a binary string $(0),(1,0),(1,1,0)$ respectively, write it as $f(i)$. Then $\langle a_i\mid i\in\Bbb N\rangle$ is mapped to $f(0)f(1)f(2)\ldots$.
To see that this is injective, simply show that you can reverse this function. By considering the intervals between $0$'s in a sequence in $\{0,1\}^\Bbb N$ we can decode the values of the original sequence.
A: To prove it injective, you have to show that there are not two input strings that map to the same output string.  Because each character in the input becomes two characters in the output you are OK.  If you did $0 \to 0, 1 \to 01, 2 \to 10$ you would have a problem:  $010$ could come from $02$ or from $10$.
