Show set of functions mapping $\mathbb N$ to $\{0, 1\}$ has same cardinality as set of functions mapping $\mathbb N$ to $\{0, 1, 2\}$ I know $\{0, 1\}^{\mathbb N}$ has same cardinality as power set of $\mathbb N$, so maybe it's possible to prove $\{0, 1\}^{\mathbb N}$~$\{0, 1, 2\}^{\mathbb N}$?
BTW, what is the relationship between $\{0, 1\}^{\mathbb N}$ and binary representation? Can this be used for the proof?
 A: There are many ways to do this. 
Here is one. Show that cadinality of maps to 0,1 is at most cardinality of maps to 0,1,2, which is at most cardinality of maps to 0,1,2,3, which is equal to cadinality of maps to 0,1. 

Here is another using your idea. By the bijection, the cardinality is equal to the cardinality of the reals. 
A: Often the easiest way to show that two sets have the same cardinality is show that there is a bijection between the two sets.  The following function is an injection from $\{0,1,2\}^\mathbb{N} \to \{0,1\}^\mathbb{N}$.
$f \mapsto \left( n \mapsto \begin{cases}\frac{f(\frac{n}{2})}{2} & \text{ where } n \mod 2 = 0 \\
f(\frac{n}{2}) \mod 2 & \text{ where } n \mod 2 = 1
\end{cases}\right)$
Where all the divisions are integer divisions.
The idea is that if $f$ is a function of the type $\mathbb{N} \to \{0,1,2\}$ then it can be encoded as a sequence of values in a function $g(n) : \mathbb{N} \to \{0,1\}$.
Showing that there is a reverse injection can be done by writing an inverse of this mapping.
As you spotted we can do this because any number can be represented by a binary string. 
