Describe explicitly the bijection between $2^\Bbb R$ and $\Bbb N^\Bbb R$ I have a problem from elementary set theory. I don't know, how to solve it and from what start.
How to describe explicitly the bijection between $2^\Bbb R$ and $\Bbb N^\Bbb R$?
 A: You can build a bijection between $2^{\mathbb Z}$ and ${\mathbb N}^{\mathbb Z}$ by :


*

*Let $(a_z)_{z\in\mathbb Z}$ an element of  ${\mathbb N}^{\mathbb Z}$.

*Define $b_z$ by :

*

*Let $S_n=\left(\sum_{k=0}^n(a_k+1)\right)-1$ for $n\ge 0$

*Let $S_n=-\left(\sum_{k=n}^{-1}(a_k+1)\right)$ for $n<0$

*Then for any $n\in \mathbb Z$ define $b_i=0$ if $\exists n, i=S_n$, otherwise $b_i=1$. 



In fact it just encoding the sequence $a_z$ into unary separated by $0$s. So it's easy to verify that we just define a bijection that we now name $\Psi$. (So here $b_z=\Psi(a_z)$)
Example :
$$\begin{array}{|c|c|}z&\dots&-4&-3&-2&-1&&0&1&2&3&4&5&6&7&8&9&\dots\\
\hline
a_z&&&&2&1&&1&2&0&3&\\
\hline
b_z&0&1&1&0&1&&1&0&1&1&0&0&1&1&1&0\\
\end{array}$$
So if you count the number of $1$s between each $0$s in $b_z$, you find back $a_z$. The origin point was chosen between $-1$ and $0$.

Now, an element $f$ of $\mathbb N^{\mathbb R}$ is just a function from $\mathbb R$ to $\mathbb N$.
It can be seen also as a function  $\phi$ from $[0,1)$ to $\mathbb N^{\mathbb Z}$ such that $$\phi(x)= (z\mapsto f(z+x))$$
and ($x=\lfloor x\rfloor+\{x\}$) 
$$f(x)=\phi(\{x\})(\lfloor x\rfloor)$$
By using the previous bijection on all $\phi(x)$, you obtain the bijection you want and the function $g$ from $\mathbb R$ to $\{0,1\}$ by :
$$g(x)=\Psi\left(\phi(\{x\})\right)(\lfloor x\rfloor)$$
