# Bijection between $\mathbb{N}\times \mathbb{R}$ and $\mathbb{R}$

How to exhibit a bijection between $\mathbb{N}\times \mathbb{R}$ and $\mathbb{R}$.
I think first we try to find bijection from $\mathbb{N}\times \mathbb{R}$ to (0,1) then bijection between (0,1) to (-π/2,Π/2) then from that to R by f(x)= tan(x). by how I define the bijection from $\mathbb{N}\times \mathbb{R}$ to (0,1)?

Hint: How many intervals of the form $(n, n+1)$ are there? Can you use that and then reuse the same bijection?

• there are infinite intervals of this form. but I can't find the actual bijection between them – Swarnadeep Bagchi May 7 '18 at 8:24
• Can you map each $(i, \mathbb{R})$ in $\mathbb{N}$ x $\mathbb{R}$ to (i,i+1) – E-A May 7 '18 at 9:01

First take bijection

$$f:\mathbb{R}\to(0,1)$$ $$f(x)=\frac{\tan(x-\pi/2)}{\pi}$$

Now define the bijection

$$F:\mathbb{R}\to[0,1)$$ $$F(x)=\begin{cases} 0 & x=0 \\ f(n-1) & x =n\text{ for }n\in\mathbb{N}, \ n\geq1 \\ f(x) &\text{otherwise} \end{cases}$$

With that we can construct the bijection:

$$G:\mathbb{Z}\times\mathbb{R}\to\mathbb{R}$$ $$G(n, x)=F(x)+n$$

For the final bijection just compose it with any bijection $\mathbb{N}\times\mathbb{R}\to\mathbb{Z}\times\mathbb{R}$.

• Ok I understand – Swarnadeep Bagchi May 7 '18 at 15:11