# Bijection between $\mathbb N^+ \times \mathbb R^+$ and $\mathbb R^+$

Let $\mathbb N^+$ denote the set of natural numbers bigger than $0$ and let $\mathbb R^+$ denote the set of real numbers bigger than $0$.

Is there a way to write down an explicit bijection between $\mathbb N^+ \times \mathbb R^+$ and $\mathbb R^+$?

You can replace $\mathbb{N}^+$ by $\mathbb{N} = \{ 0 , 1 , 2 ... \}$. Also $\psi : \mathbb{R}^+ \cong [0,1)$. Then the required bijection can be $$\mathbb{N} \times \mathbb{R}^+ \ni (n,r) \longmapsto n+\psi(r) \in \mathbb{R}^+ \ .$$

• Thank you, I corrected the map. – LinAlgMan Jul 24 '14 at 12:35

Yes:

Consider $f(x)=\dfrac{x}{1+x}$, it is a bijection $\mathbb R^+\to\mathbb [0,1[$. Now, $g(n,x)=n+x$ is a bijection like you want.

• This is actually a bijection $\mathbb{R}^+ \to (0,1)$, not $[0,1)$, since according to the OP's notation $0 \not \in \mathbb{R}^+$. What you get is almost a bijection, but there are countably many points not in the image of the function. – Clive Newstead Jul 24 '14 at 12:31
• it depends if "bigger" in the question means strictly bigger or $\geq$. Anyway it is not hard to go from one to the other with bijections. – Denis Jul 24 '14 at 12:35
• No element of $\mathbb N$ is in the range of that function... – Thomas Andrews Jul 24 '14 at 12:39
• Finding a bijection $\mathbb{R} \setminus \mathbb{N} \cong \mathbb{R}$ is pretty hard if you want a neat explicit description. (FWIW, "bigger than" unambiguously means $>$ and not $\ge$.) – Clive Newstead Jul 24 '14 at 12:40
• You just need bijections $\mathbb N\to \mathbb N^+$ and $\mathbb R^+\cup\{0\}\to \mathbb R^+$ (and the first gives the second by the way). – Denis Jul 24 '14 at 13:28

Well, there exist the following bijections:

1. An explicit bijection between $\mathbb R$ and $(0,1)$ (namely, scaled the co-tan function)
2. An explicit bijection between $\mathbb R^+$ and $\mathbb R$ (the $\exp$ function)
3. An explicit bijection between $(0,1)$ and $(0,1]$

This means there exists an explicit bijection between $\mathbb R^+$ and any semiclosed interval of $(a,a+1]$ length $1$, call that $f_{a}$. Now just create a bijection from $\mathbb N^+\times \mathbb R^+$ to $\mathbb R^+$ from the individual $f_a$s

There are many bijections between $(0,1)$ and $\mathbb R$. Using them you can get easily bijection between $\mathbb N\times\mathbb R$ and $\mathbb R\setminus\mathbb Z = \bigcup_{n\in\mathbb Z} (z,z+1)$.

So it only remains to modify the bijection to include integers. This can be done using "Hilbert hotel type" argument. (Something similar to this, with the difference that now you want to add infinitely many elements.)

EDIT: I have worked with $\mathbb R$ and not $\mathbb R^+$. As other answers explain, there is a bijection between $\mathbb R$ and $\mathbb R^+$. So this problem can be solved, too.

• Might as well use $(0,1]$ so you can get all of $\mathbb R$... I guess that's just using Hilbert's hotel one step earlier... – Thomas Andrews Jul 24 '14 at 12:38