# Find a bijection from $\mathbb R$ to $\mathbb R-\mathbb N$

Find a bijection from $\mathbb R \to \mathbb R-\mathbb N$. I want to set my function up such that all natural numbers get mapped to $n+.1$ and reals of the form $n+.1$ to $2n+.1$ is this correct?

• No, this won't be well-defined, since both, $2n$ (as a natural) and the real $n+0.1$ will be mapped to the same value. – AlexR Dec 15 '14 at 9:52
• so how should I map it? I wanted to assign some to odd and the other to evens. I am really confused haha – user201288 Dec 15 '14 at 9:53
• Try to find a bijection $[0,1)\leftrightarrow(0,1)$ first. – AlexR Dec 15 '14 at 10:10
• – Martin Sleziak Jul 15 '15 at 13:23

That won't quite work - as described, $n+0.1 \mapsto 2n+0.1$ and $2n \mapsto 2n+0.1$, so it isn't a bijection.

The easiest way I can think of for $\mathbb{R} \rightarrow \mathbb{R} \setminus \mathbb{N}$ would be to biject $\mathbb{Z} \rightarrow \mathbb{Z} \setminus \mathbb{N}$ and let all non-integers remain fixed. (For example, map $\mathbb{N}$ to negative odd integers and $\mathbb{Z}_{<0}$ to negative even integers.) This, of course, doesn't help if you then want a bijection $\mathbb{R} \rightarrow \mathbb{R} \setminus \mathbb{Z}$.

As other answers argue, what you describe won't be a bijection, but we can fix that.

You want to map $n$ to $n+0.1$, and you correctly observe that you need to map that number somewhere else. What about $n+0.01$? Then we need to move that number too, this time to $n+0.001$ an so on. In short, if fractional part (of a positive number) is of the form $\frac{1}{10^k}$ we put it at $\frac{1}{10^{k+1}}$.

I hope that helps $\ddot\smile$

The function that you've designed will map $1.1\rightarrow 2.1$ and $2\rightarrow 2.1$, so it's not bijective. One thing we can exploit here is that we still have all of the negative integers left to play with, so we can try to bring them in. For $n\in\mathbb{N}$, we can map $n\rightarrow-2n$, so we're basically hitting $\{-2, -4, -6, \dots\}$. Then for $m\in\mathbb{Z}_{<0}$ we can map $m\rightarrow2m+1$, which would hit $\{-1, -3, -5, \dots\}$. Then fix everything else in $\mathbb{R}$ and this should work.

If we found a bijection $[0,1)\to(0,1)$, then you could extend it to a bijection of $\mathbb R\to\mathbb R\setminus \mathbb Z$ trivially by shifting. Note that the function cannot be continuous because the preimage of an open set would not have to be open.

For finding a bijection of $[0,1)$ onto $(0,1)$ I don't have any idea at the moment, unfortunately.

I think you can do it this way.

We define a function $f(x) : \mathbb R \to \{\mathbb R - \mathbb N \}$ as $$f(x) = \begin{cases} \tan^{-1}x & \text{when }x \in \mathbb R-\{\tan(-2),\tan(-1),\tan(0),\tan(1),\tan(2)\} \\ x + \frac{7} {2} & \text{when } x\in \{-2,-1,0,1,2\} \end{cases}$$

Now we see that the range of the above function is $$\left\{\left\{-\frac{\pi}{2} , \frac{\pi}{2}\right\} - \left\{-2,-1,0,1,2\right\}\right\} \cup \left\{\frac{3}{2},\frac{5}{2},\frac{7}{2},\frac{9}{2},\frac{11}{2}\right\}$$

while the domain is $\mathbb R$

This may help.

Assuming natural numbers not containing zero: $\Bbb N = \{1,2,\dots\}$.

Choose any countably infinite subset of $\Bbb R$ disjoint with $\Bbb N$, for example: $$A=\{n+0.1 | n\in\Bbb N\}$$ and order it into a sequence $(a_n)_{n\in\Bbb N}$ such that $$a_n = n+0.1$$ Then $$x\mapsto \begin{cases} a_{2n} & \text{if } x=a_n \\ a_{2n-1} & \text{if } x=n \\ x & \text{otherwise} \end{cases}$$ maps $\Bbb R$ onto $\Bbb R \setminus \Bbb N$.