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Let $\mathbb{R}_{\geq 0}$ be the set of nonnegative numbers and $\mathbb{R}_{>0}$ the set of positive numbers, that is

$$ \mathbb{R}_{\geq 0} = \{\,x \geq 0 \mid x \in \mathbb{R} \,\} $$

and

$$ \mathbb{R}_{> 0} = \{\,x > 0 \mid x \in \mathbb{R} \,\} $$

Is it possible to define a bijection $f$ between these two sets?

  • Please show what you have tried. People will likely not respond with an answer because threads without shown attempts get deleted. I've had several of my answers disappear this way. – Favst Nov 19 '20 at 16:10
  • Why not? A simple example would be $f(x) = x + 1$..from the set of non-negative numbers $\to$ positive numbers... – Aman Kumar Nov 19 '20 at 16:11
  • @ProfessorofStupidity But this map is not bijective in the non-negative reals. – Peter Nov 19 '20 at 16:14
  • @Peter But we have to find bijection from non-negatives to positives. In that case this holds? Right? Or I am not able to understand your point... – Aman Kumar Nov 19 '20 at 16:19
  • We have no solution for $f(x)=\frac{1}{2}$ for example with a non-negative $x$. Hence this map is not surjective. – Peter Nov 19 '20 at 16:21
  • @Favst Well, I agree with the principle, but showing functions that surely aren't bijections didn't seem very smart to me. –  Nov 19 '20 at 16:38

4 Answers4

9

Yes, of course. First map every number that is not a non-negative integer to itself. Then map every non-negative integer n to n+1.

Acccumulation
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user247327
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    Clever solution (+1) – Peter Nov 19 '20 at 16:15
  • Thanks! I was trying to define the bijection without mapping any element to itself (for reasons that remain mysterious, I thought it would break injectivity). So, now I'm curious if there is a solution with the extra constraint that $f(x) \neq x$ for each x. –  Nov 19 '20 at 16:52
  • @BlackBrain doesnt $x \mapsto x+1$ work there to? – D. Ben Knoble Nov 20 '20 at 00:40
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You can also take every $[n,n+1)$ interval and map it into $(n,n+1]$ by reflecting it.

The idea here being mapping $[0,\infty)$ into $(0,\infty]$ via $\frac{1}{x}$, but we can only do it if the right extreme of the second interval is included. Luckily, we can cover $\mathbb{R}$ with intervals of such form.

rod
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  • If we choose $f(x)=\frac{1}{x}$ , we cannot map $0$. But the reflecting idea should work. – Peter Nov 19 '20 at 16:25
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Of course. Let f be from the set with 0 to the set without 0:

f(x) = x when x is not integer;

f(0)= 1

f(1) = 2

f(2) = 3

etc.

robin3210
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The have the same cardinality, so a bijection exists. You mentioned in a comment that you're also interested in a function that doesn't have any fixed points. You can adapt roddick's answer by just mixing up the intervals. For instance, you could send $[0,1)$ to $(10,11]$, $[1,2)$ to $(11,12]$, $[2,3)$ to $(0,1]$, and so on.

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