I tried $f(x)=\frac1x$, but then domain is $\Bbb R-\{0\}$. When we divide functions into two parts, for example, when domain is $\{0\}$, $f(x)$ is something, but we have to consider that function must be bijective. What do you think?
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$f(x) = x+1$ if $x$ is a non-negative integer, and $f(x)=x$ otherwise. – Ned Jan 07 '22 at 16:40
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what about range (0,1) though, range must be R-{0} – mertk Jan 07 '22 at 17:08
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$f(0.367)=0.367$ etc, all reals except $0$ are in the range – Ned Jan 07 '22 at 17:11
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There is a simple way to find a bijection from an infinite $A$ to $A -{x}$. Do you know it? The identity map is a bijection. But it maps $0$ to $0$ and we don't have $0$ in the codomain... so we must map $0$ to somewhere else. So map $0\mapsto 1$. But now we must map $1$ to somewhere else.... So map it to $2$. Now we must map $2$ to somewhere else.... so map it to $3$ and so on. For every $n\in \mathbb N$ map it to $n+1$. Everything else .... just map to itself. That'll do $f(x)=\begin{cases}1&x=0\ x+1&x\in\mathbb N\ x&\text{otherwise}\end{cases}$. – fleablood Jan 07 '22 at 20:42
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If you think about this this is just the hilbert hotel. The only difference is that all the people who are not in a natural numbered room are allowed to stay put. – fleablood Jan 07 '22 at 20:46
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"what about range (0,1) though, range must be R-{0}" Not a problem. If $x\in(0,1)$ then $f(x) = x$ and the function is still surjective. And if $f(w) =x \in (0,1)$ then $x\not\in \mathbb N$ so $w\not \in \mathbb N\cup{0}$ so $x=f(w)=w$ so $x=w$ so the function is still injective. – fleablood Jan 07 '22 at 20:52