It is well-known that the function $f (x) = \dfrac{1}{x}$ is not bijective on the domain of non-negative real numbers (that is, $[0, \infty)$.) , since $f( \cdot)$ may not be well-defined at zero point.

I am curious that what if we admit the convention that $0^{-1} = \infty$ and $\infty^{-1} = 0$.

Then, under this setting, is $f(x)$ bijective on $[0, \infty]$ now?

Could anyone help me out please? Thank you very much in advance!

  • $\begingroup$ Whether $\infty$ is a symbol or a number, it's still an element of your set $[0,\infty]$. The function $f:[0,\infty]\to [0,\infty]$ is bijective because no two elements of the set map to the same element, and every element of the set is mapped to by some element. $\endgroup$ – gj255 Jul 4 '18 at 23:21
  • $\begingroup$ Thank you so much @gj255 . I got it now :) $\endgroup$ – Paradiesvogel Jul 4 '18 at 23:29
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    $\begingroup$ "as far as I know, $\infty$ is just a symbol rather than a number." It is no more or less of a number than $\sqrt{2}, i,$ or $j$ are. True, $\infty$ is not a real number as in an element of $\Bbb R$, nor is it a complex number or a quaternion, etc... but it is an "extended real number" or "extended complex number" etc... Anything could be a "number" in the proper context, including infinity. It just so happens that in most contexts it is not, but that is not to say that there aren't some contexts where it is, the scenario you describe being one of them. $\endgroup$ – JMoravitz Jul 4 '18 at 23:36
  • $\begingroup$ @JMoravitz Thanks for your explanations. I learned it :) and really appreciate it $\endgroup$ – Paradiesvogel Jul 4 '18 at 23:42
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    $\begingroup$ We could, as far as the rules of what you are allowed to do define $\frac 1{pinkhonkhonk} = Babar$ and $\frac 1{Babar}=Tantor$ and $\frac 1{Tantor} = pinkhonkhonk$ as say $f:(0,\infty)\cup\{pinkhonkhonk,pinkhonkhonk,Tantor, Babar\}\to(0,\infty)\cup\{pinkhonkhonk,pinkhonkhonk,Tantor, Babar\}$ via $f(x)=\frac 1x$ and that is a bijection. So defining the extended reals =$\Bbb R\cup\{\pm\infty\}$ is certainly acceptable. Just don't assume it means anything more than it does or behaves as the reals do. $\endgroup$ – fleablood Jul 5 '18 at 0:34

Yes, if you define $f(0)=\infty$ and $f(\infty )=0$, then your function

$$ f(x)=1/x \text { if $x\in(0,\infty)$ }$$ along with $f(0)=\infty$ and $f(\infty )=0$ will be bijective on $[0,\infty]$

It will map $(0, \infty)$ to $(0,\infty)$ and maps the endpoints to the endpoints.


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