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Is there any (not necessarily continuous) function $f:[0,1)\to \mathbb{R}$ such that $f$ is invertible, $f(x)$ has a vertical asymptote at $x=1$ with $f(x)$ approaching $+\infty$ as $x$ moves closer to $1$, and that $\mathrm{dom}(f^{-1})=\mathbb{R}$? I've the feeling that such a function can't actually exist. Any idea?

W2S
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    Hint: both spaces can be broken into infinitely many half-open intervals. For example,$$[0,1)=\left[0,\frac12\right)\cup\left[\frac12,\frac23\right)\cup\left[\frac23\cup\frac34\right)\cup\cdot$$and$$\Bbb R=[0,\infty)\cup[-1,0)\cup[-2,-1)\cup[-3,-2)\cup\cdots.$$ – Akiva Weinberger Sep 13 '22 at 17:31
  • Does this answer your question? [How to define a bijection between $(0,1)$ and $(0,1]$?](https://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1) – Anne Bauval Sep 13 '22 at 17:35
  • Sorry I forgot one of the constraints... :( – W2S Sep 13 '22 at 17:37
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    It is possible even with the extra constraint. – Akiva Weinberger Sep 13 '22 at 17:40
  • Indeed, e.g. by choosing some well-known functions with asymptotes (like in my answer) and there are many of those – LegNaiB Sep 13 '22 at 17:42
  • @AkivaWeinberger yes, your example works even with the restriction. Thank you! – W2S Sep 13 '22 at 17:44
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    See my comments under my answer for more. (If the two sets have different Euler characteristics - defined as "vertices minus edges" - then the graph must use infinitely many pieces.) – Akiva Weinberger Sep 13 '22 at 17:56
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    In two dimensions, the equivalent notion is "vertices minus edges plus faces", which you may have heard of before. (I won't define "tame" here, but full theorem is, two tame sets have a tame bijection iff their Euler characteristics are the same and their highest dimensions are the same.) – Akiva Weinberger Sep 13 '22 at 18:08

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Here is one example. Note that the graph has infinitely many pieces; in fact, this is necessary for this problem.enter image description here

Akiva Weinberger
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  • Interesting. Why are infinitely many pieces necessary? Similarly to how one can extend $(0,1)$ to $[0,1)$? – LegNaiB Sep 13 '22 at 17:48
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    If a graph has finitely many pieces, then you may compute its Euler characteristic as follows: an open curve (without endpoints) has value $-1$; a closed curve (with both endpoints) or an isolated point has value $1$; a half-open curve has value $0$. Add these up to get the graph's Euler characteristic. It turns out that this always equals the Euler characteristic of both the domain and the codomain. Since $[0,1)$ and $\Bbb R$ have different Euler characteristics ($0$ and $-1$), the graph of a bijection must use infinitely many pieces. – Akiva Weinberger Sep 13 '22 at 17:50
  • Another way to think of the same calculation: it's vertices minus edges (where vertices are isolated points or endpoints, and edges are intervals). A closed interval has two vertices and one edge so it has Euler characteristic $-1$. My graph has Euler characteristic $\infty-\infty$, which is undefined. – Akiva Weinberger Sep 13 '22 at 17:55
  • Thank you very much. This is actually extremely useful and interesting. Thanks – W2S Sep 13 '22 at 17:57
  • Shouldn't the Euler characteristic should be 0 of your graph? Because each piece has euler char 0? – LegNaiB Sep 13 '22 at 17:59
  • @LegNaiB This only applies to so-called "tame sets", which have finitely many pieces. (The "vertices minus edges" definition is probably clearer here.) – Akiva Weinberger Sep 13 '22 at 18:07
  • ah thanks for your explanation! – LegNaiB Sep 13 '22 at 18:08
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We know that $[0,1)$ and $\mathbb{R}$ have the same size (both are uncountably infinite), so there actually exists a bijection between those two sets.

For matching $[0,1)$ to $\mathbb{R}_{\geq 0}$ you can simply take an asymptotic function and change it so that it fits into the interval, e.g. $\frac{1}{1-x} - 1$.

To get all $\mathbb{R}$ as the image, you can just define the function stepwise by going to infinity on $[0.5, 1)$ and to negative infinity on $(0, 0.5)$. As the function doesn't need to be continuous, you can just add some extra value for $x=0$:

$$ x \in [0.5, 1): f(x) = \frac{1}{1-x} - 2, \\ x\in (0,0.5): f(x) = -\frac{1}{x} + 2, \\ x = 0: f(x)= 0 $$

LegNaiB
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