# $f:[0,1)\to \mathbb{R}$ invertible and surjective (onto)?

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?

• 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$$\Bbb R=[0,\infty)\cup[-1,0)\cup[-2,-1)\cup[-3,-2)\cup\cdots.$$ Sep 13, 2022 at 17:31
• Does this answer your question? How to define a bijection between $(0,1)$ and $(0,1]$? Sep 13, 2022 at 17:35
• It is possible even with the extra constraint. Sep 13, 2022 at 17:40
• 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.) Sep 13, 2022 at 17:56
• 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.) Sep 13, 2022 at 18:08

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

• Interesting. Why are infinitely many pieces necessary? Similarly to how one can extend $(0,1)$ to $[0,1)$? Sep 13, 2022 at 17:48
• 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. Sep 13, 2022 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. Sep 13, 2022 at 17:55
• Thank you very much. This is actually extremely useful and interesting. Thanks
– W2S
Sep 13, 2022 at 17:57
• Shouldn't the Euler characteristic should be 0 of your graph? Because each piece has euler char 0? Sep 13, 2022 at 17:59

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 $$x=0$$:

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

• But then $f(0)=f(0.5)=0$ right? So it's not invertible
– W2S
Sep 13, 2022 at 17:43
• This is not bijective (invertible) because $f(x)=f(.5)$. Sep 13, 2022 at 17:43
• Oh right, then add the example of the link with the bijection between $(0,1)$ and $[0,1)$ Sep 13, 2022 at 17:45