Example of $f:\mathbb{R}\to\mathbb{R}$ injective and bounded, but with inverse not bounded or injective. I am trying to come up with an example of a bounded and injective function $f:\mathbb{R}\to\mathbb{R}$ such that $f^{-1}$ is not injective or bounded.
What are examples that could apply in this situation?
Thanks for helping! :D
 A: As a first thought, I would say $f^{-1}$ is always injective, since a function is invertible if and only if it is injective and surjective. $f^{-1}$ is certainly invertible since its inverse is $f$, thus it is also injective.
For the bounded question, consider the function $$
f(x) = \left\{
\begin{aligned}
&\dfrac{1}{1+(\frac{1}{x})^2} &&: x > 0\\
&0 &&: x=0 \\
&-f(-x) &&: x<0
\end{aligned}
\right.$$
This function is bounded between -1 and 1 and one-to-one on $\Bbb R$. It has an inverse, at least for its image (which is not all of $\Bbb R$, since it's bounded), $f^{-1}: (-1,1) \longrightarrow \Bbb R$. If you look at the graph of $f$, you can see that this inverse will not be bounded.
A: Consdier the function $f: \{0,1,2,3,\dots\}  \to \Bbb R$ defined by
$\tag 1 n \mapsto {\displaystyle \sum _{i\mathop {=} 1}^{n} 2^{-i}}$
By convention/definition, $f(0) = 0$.
The function $f$ can be extended to all of $\Bbb Z$ by requiring that $f(-x) = -f(x)$.
So we have an injective and increasing function $f: \Bbb Z \to (-1,1)$.
Using linear interpolation (see also the wiki piecewise linear function article) we have an injective extension (might as well keep abusing the $f$ setting),
$\tag 2 f: \Bbb R \to (-1,1)$
By design, $f$ is a bijective function.
