Proving all 1-dimensional isometries can be expressed as two or less reflections As an assignment, we are asked to prove that the group of all discontinous (having no accumulation point) isometries without fixed points in one dimension are generated by only one element (which is supposed to be a translation).
As a part of this proof, I want to show that all 1-dimensional isometries can be expressed as two or less reflections. I want to do this to rule out any other thinkable or unthinkable isometry as the generating element.
I thought I could use the fact that one 1-dimensional isometry is uniquely defined by two points. I.e., let $\text{Iso}(\mathbb{R})$ be the group of all 1-dimensional isometries and $T,S \in \text{Iso}(\mathbb{R})$ and $Tx = Sx, Ty = Sy$ then $Tz = Sz$ for all $x,y,z\in\mathbb{R}$. Then, in one dimension, there are only three options:
i) $Tx = x$, $Ty = y$ for some $x,y\in\mathbb{R}$. In this case we obviously have $T = I$ (the identity).
ii) $Tx = x$ but $Ty \neq y$ for all $y \neq x$. I have proved that $T$ is a reflection around $x$ in this case.
iii) $Tx \neq x$ for all $x$. This is where I am stuck.
It feels very intuitive that iii) will lead to a translation, which in turn may be expressed as two reflections. However, I have yet to prove this. I have tried using that $|x - y| = |Tx - Ty|$ for all isometries and tried introducing a point $y$ and a reflection $R_y$ around that point to show that $TR_y$ has to be a reflection, and deduce from this that $T$ has to be a translation, but without success.
Any help would be appreciated.
 A: Here is a proof of iii):
Since $T$ is an isometry, for any $x$ we have $|T(x)| = |x - T^{-1}(0)|$. Since the space is $\mathbb{R}$ this means there are only two options: either $T(x) = x-T^{-1}(0)$ or $T(x) = T^{-1}(0) - x$. The difference between these two options is $2(x-T^{-1}(0))$ so since $T$ is continuous we must choose one option consistently on each of the two connected components of $\mathbb{R}\setminus\{T^{-1}(0)\}$. Choosing $T(x) = T^{-1}(0) - x$ wouldn't work on the component containing $\frac{1}{2}T^{-1}(0)$ because $\frac{1}{2}T(0)$ would be a fixed point (which we assume $T$ does not have) so we must choose $T(x) = x - T^{-1}(0)$ on that component.
Since $T$ should also be smooth, this means we must also choose $T(x) = x - T^{-1}(0)$ on the second component since otherwise the limit $\mbox{lim}_{x\to T^{-1}(0)}T'(x)$ would not exist.
We would have $\mbox{lim}_{x\to T^{-1}(0)}\frac{d}{dx}(x-T^{-1}(0)) = 1$ whereas $\mbox{lim}_{x\to T^{-1}(0)}\frac{d}{dx}(T^{-1}(0) - x) = -1$.
Hence $T$ is the translation $T(x) = x - T^{-1}(0) = x + T(0)$.
