In $\mathbb{R^2}$, taking the subset $S=\{\sigma^n(1,0) |n\geq 0\}$ where $\sigma$ is a rotation by an irrational angle, I see that $S$ is isometric to $S-\{(1,0)\}$. How do I show that this can not be done in $\mathbb{R}$?
So far I have: (Let $H$ be the bounded set.)
- No translation of $H$ can be contained in a proper subset.
- Any reflection of $H$ whose reflection is contained in $H$ must contain all of $H$. (I'm unsure about this)
Here is a start. If the bounded set contains both its inf and sup, then so must an isometric subset for diameter considerations. Furthermore any isometry of the subset onto the full space must send the boundary to the boundary. Every point in the bounded set is determined uniquely by its distance from the ends and thus the map of the boundary determines the isometry. Thus, the map must be a reflection. Some point of the subset p must reflect onto a point in the complement q. But then, no point in the reflection of the subset covers p (as q was the only candidate). Hence, not onto, a contradiction.
If you have only either the inf or sup the argument is even easier. However, if you have neither I suspect you would want to construct them in some manner. Maybe by considering equivalence classes of pairs of Cauchy sequences who term-wise distance approaches the diameter.
Hope this helps.