A challenge problem from Sally's Fundamentals of Mathematical Analysis.
Problem reads: Suppose $A$ is a subset of $\mathbb{R}^2$. Show that $A$ can contain at most one point $p$ such that $A$ is isometric to $A \setminus \{p\}$ with the usual metric.
I'm really not sure where to begin. I've found a fairly trivial example of a set for which this is true: let $A$, for example, be $\{(n,0) : n \in \{0\} \cup \mathbb{Z}^+\}$. Then we may remove the point $(0,0)$ and construct the isometry $f(n,0) = (n+1, 0)$. This is clearly an isometry because $d((n,0),(m,0)) = d((n+1,0),(m+1,0))$, in other words, we are just shifting to the right. But now suppose we remove some $(p,0) \neq (0,0)$. Then we must have $d((m,0), (m+1,0)) = 1$ for all points $(m,0), (m+1,0)$, but since $(p,0)$ was removed we will always have a "jump" point where the distance between two successive points is $2$.
But I'm not sure where to proceed. Isometries are equivalence relations, so maybe we can show that if $A \setminus \{p\}$ is isometric to $A \setminus \{q\}$, then $p = q$?
I will say that given how often Sally's errant in his book and that some of the other challenge problems are open problems, this might not have a reasonable solution (if it's even true).
Any ideas?
To avoid any confusion, the problem isn't asking a proof for the set not being isometric to itself minus two points at the same time. It's asking for a proof that there is at most one unique point that you can remove from the set and then create an isometry. This was something I misinterpreted for a while.
Edit: This is still stumping me. I'm beginning to wonder whether it's even true at all. Well, I've put a bounty on it, which hopefully serves as bit more incentive to try this problem out!
Best Answer
This is not a formal answer to question. But just to let future readers and the OP know that a brilliant and detailed solution to this problem can be found Here in mathobverflow. This was prosed by @Ycor.