Let, A⊂R2A\subset\mathbb{R}^2. Show that AA can contain at most one point pp such that AA is isometric to A∖{p}A \setminus \{p\}.

A challenge problem from Sally’s Fundamentals of Mathematical Analysis.

Problem reads: Suppose A is a subset of R2. Show that A can contain at most one point p such that A is isometric to A{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{0}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)(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{p} is isometric to A{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!


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.

Source : Link , Question Author : David Bowman , Answer Author : Guy Fsone

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