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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!

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    $\begingroup$ Good example, well thought through. This looks like a toughie. $\endgroup$ – Ben Grossmann Nov 22 '16 at 4:22
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    $\begingroup$ @Omnomnomnom: A need not have any isolated points: let $A=\{\langle 1,n\rangle:n\in\Bbb N\}$, where the ordered pairs are polar coordinates. An anticlockwise rotation about the origin through $1$ radian is an isometry of $A$ to $A\setminus\{\langle 1,0\rangle\}$, and $A$ is dense in $S^1$. $\endgroup$ – Brian M. Scott Nov 22 '16 at 23:32
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    $\begingroup$ Question is answered here: mathoverflow.net/questions/255448/… $\endgroup$ – Ben G. Dec 26 '16 at 8:24
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    $\begingroup$ @BenG. is there not a purely analytic solution? I would be disappointed if this is true (and disgusted by Sally) $\endgroup$ – David Bowman Dec 27 '16 at 23:42
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    $\begingroup$ @GuyFsone: Why did you assign a bounty here, at MSE? It looks like an invitation to post incorrect solutions. The problem was solved by YCor in his answer here: mathoverflow.net/questions/255448/…. $\endgroup$ – Moishe Kohan Nov 13 '17 at 20:37
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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.

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This is an elementary proof in the case that neither isometry is a rotation. Suppose there to be isometries $P$ mapping $A$ to $A \setminus \{p\}$ and $Q$ mapping $A$ to $A \setminus \{q\}, p≠q$. $P$ and $Q$ cannot be self-inverse and so are either translations, glide reflections or rotations. Unless one or both of the isometries are rotations we therefore know that $P^2$ and $Q^2$ are both translations.

If $P^2$ and $Q^2$ are not parallel translations, the effect on a point of successively performing $Q^2$ $n$ times, then $P^{-2}$ $ n$ times, then $Q^{-2}$ $ n$ times and finally $P^2 $$n$ times is to make the point traverse $4n$ points of a parallelogram. Furthermore, by increasing the value of $n$, we can make the second and third edges of this parallelogram as far from the points $p,P(p),q$ and $ Q(q)$ as we wish. Starting from $P(p) ∈A$ we can therefore assume that all of the $4n$ points are also in $A$. In particular,$$P^{-1} (p)= P^{2(n-1)} Q^{-2n} P^{-2n} Q^{2n} P(p)∈A,$$ a contradiction.

If $P^2$ and $Q^2$ are parallel translations then first suppose $P$ and $ Q $ are glide reflections with parallel but not equal axes. Then $ PQ $ can be used in place of $ Q^2$ in the above case. Otherwise, $P$ and $Q$ commute and then $Q^{-1} (p)∈A$ since $p≠q$. Therefore $P^{-1} Q^{-1} (p)∈A$ since $Q^{-1} (p)≠p$. Then $P^{-1} (p)=QP^{-1} Q^{-1}(p)∈A$, a contradiction.

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  • $\begingroup$ Why should an isometry of A with A-p necessarily extend to an isometry of R^2? (Which is what I presume you are using in order to classify isometries into translations, rotations, etc.) $\endgroup$ – Bruno Joyal Nov 16 '17 at 16:37
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    $\begingroup$ I was assuming from the outset that we were dealing with isometries of the Euclidean plane. $\endgroup$ – S. Dolan Nov 16 '17 at 17:03
  • $\begingroup$ $A$ is a metric space in its own right, and the notion of "isometry of $A$" makes no reference to its living in $\mathbb R^2$... $\endgroup$ – Bruno Joyal Nov 16 '17 at 17:50
  • $\begingroup$ The question does not use the phrase "isometry of A" . I chose to use isometries of the Euclidean plane for "an elementary proof" in some cases. $\endgroup$ – S. Dolan Nov 16 '17 at 19:43

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