Is there a subset $X$ of plane with two points $x, y$ such that each one of $X \setminus \{x\}$, $X \setminus \{y\}$ is isometric to $X$? I tried hard to construct a counterexample but failed.

Sorry if this is too easy for mathoverflow.

Edit: The question was initially asked on MathSE on Nov. 22 '16, with a reference to the book by Paul Sally, Fundamentals of Mathematical Analysis, Pure and Applied Undergraduate Texts 20. Amer. Math. Soc. 2013 (Amazon link). It appears as Problem 3.2 p.77, stated precisely as

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 $A\smallsetminus\{p\}$ with the usual metric.

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    $\begingroup$ you mean "an example". And you also mean $x\neq y$ $\endgroup$
    – YCor
    Nov 23, 2016 at 21:58
  • $\begingroup$ Simple and nice! (Yes, we should have $\ x\ne y\ $ assumed). $\endgroup$ Nov 23, 2016 at 22:07
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    $\begingroup$ It's not possible using 2 translations. Indeed, if say $X$ is stable under $+a$ and $+b$, and stable under $-a$ except at $x$ and $-b$ except at $y$, then one can perform $y\mapsto y-a\mapsto y-a-b\mapsto y-b$ and get a contradiction (this "path" is valid because $y\neq x$ and $a\neq 0$) $\endgroup$
    – YCor
    Nov 23, 2016 at 22:28
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    $\begingroup$ I've added the tag group theory because it's related to group or semigroup action of isometries on the plane. $\endgroup$
    – YCor
    Nov 23, 2016 at 22:29
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    $\begingroup$ @JoelDavidHamkins It follows from my comment that there's no 1-dimensional example. Indeed, since in dim 1 non-translations have order 2, both $x,y$ should be translations. $\endgroup$
    – YCor
    Nov 24, 2016 at 7:08

3 Answers 3


(Initial post November 24, 2016, edited November 27, 2016) This does not exist.

The proof that $X$ doesn't exist is a bit elaborate and makes use of ends of coset spaces. I will prove:

(a) Let $\Gamma$ be a finitely generated subgroup of the group of isometries of the plane. If $\Gamma$ is not virtually infinite cyclic, then every $\Gamma$-commensurated subset $X$ of the plane is $\Gamma$-transfixed (see terminology below).

(b) For $\Gamma$ virtually infinite cyclic (that is, with an infinite cyclic finite index subgroup), a subset of the plane as prescribed does not exist.

Terminology (borrowed from here): if a group $\Gamma$ acts on a set $W$ (here, the plane), a subset $X\subset W$ is $\Gamma$-commensurated if $X\Delta\gamma X$ is finite for all $\gamma\in\Gamma$, and $\Gamma$-transfixed if there exists a $\Gamma$-invariant subset $X'$ such that $X'\Delta X$ is finite (clearly this implies that it is $\Gamma$-commensurated and furthermore that the cardinal of $X\Delta\gamma X$ is bounded independently of $\Gamma$).

(The link with ends of coset spaces is that for a finitely generated group $\Gamma$ with subgroup $\Lambda$, the Schreier graph has $\ge 2$ ends if and only if $\Gamma/\Lambda$ has a non-$\Gamma$-transfixed $\Gamma$-commensurated subset. Here restricting to transitive actions would be inconvenient, so I'm using the above language.)

First (a) and (b) imply the non-existence of $X$ as required.

We start from the classical fact that in a Euclidean space, any isometry between any two subsets has an isometric extension to the whole plane (see appendix C.2 in the Bekka-Harpe-Valette book, or check directly). So the question can be reformulated as follows:

(*) Does there exist a subgroup $\Gamma$ of the group of isometries of the Euclidean plane, generated by two isometries $\alpha,\beta$, such that, for some subset $X$ of the plane and $x,y\in X$ with $x\neq y$, we have $\alpha(X)=X\smallsetminus\{x\}$ and $\beta(X)=X\smallsetminus\{y\}$?

Indeed, take (a), (b) for granted, and assume by contradiction that $\Gamma$ exists. Then both generators commensurate $X$ and hence $\Gamma$ commensurates $X$. Since the first of this isometries maps $X$ to $X\smallsetminus\{x\}$, its $n$-th power maps $X$ to the complement of $n$ points in $X$: in particular the cardinal of $X\Delta\gamma X$ is unbounded and $X$ cannot be transfixed. So we get a contradiction, unless $\Gamma$ is virtually infinite cyclic, which is ruled out by (b).

Finally I'll justify that such "paradoxical" decompositions can occur without any reference to free subgroups:

(b') In the isometry group of the 4-dimensional Euclidean space, there exists $\Gamma$ and $X$ as in (*), with $\alpha,\beta$ generating a free abelian group of rank 2.

Let us first prove (a). The proof of the easier (b) is below.

Let $T$ be the group of translations in $\Gamma$. We distinguish two cases (the main one being the third).

1) $T$ is cyclic (either trivial or infinite).

Assuming that $X$ is not transfixed, since $\Gamma$ is finitely generated, by (Prop. 4B2 here), there exists a $\Gamma$-orbit $\Omega$ such that $\Omega\cap X$ is non-transfixed. Fixing a point $\xi$ in $\Omega$, write $\Omega=\Gamma/\Lambda$; that $\Omega\cap X$ is not transfixed implies that the number $e(\Gamma,\Lambda)$ of ends of the coset space $\Gamma/\Lambda$ is $\ge 2$ (by definition $e(\Gamma,\Lambda)>1$ means that $\Gamma/\Lambda$ has a $\Gamma$-commensurated subset that is not transfixed, or equivalently that is neither finite nor cofinite).

Then $\Gamma$ is a polycyclic group. Houghton (1982) proved that for a polycyclic group $\Gamma$, a subgroup $\Lambda$ satisfies $e(\Gamma,\Lambda)>1$ iff $\Lambda$ has Hirsch length 1 less than $\Gamma$ and has normalizer of finite index in $\Gamma$. Hence here, let $P$ be this normalizer and $P_+$ the set of motions in $P$ (of index at most 2 in $P$). Let $Y$ be the set of points in the plane fixed by $\Lambda$. Then $\xi\in Y$ and $Y$ is $P$-invariant.

Then $Y$ is an affine subspace. If $Y=\{\xi\}$, then $\xi$ is fixed by $P_+$, which is thus a group of rotations. Since $\Gamma$ is infinite, so is $P_+$, and so is some finite index subgroup $Q$ of $P_+$ that is normal in $\Gamma$; then $\xi$ is the unique fixed point of $Q$, and hence is fixed by $\Gamma$, a contradiction since $\Omega=\{\xi\}$ was supposed to contained a non-transfixed subset. If $Y$ is the plane, then $P_+$ is trivial, so $\Gamma$ is finite, a contradiction.

If $Y$ is a line, then $P_+$ acts by translations parallel to this line. Since $\Gamma$ is not virtually cyclic, $P_+$ is a finitely generated abelian group of $\mathbf{Q}$-rank $\ge 2$ and thus is 1-ended; it acts freely on the plane and hence transfixes every subset it commensurates, again a contradiction.

(Edit: see below for a more direct approach to Case (1))

2) $T$ is not cyclic (hence either it contains a free abelian group of rank 2, or a rank 1 infinitely generated abelian group of rank 1). A result of Scott-Sonneborn (Coll. Math., 1963), later rediscovered by Oxley (Math Z, 1972) implies that $T$ is 1-ended, in the sense that every subset of $T$ commensurated by $T$-translations is finite or cofinite [*]. Hence the intersection of $X$ with every $T$-orbit is either finite or cofinite. Actually this intersection is empty or the whole orbit for all but finitely many orbits [**]. Hence we can find a subset $X'$ with $X'\Delta X$ finite, that is $T$-invariant.

Since $T$ is normalized by $\Gamma$, $\Gamma$ permutes the $T$-cosets. Hence $X'\Delta\gamma X'$, for $\gamma\in\Gamma$, is a union of $T$-cosets, and thus either infinite or empty. Since $X'$ is commensurated by the $\Gamma$-action, it is thus empty and $X'$ is $\Gamma$-invariant, so $X$ is $\Gamma$-transfixed.

[*] From Scott-Sonneborn's paper, use Theorem 5, or from Oxley's paper, use Proposition 3.3 in rank $\ge 2$ and Proposition 3.11 in the infinitely generated rank 1 case.

[**] Suppose the contrary. Consider a countable infinite disjoint union $T\times I$ of copies of $T$ and for all $i$ a subset $F_i$ of $T$ (nonempty finite or a proper cofinite subset) such that the disjoint union $\bigcup F_i\times\{i\}$ is commensurated by $T$. Passing to complements, we can suppose that all $F_i$ are nonempty finite. Fix an infinite cyclic subgroup $Z$ of $T$. We can find elements $t_i\in T$ such that the $F_i+t_i\subset T$ are pairwise disjoint, and disjoint of $Z$. Then $\bigcup F_i+t_i$ is commensurated, infinite, disjoint from $Z$ and hence with infinite complement. This contradicts that $T$ is 1-ended.

Edit: here's a more direct approach to Case (1) in the setting of the question:

(1a) If $\Gamma$ has no nontrivial translation, then it's embeddable into $\mathrm{O}(2)$ hence virtually abelian. Then $\alpha^n$ is an element of infinite order and not a translation, hence a rotation of infinite order. Hence its fixed point is unique, and hence is fixed by some finite index subgroup of $\Gamma$, and hence by uniqueness by $\Gamma$. So the action fixes a point, say $O$. By removing entire $\Gamma$-orbits in $X$, we can suppose that $X$ is reduced to the (possibly equal) orbits of $x$ and $y$. One boils down to $\|x\|=\|y\|$ by possibly performing a similarity fixing $0$ to the orbit of $y$ (mapping this orbit to a subset disjoint from the orbit of $x$). Eventually $X$ lies in a cercle centered at $0$, and we can use the abelian case as in (2) unless $\Gamma$ is virtually cyclic.

(1b) the group of translations in $\Gamma$ is infinite cyclic: then it is normal, and hence centralizes a subgroup of index 2; since the centralizer of a nontrivial planar translation is reduced to the group of rotations, we deduce that $\Gamma$ has a subgroup of index 2 consisting of translations so $\Gamma$ is virtually cyclic.

Let us check (b). Then the "index character" $\tau:\Gamma\to\mathbf{Z}$ mapping $\gamma$ to $\#(X\smallsetminus\gamma X)-\#(\gamma X\smallsetminus X)$ is a homomorphism (this is a general fact). Here $\tau$ maps both $\alpha,\beta$ to 1; it is thus a surjective homomorphism and since $\Gamma$ is virtually cyclic, its kernel is finite, i.e. $\Gamma$ is finite-by-(infinite cyclic).

It is then easy to see that either $\Gamma$ fixes a unique point $\xi$ (which we can suppose to be 0), or fixes no point and preserves a line.

In the first case, $\Gamma$ acts freely on the complement of $\{0\}$. Being generated by two elements $\alpha,\beta$ of infinite order, it acts by rotations and hence is abelian. So we can get a contradiction exactly as in my comment for two translations (we can perform $y\mapsto\alpha^{-1}y\mapsto\beta^{-1}\alpha^{-1}y\mapsto\beta^{-1}y$).

In the second case, it preserves a line; since it is generated by two elements of infinite order, these act as nontrivial translations on this line and we deduce again that $\Gamma$ is abelian (action on this line, and hence on this coordinate is by translation, and action on the orthogonal coordinate is just by possible change of sign). We conclude again with the same trick.

Proof of (b'): for convenience, we use complex coordinates and work in $\mathbf{C}^2$. Fix an element $\zeta$ in the unit circle of infinite multiplicative order. Define the diagonal $\mathbf{C}$-linear isometries $\alpha(z,z')=(\zeta z,z')$ and $\beta(z,z')=(z,\zeta z')$. Define $$X=\{(\zeta^n,0):n\ge 0\}\cup\{(0,\zeta^n):n\ge 0\},\quad x=(1,0),\;y=(0,1).$$ It is immediate that they satisfy the required hypotheses of (*).

Edit (August 25, 2018)

This (the non-existence of a planar subset as in the OP's question) was originally proved by E.G. Straus, On a problem of W. Sierpinski on the congruence of sets, Fundam. Math. 44 (1957), p. 75-81; available without restriction (at this time) here (Biblioteka Nauki); see also the Math Review link. The proof seems completely by hand and takes 4 pages.


I don't think there is such a set for the plane, but I'll point out that there is one for the sphere $\mathbb S^2$. Namely, let $S$ and $T$ be two members of $SO_3$ such that the group $G$ they generate is free. Take any $x, y \in \mathbb S^2$ such that $y \notin G(x)$. Each member of $G$ can be uniquely written as a "reduced word" $U_1 \ldots U_n$ where each $U_i$ is one of $S$, $T$, $S^{-1}$ or $T^{-1}$, no $S$ and $S^{-1}$ are adjacent and no $T$ and $T^{-1}$ are adjacent; the empty word corresponds to the identity element $e$. Let $G_S$ be the subset of $G$ corresponding to reduced words that do not end in $S^{-1}$ and $G_T$ the subset corresponding to reduced words that do not end in $T^{-1}$. Then take $X = G_S(x) \cup G_T(y)$. We have $S(X) = X \backslash \{x\}$ and $T(X) = X \backslash \{y\}$.

  • $\begingroup$ Nice idea @Robert! I thought of disproving the original question by showing that: (1) any isometry $X\to X\setminus\{x\}$ must be affine. (2) If there were such an $X$, the free group would sit inside the affine group on the plane. However I was not able to fill in the details :-( $\endgroup$
    – Ruy
    Nov 24, 2016 at 11:35
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    $\begingroup$ @Ruy This approach with free groups is hopeless. For instance, consider the metric space consisting of two copies of the unit circle, where the distance in each circle is the standard one, and the distance between two points of each circle is 2. Then the isometry group has no free subgroup (nor even free subsemigroup): it has an abelian normal subgroup of index 8. But in this space, it is easy to find a subset $X$ as in the question. $\endgroup$
    – YCor
    Nov 24, 2016 at 17:33
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    $\begingroup$ (I mean: free groups allow to produce subsets $X$ as in the question, but absence of free groups will not help proving there's not any $X$.) $\endgroup$
    – YCor
    Nov 28, 2016 at 8:30

In order to prove that $X$ cannot exist, I think that you can argue in the following way: for $x$ to satisfy this condition in $X$, the set $X$ should contain a straight or a zig-zag "discrete ray" with the "top" at $x$, and all other elements should split into unions of either straight of zig-zag `discrete lines' in the same direction, depending on whether the isometry of $X$ and $X\backslash\{x\}$ is achieved by translation or a glide reflection. I use the terminology from here. (One can see that other types of isometry cannot map $X$ onto $X\backslash\{x\}$.) Now you show that none of the points can play the role of $y$.

Edit: After reading Yves's comment and Robert's answer, I realized that one has to go deeper and to use the fact that the group of isometries of the plane does not have free subgroups.

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    $\begingroup$ The isometry from $X$ to $X-\{x\}$ might be achieved by a rotation: for instance, consider in the complex plane the multiplication by $z$ of modulus 1 and infinite multiplicative order, and $X=\{z^n:n\ge 0\}$, which is mapped to $\{z^n:n\ge 1\}$ by the rotation multiplying by $z$. $\endgroup$
    – YCor
    Nov 24, 2016 at 0:34
  • $\begingroup$ You are right, I overlooked that. So one has actually to consider such "circular rays", too. $\endgroup$ Nov 24, 2016 at 0:44
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    $\begingroup$ @YCor, the example which you have mentioned was given a long time ago by Wacław Sierpiński. He published some notes about metric spaces (of a purely metric character). Perhaps Sierpiński was the first one but I don't know. $\endgroup$ Nov 24, 2016 at 1:45
  • $\begingroup$ @YCor, OK, I am glad. $\endgroup$ Nov 24, 2016 at 4:45
  • $\begingroup$ Free subgroups is a recipe to produce such things, but absence of free subgroups will not help much. $\endgroup$
    – YCor
    Nov 24, 2016 at 7:10

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