In the 1950s, Steinhaus posed the following problem. Is there

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1 Sets meetig isometric coies of the lattice Z 2 i exactly oe oit Steve Jackso ad R Daiel Mauldi Deartmet of Mathematics, Uiversity of North Texas, Deto, TX Commuicated by Robert Solovay, Uiversity of Califoria, Berkeley, CA, Setember 11, 2002 (received for review Aril 16, 2001) The costructio of a subset S of 2 such that each isometric coy of 2 (the lattice oits i the lae) meets S i exactly oe oit is idicated This rovides a ositive aswer to a roblem of H Steihaus [Sieriński, W (1958) Fud Math 46, ] I the 1950s, Steihaus osed the followig roblem Is there a set S i the lae such that every set cogruet to 2 has exactly oe oit i commo with S? The roblem seems to have first aeared i 958 aer of Sieriński (1) Steihaus also asked several related questios that have bee stated ad studied i refs 1 3 This secific roblem has bee widely oted (see, eg, refs 4 ad 5) but has remaied usolved util ow Here by usig a combiatio of techiques from aalysis, set theory, umber theory, ad lae geometry we show the aswer is i the affirmative Theorem 1 There is a set S 2 such that for every isometric coy L of the iteger lattice 2 we have S L 1 We call a set S as i Theorem 1 a Steihaus set ad ote that whether there ca be a Lebesgue measurable Steihaus set remais usolved This roblem has bee the origi of may aers icludig those of Beck (6), Croft (7), Komjáth (3), ad Koloutzakis (8) Koloutzakis ad Wolff (9) showed that there is o measurable Steihaus set for the higher-dimesioal versio of Steihaus s roblem for the stadard lattice Steihaus s roblem ad variats were discussed i some detail by Croft (7) ad have bee udated i sectios E10 ad G9 of ref 4 A straightforward iductio argumet quickly rus ito roblems, as oted i refs 7 ad 10 We avoid this by usig a hull costructio that we describe shortly Let us say a lattice distace is a real umber of the form 2 m 2, where, m Let deote Euclidea distace i 2 Our methods allow us to rove a stregtheig of Theorem 1 Theorem 2 There is a set S 2 satisfyig: 1 For every isometric coy L of 2 we have S L 2 For all distict z 1, z 2 S, (z 1, z 2 ) 2 The Steihaus roblem has a atural iterretatio for smaller sets of lattices Namely, give a arbitrary set L of lattices (each of which is a isometric coy of 2 ), we may ask whether there is a set S satisfyig art 2 of Theorem 2 ad such that S L for all L L We call a set S 2 satisfyig art 2 a artial Steihaus set Our first ste toward rovig Theorem 2 is establishig this restricted versio of the roblem for the case where L is the (coutable) family of ratioal traslatios of 2 The roof ivolves oly elemetary umber theory ad combiatorics Lemm Let L deote the set of ratioal traslatios of 2, that is, lattices of the form 2 (r, s) where r, s The there is a set S 2 satisfyig the followig: 1 For every lattice L L, S L 2 For all distict z 1, z 2 S, (z 1, z 2 ) 2 Proof: We wat to show there is a ma f :( ) ( ) 3 such that 2 (f(x), f(y)) holds for x y ad f is a selector: f([r], [s]) [r] [s] The set S is the image set of the selector For this we defie a grah o the Abelia grou by joiig x with y if ad oly if there are elemets g(x) x ( ) ad g(y) y ( ) such that the square of the distace betwee g(x) ad g(y) is a iteger Whe costructig the selector we oly have to worry about oits that are joied Let H be the subgrou of cosistig of elemets with deomiators oly divisible by rimes of the form 4k 1 Note that if (r, s) ad r 2 s 2, the (r, s) H For, suose r a b, s c d (writte i lowest terms), ad r 2 s 2 e Suose 2or is a rime cogruet to 3 mod 4, ad divides b or d The the exact ower of dividig b, say k, must be the exact ower dividig d, as otherwise multilyig through by the square of the least commo multile m of b ad d would give r 2 m 2 s 2 m 2 em 2 where the right-had side ad exactly oe of the left-had side terms are divisible by, a cotradictio Cosider the case 3 mod 4 Sice these exact owers are the same, either term o the left is divisible by, ad both are ozero mod This would give 1 is a square mod, a cotradictio, as 3 mod 4 The case 2 is left to the reader Thus, if [x], [y] are joied, the x y H; that is, o edges go betwee distict cosets of H, so it suffices to make the costructio er cosets, ad this agai reduces to makig the costructio o H To motivate the followig argumet, let x (i 1 d, j 1 d), y (i 2 d, j 2 d), where d is oly divisible by rimes cogruet to 1 mod 4, ad suose f([x]) x (k 1, l 1 ), f([y]) y (k 2, l 2 ) ad (f([x]), f([y])) 2 Multilyig through by d 2 this becomes i 1 i 2 2 j 1 j 2 2 2d i 1 i 2 k 1 k 2 2d j 1 j 2 l 1 l 2 d 2 [1] Suose a is oe of the rime comoets of d, ad e, f are the exact owers of dividig i 1 i 2, j 1 j 2, resectively If e a, the easily f a as well, ad coversely If e a, the easily e f This gives (i 1 i 2 e ) 2 (j 1 j 2 e ) 2 0 mod a e Recall that for each rime ower a, where 1 mod 4, there are exactly two square roots, say a, a,of 1 mod a, ad for ay a a, a mod a, a mod a are the roots mod a So we must have (j 1 j 2 e ) a(i 1 i 2 e ) mod a e, where a deotes oe of the roots So, j 1 j 2 a(i 1 i 2 ) mod a holds i either case (e a or e a) Sice this holds for each rime ower a of d, we coclude that for (f([x]), f([y])) 2 to hold, we must have j 1 j 2 (i 1 i 2 ) mod d, where 2 1 mod d To further motivate the argumet, suose ow that d a is a rime ower ad j 1 j 2 (i 1 i 2 ) where is chose so that 2 1 mod 2a Substitutig i Eq 1 ad dividig through by 2d(i 1 i 2 ) gives k 1 l 1 k 2 l 2 mod a b, where b is the exact ower dividig i 1 i 2 Note, for examle, that if a 1 (ie, d is a rime), the this equatio would ot hold if we had k 1 k 2 0 ad l 1 i 1, l 2 i 2 This suggests we somehow write each x (i d, j d) i terms of the basis elemets To whom corresodece should be addressed jackso@utedu wwwasorg cgi doi as PNAS December 10, 2002 vol 99 o

2 (1, a), (1, a) for each a dividig d ad use this to defie the k, l values [where f([x]) x (k, l)] We ow carry out the details of this argumet Let P 1, P 2, eumerate the ositive owers of rimes of the form 4k 1 such that if divides P j, the i j By recursio, for each i fix the distict mod atural umbers i, i 0 such that 2 i 2 1, i 2 1 ad 2 j i, j i, where P j is the ext member of the sequece that is a ower of the same rime as Note that i ad i 0 are the distict square roots of 1 mod 2 Let B( ) be a iteger divisible by every P 1,, 1 but ot by, ad, if, the let A( ) be a iteger divisible by each of P 1,, 1, which are ot owers of, but (A( ), ) 1 If (x, y) is a air with x, y ratioal umbers, 0 x, y 1, ad the deomiators of x, y are oly divisible by rimes of the form 4k 1, the (x, y), we claim, ca uiquely be writte mod1as A i 1, i i 1, i, [2] with 0 i, i ; here is the rime whose ower is To see this, first write (x, y) [(e 1, f 1 ) 1 ] [(e k, f k ) k ] mod 1, where the e i, f i are itegers ad d 1 k Each term i this sum is of the form (e, f) P j for some P j, ad it is eough to show that this term ca be writte as [A( ) ]( i (1, i ) i (1, i )) where the rage over the divisors of P j, ad 0 i, i Fid j, j i the desired rage with A(P j )( j (1, j ) j (1, j )) (e, f) mod, where P j is a ower of This is ossible as (A(P j ), ) 1, ad the two-by-two system is osigular mod The (e, f) P j A(P j ) P j ( j (1, j ) j (1, j )) is of the form (e, f ) P k where P j P k Cotiuig, we fiish The uiqueess art of the claim is easily checked We ow add (0, t i B( )) to the oit defied by Eq 2, where t i ( i i ) This will be f([x], [y]) Assume that the square of the Euclidea orm of the differece of two such oits is a iteger The differece of the two oits is of the form A u i 1, i v i 1, i 0, S, [3] where S (u i v i )B( ) with ( 1) u i, v i ( 1) for, ad the sum here is take over all i such that ot both u i ad v i are zero If the oit give by Eq 3 is (a d, b c S), where (a, d) (b, c) 1, the d c, as otherwise the square of the orm could ot be a iteger From this we get that b a (mod d), where (mod d 2 ) Next, ote that all the that occur i the sum i Eq 3 divide d Thus, for every i, either i (mod 2 ) or i (mod 2 ), ad the same case must hold for owers of the same rime By reamig, if eeded, we assume that i (mod 2 ) holds for every i Sice b a (mod d) ad i (mod 2 ), we have A v i i i 0 mod 1 Notice that does ot divide i i where is a ower of, ad from this we get by backward iductio o i that v i 0 holds for every i What we have ow is that d a 2 b 2 S d is a iteger, where ad a d u ia, b d u i i A, S u i B Let i be the first idex such that u i 0, ad let The the first ozero term of S, that is, u i B( ), is divisible by 1 but ot by ad all later terms are divisible by [because of the factors B(P j )] So S is divisible by 1 but ot by Ifwe relace b d by a d the we ca easily coclude, usig 2 1 (mod d 2 ), d a 2 a 2 S 2 a S mod 1, d d which is certaily ot a iteger, as d is divisible by while S is ot We ow argue that the same holds for (a d, b d) istead of (a d, a d) To this ed, it suffices to show that the iteger a d b d is divisible by Ideed, if X is that last differece, the we ca reeat the above argumet with S X i lace of S To show the last claim, decomose it as a d b d u j j A P j P j, where cotais the terms with P j ad cotais the other terms I the first sum, usig the fact that j (mod P j 2 ), every term is a iteger divisible by The secod sum is a iteger of the form B C where B is divisible by [because of the factors A(P j )] but C is ot divisible by Note that for ay choice of the A( ), B( ) we have f([0], [0]) (0, 0) i the above costructio However, for ay (e, f) 2 we could add (e, f) to the value of f([x], [y]) for all x, y ad still kee the squared distaces betwee distict oits i the rage of f oiteger Thus, we are free to make f([0], [0]) ay oit i 2 Actually, we require a slightly stroger form of Lemm To state it we call sets of the form x ( ), (x ) -subsets of Also, a subset E is small, if for every -subset D, D E is cotaied i the uio of fiitely may lies If L is isometric to 2 ad Q is the set of oits havig ratioal coordiates with resect to L, the we defie i a aalogous maer the otio of E beig small relative to L Let d be a ositive iteger Let R d be the subgrou of of oits ([x], [y]) where x, y ca be writte with deomiator d Let H d H R d Ifd 1 k q 1 where each i 1 (mod 4) ad each q j 2orq j 3 mod 4, the H d are those ([x], [y]) where x, y ca be writte with deomiator 1 k Note that R d is isomorhic to the direct sum H d Q K d, where K d is the subgrou of ([x], [y]) where x, y ca be writte with deomiator q 1 I articular, the distict cosets of H d i R d are give by ([x], [y]) H d, where ([x], [y]) K d Suose ow f is a selector o R d We say f is good if o each coset of H d i R d, f is defied as i the costructio of Lemm More recisely, we mea the followig Let d 1 k q 1 as above (if d 1, we declare f to be good) Let d 1 k The for each coset ([x], [y]) H d, where ([x], [y]) K d, there is a sequece of rime owers P 1, P 2,, 0 that build u d (that is, d is the roduct of the that do ot divide aother term i the sequece) ad itegers A( ), B( ), i i 0, as i the roof of Lemm, ad a iitial wwwasorg cgi doi as Jackso ad Mauldi

3 traslatio (e, f) such that for all ([r], [s]) H d, f(([x], [y]) ([r], [s])) (e, f ) f ([r], [s]), where f is defied o H d as i Lemm usig the, A( ), ad B( ) It was show i the roof of Lemm that if f is a good selector o R d, the the rage of f is a artial Steihaus set Lemma 2 Let d be a ositive iteger ad f be a good selector o R d Suose d d ad E is a small set missig ra(f) The f may be exteded to a good selector f o R d whose rage also misses E Proof: Usig the fact that the costructios o the differet cosets of H are ideedet (comare the first aragrah i the roof of Lemm) ad also that a ratioal traslatio of a small set is small, it is eough to show the followig (chagig somewhat the otatio from the statemet of the lemma) Let d be a roduct of ositive owers of rimes cogruet to 1 mod 4 Let P 1, P 2,, 1 build u d (as defied above) Let A(P 1 ), B(P 1 ),, A( 1 ), B( 1 ) satisfy the requiremets give i Lemm Let f be the selector o H d defied from these quatities as i Lemm Suose fially that E is a small set missig ra(f), ad is give with P 1,, buildig u d The we show that there are A( ), B( ) so that if f is defied usig these exteded sequeces, the the rage of f misses E The oit is we have eough freedom i choosig the values of A( ), B( ) We are assumig f([x], [y]) has bee defied for all (x, y) havig a reresetatio as i Eq 2 with the sum ragig over rime owers i the list P 1,, 1 Let A, B, satisfy the requiremets for A( ), B( ) give i Lemm Let U be the roduct of the P j with j i ad P j, relatively rime The A A K U, B B L U also satisfy these requiremets for ay K, L Let f be as costructed i Lemm by usig A, B For ay fixed x, y R d R d, a comutatio shows that the corresodig values of f ([x], [y]) will have the form (x KUa, y KUb LU a) for some fixed x, y, a, b with a, b 0 Sice E is small, for each [x], [y] the requiremet that f([x], [y]) E rules out oly the K, L lyig o fiitely may lies i As there are oly fiitely may [x], [y] to cosider at stage i, it is clear that we ca choose K, L so that f misses E We thak oe of the referees for oitig out the roofs of the recedig lemmas These roofs are based o ad motivated by the more comlicated roofs that we give i ref 10 for stroger results To state oe of these results we adot the followig termiology For ratioals r, s, let L r,s 2 (r, s) be the ratioal traslatio of 2 by (r, s) Let R 2 ([0, 1) [0, 1)) For each ositive iteger d, let R d R be defied by R d {(i d, j d):0 i, j d} We rove the followig i ref 10 Lemma 3 Let d 1 ad suose fuctios k, l maig R d ito have bee defied such that settig S d {(r k(r, s), s l(r, s)) : r, s R d } we have: d : for all distict z 1, z 2 S d, z 1, z 2 2 The for ay d with d d, the fuctios k, l may be exteded to R d so as to satisfy ( ) d This last lemma assures us ot oly that there is a set S satisfyig Lemma 2, but also that ay artial Steihaus set defied for the traslates i R d may be exteded We also rove similar extesio theorems where we must avoid small obstructio sets These extesio lemmas are much stroger tha is required for the roof of our mai theorem, but we feel that they may be useful i studyig aalogous roblems for other lattices, dimesios, ad grous of isometries By a ratioal traslatio of a lattice L we mea a lattice of the form L ru sv, where r, s, ad u, v are the uit basis vectors of L Defiitio: Two lattices are equivalet L 1 L 2,ifL 2 ca be obtaied from L 1 by ratioal rotatios ad traslatios I other words, L 1 L 2, if ad oly if all of the oits of L 2 have ratioal coordiates with resect to the coordiate system determied by L 1 (ad vice versa) This is easily a equivalece relatio, ad each equivalece class is coutable We also ote that if two lattices are ot equivalet, the there ca be at most oe oit with ratioal coordiates with resect to both of them A imortat asect of the costructio is that if oe has a Steihaus set for all the ratioal traslates of a give lattice L, the it is a Steihaus set for the equivalece class of L: Lemma 4 Let L 1 be a lattice ad suose S 2 satisfies the followig: 1 For every lattice L that is a ratioal traslatio of L 1, S L 2 For all distict z 1, z 2 S, (z 1, z 2 ) 2 The for every lattice L that is equivalet to L 1, we have S L Naturally, this last observatio suggests costructig a full Steihaus set by iductio o the equivalece classes of lattices This leads us to the set-theoretic art of the roof The trasfiite iductio cosiders collectios of lattices that are sufficietly closed The mai closure roerty we eed arises from Lemma 6, ad the corresodig argumet is give i the Claim below Rather tha secify at the outset exactly what closure roerties we wish our sets at each stage to satisfy, it is more coveiet to follow a stadard set-theoretic ractice If CH held, we would at stage cosider those equivalece classes of lattices that lie i a elemetary substructure (or hull) M of V for some large (though actually will suffice) Here V deotes the iitial segmet of the uiverse of sets of rak less tha, ad a elemetary substructure deotes a subset closed uder the fuctios f: (V ) 3 V, for some, which are defiable i V (the so-called Skolem fuctios of V ) For the beefit of the reader ufamiliar with set-theoretic termiology, we ote that this is merely a coveiet shorthad for requirig that the sets we cosider be sufficietly closed without havig to secify i advace exactly which fuctios we wat them to be closed uder Thus, i lace of M the reader could substitute a sufficietly closed collectio of oits ad lattices (the closure roerties eeded will be aaret from the followig argumet ad could be secified i advace) I the geeral case (ot assumig CH) the costructio is essetially a iteratio of this substructure costructio Although the set-theoretic termiology could be largely elimiated (see also the commets at the ed of this article), we believe doig so would lesse the geerality of the method ad hide our motivatio (for examle, it was by ursuig the hull strategy just outlied that we were led to cosider Lemma 6) We ow describe the articular well-orderig of the equivalece classes that we will use First, some otatio If L 2 is a isometric coy of 2, let [L] deote the equivalece class of L uder the equivalece relatio of the defiitio Let deote the family of all equivalece classes Let L 3 L(L) be a fuctio that icks for each equivalece class L a member L(L) L The costructio of the eumeratio begis by lettig {M 0 : 0 2 } be a icreasig cotiuous (ie, at limit stages we take uios) chai of elemetary substructures of a large V with M 0 2 for all 0 2 ad such that each lattice ad each equivalece class of lattices is i oe of these substructures Assume also that M 0 (The startig set M 0 is a excetio i that it is ot a elemetary substructure) Let N 0 M 0 1 M 0 By simultaeous recursio we defie a subtree T of ON ad a ordial ( ) ad sets M, N for T Set ( ) 2 I geeral, suose that M is defied for i a certai subtree of ON IfM 0,, k is defied, we assume also that ( 0,, Jackso ad Mauldi PNAS December 10, 2002 vol 99 o

4 k 1 ) has bee defied ad is a ifiite cardial Furthermore, we assume i this case that M 0,, k 1, is defied if ad oly if ( 0,, k 1 ) We let N 0,, k deote M 0,, k 1 M 0,, k Suose ow that M 0,, k is defied If N 0,, k cotais oly coutably may equivalece classes of lattices, let {L 0,, k; } s, where s, eumerate these equivalece classes I this case, ( 0,, k ) is a termial ode i the tree, T, of idices for which M is defied Otherwise, let ( 0,, k ) N 0,, k ad exress N 0,, k k 1 0,, k M 0,, k, k 1 as a cotiuous icreasig uio, where each M 0,, k, k 1 is the itersectio of N 0,, k with a elemetary substructure of V, ad each M 0,, k, k 1 has size ( 0,, k ) Assume also M 0,, k,0 We ote two facts Easily, the tree of idices is well fouded (sice the cardials are strictly decreasig alog ay brach) Also, if,,a m M 0,, k, k 1 ad f is a Skolem fuctio of V ad f(,, a m ) N 0,, k, the f(,, a m ) M 0,, k 1 Notice if is icomatible with, the N ad N have o equivalece class i commo Furthermore, every equivalece class occurs as some L 0,, k ; Thus, the L 0,, k ; recisely eumerate the equivalece classes of lattices By a -block of lattices we mea the (equivalece classes of) lattices of the form L 0,, k ; for some fixed termial idex 0,, k We cosider the idices to be (well) ordered lexicograhically The followig easy lemma will be used Lemma 5 Suose is a idex for which M is defied Let,,a m M ad suose b is defiable from,,a m i V The b M The idea for costructig a Steihaus set is by trasfiite iductio alog the termial odes of the tree T However, it turs out we eed aother lemma i ure lae geometry to make a iductive extesio Lemma 6 Let c 1, c 2, c 3 be three distict oits i the lae, ad let r 1, r 2, r 3 0 be real umbers Let C 1 be the circle i the lae with ceter at c 1 ad radius r 1, ad likewise for C 2 ad C 3 Let a, b, c be three distict oits i the lae The, excet for the excetioal case described afterward, there are oly fiitely may triles of oits ( 1, 2, 3 ) i the lae such that 1 1 C 1, 2 C 2, ad 3 C 3 2 The triagle is isometric with the triagle abc (we allow the degeerate case where the oits a, b, c are colliear) The excetioal case is whe r 1 r 2 r 3 ad the triagle abc is isometric with c 1 c 2 c 3 This lemma seems to be kow withi egieerig mathematics secifically regardig the geometry of mechaisms (11) Although ot exlicitly stated, Lemma 6 follows from the aalysis of Gibso ad Newstead i ref 12 The Gibso Newstead argumet uses a sigificat amout of algebraic geometry We also give two elemetary roofs of Lemma 6 i ref 10, a algebraic oe usig Gröber bases ad comuter algebra, ad the other a urely geometric roof Fix ow a termial idex ( 0,, k ) Assume iductively we have defied for each termial idex a set S 2 that satisfy the followig: 1 If 1 2, the S 1 S 2 2 For every termial idex less tha, S meets every lattice i every equivalece class L ; 3 Every oit of S S lies o some lattice of the form L ; 4 For all distict z 1, z 2 S, (z 1, z 2 ) 2 5 Suose 1 2, x S 1, ad y S 2 2 S The if (x, y) 2, the x, y both have ratioal coordiates with resect to some lattice of the form L 2 ; Let S S We show how to exted S to a set S also satisfyig 4, 5 ad such that S meets every lattice i each equivalece class L, This suffices to rove Theorem 2 To ease otatio, let L L ;, ad let L L(L ) From Lemma 4, it suffices to maitai roerty 4, have roerty 5 whe 2, ad have S meet every ratioal traslatio of each L (recall a ratioal traslatio of L refers to a motio that is a traslatio i the coordiate system of L ) For itegers, d, i, j, let L d,i,j deote the traslatio of L by the amout (i d, j d) (i the coordiate system of L ) We also eed the followig easy lemma Lemma 7 Let L be a lattice ad z 2 Suose z has coordiates (x, y) with resect to the lattice L, where at least oe of x, yis irratioal The there is a lie l l(z, L) such that if w has ratioal coordiates with resect to L ad w l, the (w, z) 2 These last two geometric lemmas give the followig claim (see ref 10 for details) Claim: For each ad ratioals i d, j d, there is a fiite set of lies G (i d, j d) with the followig roerty: if c S does ot have ratioal coordiates with resect to L,ifz L d,i,j, ad if (c, z) 2, the z G (i d, j d) Fix a bijectio (, m), m betwee 2 ad We may assume this bijectio is icreasig i each coordiate Let Q be those oits havig ratioal coordiates with resect to L We costruct artial Steihaus sets T m i stages, with T 0 T 1 Q We will arrage it so that if T m T m, the T meets every lattice i L At stage i, m we exted T m 1 to T m (or defie T 0 if m 0) We will follow Lemma 2 i doig each extesio Fix i, m, ad assume that for all, q i that T q is defied Assume iductively that the T q so far defied satisfy the followig (below, iterret T q 1 as if q 0): (a) T q cotais T q 1, ad each T q is a artial Steihaus set (that is, (z 1, z 2 ) 2 for all distict z 1, z 2 T q ) (b) T q is the rage of a good selector f q o R dq Here R dq is the set of oits havig coordiates with resect to L that ca be writte with deomiators d q, ad d 0, d 1, is a sequece (deedig o ) with d j d j 1 for all j ad such that every iteger divides some d j (c) Let E i,j,d (G (i d, j d) L d,i,j ) Note that E is small relative to Q The (T q (S a,b,q T b a )) E (d) Suose z S, ad i 2 2, q 2 is least so that z T 2 If i 1 1, q 1 i 2 ad 1 2, the z does ot have ratioal coordiates with resect to L 1 (e) Suose z S, ad i 2 2, q 2 is least so that z T 2 If i 1 1, q 1 i 2 ad y T 1 q1 does ot have ratioal coordiates with resect to L 2, the z l(y, L 2 ), where l(y, L 2 )isasilemma 7 Cosider ow i, m, ad we defie T m so as to also satisfy the above roerties Cosider first the case m 0; that is, we are at the begiig stage i the costructio of T We claim that there is at most oe oit i S,q i T q that is also i Q For, suose y, z were two such oits Note that (y, z) 2 Suose first that y, z S Say y S 1 1 S, z S 2 2 S where 1 2 If 1 2, the each of y, z lies o a lattice i N 2 Sice L is defiable from y ad z, L is defiable from two lattices i M for some From Lemma 5 it follows that L M, a cotradictio If 1 2, the from iductive roerty 5, y, z both have ratioal coordiates with resect wwwasorg cgi doi as Jackso ad Mauldi

5 to some lattice L i N 2 This would agai imly that L M, a cotradictio Suose ext that y S ad z S Let, q be least so that z T q (so ) Sice by c, z G (i d, j d), we must have that y is ratioal with resect to L [as otherwise (y, z) 2 ] Thus, both y ad z have ratioal coordiates with resect to both L ad L, a cotradictio of the fact that there ca be at most oe oit with ratioal coordiates with resect to both lattices Suose ow y, z S Let y T 1 q1, z T 2, with i 1 1, q 1, i 2 2, q 2 chose miimally, so 1, 2 Assume without loss of geerality that i 1 i 2 Ifi 1 i 2, the y, z are ratioal with resect to both L 2 ad L, a cotradictio If i 1 i 2, the from e, y is ratioal with resect to L 2 Soy, z are both ratioal with resect to L 2 ad L, agai a cotradictio Let E 0 be the uio of i,j,d (G (i d, j d) L d,i,j ) together with z l(z, L ) where z rages over the oits i,q i T q ot havig ratioal coordiates with resect to L, together with the (fiitely may) oits of Q that are ratioal with resect to oe of the lattices L with, ad, q i for some q (for m 0 this is equivalet to ) Clearly E 0 is small with resect to Q Let w be the uique oit i (S,q i T q ) Q if it exists, ad otherwise let w be ay oit of L E 0 Aly ow Lemma 2 to get T 0 avoidig E 0 {w } ad with w T 0 From the defiitio of E 0 it is clear that c e are still satisfied Cosider ext the case m 0 Defie E m exactly as above ad aly Lemma 2 to get T m, addig oly oits that avoid the set E m Agai, c e are satisfied Note for the argumets below that if z S ad, m is least so that z T m, the z w This comletes the defiitio of the T m, ad we have verified a e Let T,m T m Let S S T We must show that iductive hyotheses 1 5 are satisfied Proerties 1 3 are immediate from the costructio To see hyothesis 5, let ad y S, z S S, ad suose (y, z) 2 Let, m be least so that z T m Sice z i,j,d G (i d, j d) by costructio, we must have that y is ratioal with resect to L Fially, we verify hyothesis 4; that is, we show S is a artial Steihaus set Let y, z be distict oits i S, ad assume (y, z) 2 We may assume z T S Suose first that y S Let i, m be least with z T m As i the revious aragrah we must have y ratioal with resect to L as otherwise (y, z) 2 Let i, 0, soi i I defiig T 0, y would the have bee the oit w So, y T m, ad sice this is a artial Steihaus set, (y, z) 2, a cotradictio Assume ow that y, z S Let y T 1, z T 2, with i 1 1, m 1, i 2 2, m 2 chose miimally Clearly i 1 i 2 ; i fact 1 2 Assume without loss of geerality that i 1 i 2 We must have y ratioal with resect to L 2 as otherwise from the defiitio of E 2 ad l(y, L 2 ) we would have (y, z) 2 Let i 3 2,0 (ote that i 3 i 1 ) If i 1 i 3, the y is the oit w 2, which lies i T 2, ad so (y, z) 2 If, however, i 1 i 3, the from the defiitio of E 1 we would have that y is ot ratioal with resect to L 2 [sice ito E 1 we added the (at most oe) oit of Q 1 Q 2 ] This agai imlies that (y, z) 2, a cotradictio This comletes the roof of Theorem 2 Fially, we oit out two simlificatios that could be made to the roof First, oe could make the iductive costructio more secific for this roblem Fix a trascedece basis for over, ad let be a well-orderig of this trascedece basis The the fiite -decreasig sequeces s from the basis are well-ordered lexicograhically For each such s, there are oly coutably may lattices that are algebraic over ad s Do the trasfiite costructio i the order of these sequeces s, at each stage hadlig those coutably may lattices that are algebraic over ad s but ot algebraic over ad s for ay s s The all of the argumets go through as before, if oe relaces defiable with algebraic The key oit is if three of the sequeces m all recede lexicograhically ad first differ from at the same ositio k, the the uio of these three sequeces, arraged i -decreasig order, still recedes lexicograhically Secod, the full stregth of Lemma 6 is ot eeded for the roof Let agai (usig the otatio of the Claim) E be those oits z Q (ie, havig ratioal coordiates with resect to L ) such that 2 (c, z) for some c S where c does ot have ratioal coordiates with resect to L The it suffices to show that E is semismall with resect to Q, which meas for each ratioal traslatio L L d,i,j of L there is a fiite set of lies F such that for ay lie l F, l L E is fiite This is because the roof of Lemma 2 shows that we may actually avoid ay Q semismall set i costructig the T q To see E is semismall, it suffices to show that there is a boud s such that if c 1,, c s are oits i the lae with (c i, c j ) 2 for distict c i, c j, ad if z 1,, z s are colliear oits with (c i, z i ) 2 ad (z i, z j ) 2, the the z i are defiable from {c 1,,c s }; i fact, for fixed c 1,,c s, distaces (c i, z i ) ad (z i, z j ), there are oly fiitely may such {z 1,,z s } This fact follows from lem of ref 3 We thak the referees for oitig out these ossible simlificatios as well as the simlified roof of Lemm that we reseted here We also thak Robert M Solovay for his detailed commets ad attetio to the aer SJ was suorted by Natioal Sciece Foudatio Grat DMS , ad RDM was suorted by Natioal Sciece Foudatio Grat DMS Sieriński, W (1958) Fud Math 46, Erdős, P (1985) A NY Acad Sci 440, Komjáth, P (1992) Q J Math Oxford 43, Croft, H T, Falcoer, K J & Guy, R K (1991) Usolved Problems i Geometry (Sriger, New York) 5 Erdős, P, Gruber, P M & Hammer, J (1989) Lattice Poits (Logma Sci Tech, Harlow, Essex, UK) 6 Beck, J (1991) Studia Sci Math Hug 24, Croft, H T (1982) Q J Math Oxford 33, Koloutzakis, M N (1996) i Aalytic Number Theory, eds Berdt, B C & Diamod, H G (Birkhäuser, Bosto), Vol 2, Koloutzakis, M N & Wolff, T (1999) Mathematika 46, Jackso, S & Mauldi, R D (2002) J Am Math Soc 15, Hut, K H (1990) Kiematic Geometry of Mechaisms, Oxford Egieerig Sciece Series (Claredo, Oxford) 12 Gibso, C G & Newstead, P E (1986) Acta Al Math 7, Erdős, P, Jackso, S & Mauldi, R D (1994) Fud Math 145, Erdős, P, Jackso, S & Mauldi, R D (1997) Fud Math 152, Jackso ad Mauldi PNAS December 10, 2002 vol 99 o