Model Theory 2016, Exercises, Secod batch, coverig Weeks 5-7, with Solutios 3 Exercises from the Notes Exercise 7.6. Show that if T is a theory i a coutable laguage L, haso fiite model, ad is ℵ 0 -categorical, the T is complete. Solutio: Let M, N be models of T. By assumptio both are ifiite so, by the Dowwards Löweheim-Skolem Theorem, each has a coutable elemetary substructure: M 0 M, N 0 N. By assumptio M 0 N 0, hece M 0 N 0. So we have M M 0 N 0 N ad M N, as required. Exercise 7.7. Show that T dlo is ot κ-categorical where κ is the cardiality of the cotiuum. Solutio: There are various possible solutios. We have to produce two models of cardiality 2 ℵ0 which are ot isomorphic. A useful basic costructio is to take two (or more) ordered sets ad the put oe above the other. For example, take (Q; ) ad(r ad produce two o-isomorphic models of T dlo by placig (R above (Q i the first ad placig (Q above (R i the secod. Explicitly, defie P 1 =({0} Q) ({1} R ad order this by (0,x) < (1,y) for all x, y ad, for i =1, 2, (i, x) (i, y) iffx y. The defie a orderig o P 2 =({0} R) ({1} Q i the same way. Note that the cardialities of both P 1 ad P 2 are that of the cotiuum. Also both are easily see to be desely, totally, ordered ad to have o greatest or least elemet. But they are ot isomorphic because there are elemets a P 1 such that there are oly coutably may b P 1 with b<a; whereas every elemet i P 2 has ucoutably may elemets below it. p. 29 of the Notes Show that the property ( ), which characterises the Radom Graph, is elemetary, meaig that there is some setece σ orset,σ,of seteces i the appropriate laguage L such that the L-structures with property ( ) are exactly those which satisfy σ/σ. Solutio: I fact we eed to use a ifiite set of seteces. For each m, (i fact, it would be eough to do it for m = ) we write dow a setece which says that, give two disjoit sets, the first with m elemets ad the secod with elemets, there is a elemet which is related to each poit i the first set ad to oe of those i the secod set. Formally, this is i=m,j= ( x 1,...,x m, y 1,...,y ( i=1,j=1 m x i y j ) z ( R(z,x i ) R(z,y j )) ). Exercise 8.7 Prove that if W is ay fiite subset of Ra the the iduced subgraph o Ra \ W is isomorphic to the Radom Graph. Solutio: We check property ( ). So let U, V be fiite disjoit subsets of Ra \ W. The W U, V are fiite disjoit subsets of Ra so there is z RawhichiscoectedtoeachelemetofW U adtooelemetofv. I particular, z is coected to each elemet of U ad to o elemet of V. i=1 j=1 9
Furthermore, sice z iscoectedtoeachelemetofw ad there are o loops i R, z Ra \ W. As required. Exercise 8.8 Does the coclusio of 8.6 still hold if we do t assume that the X i are disjoit? Solutio: No; we ca produce a couterexample. For example, fix z 0 Ra. Let U be the set of w Ra which are coected to z 0 ad let V be the set of those which are ot. So U,V form a partitio of Ra. Note that z V,soV does ot satisfy ( ), hece V is ot a copy of the Radom Graph. Set U = U {z}. Sicez is coected to every poit of U except itself, U also fails to satisfy ( ), hece is ot isomorphic to the Radom Graph. Exercise 8.9 Let S be a coutable model of (Zermelo-Fraekel) set theory. O S defie a graph by joiig x to y iff either x y or y x. Prove that the resultig graph is isomorphic to R. Solutio: Let U, V be fiite disjoit sets of elemets of S. Note that V is a elemet of S ad so, therefore, is Z = U {V }. If u U the u Z, soz is coected to each elemet of U. O the other had, let v V. If v were coected to Z the either v Z - but the, sice U V =, wewouldhave v = V, hece V V which cotradicts the Axiom of Foudatio - or Z v, but the v V Z v - agai cotradictig the Axiom of Foudatio. So we have ( ) as required. Exercise 8.10 R R c where, for ay graph G, we defie G c by replacig the edges of G with o-edges ad the o-edges of G with edges. Solutio: Oce agai, we check coditio ( ); but this is immediate o swappig the roles of U ad V i that coditio. Exercise 8.12 A relatioal structure is said to be ℵ 0 -homogeeous if every isomorphism betwee fiite substructures exteds to a automorphism of the structure. Prove that the Radom Graph is ℵ 0 -homogeeous. Solutio: (Sketch) Give a isomorphism betwee fiite substructures, regard this as the startig poit (a partial isomorphism with fiite domai) of buildig a automorphism of R, ad the use the back-ad-forth argumet (i the lecture otes for Sectio 7) that shows uiqueess of the Radom Graph up to isomorphism. Exercise 8.14 For some specific examples of L, write dow a formula for S(L,). Solutio: This rather depeds o the example(s). Exercise 8.15 Let L be a laguage with o fuctio symbols, oe costat symbol c ad oe uary relatio symbol R, show that S(R(c),) = S( R(c),). Solutio: We show that, for 2, half the models with uderlyig set {1,...,} satisfy R(c) ad half do t (that is, satisfy R(c)). For that it will be eough to defie a bijectio betwee those two sets of L-structures (o a give uderlyig set). That is achieved by pairig the L-structure M =(M; X), where X = R M,withtheL-structure M 0 =(M; M \ X). To see that this gives a bijectio as required, ote that (M 0 ) 0 = M ad that exactly oe of M, M 0 satisfies R(c). Exercise 8.18 Prove that, for ay seteces σ 1,...,σ k T as,wehave k i=1 σ i 10
T as. Show that T as is deductively closed, thatis T as = T as. Solutio: Give ε>0, choose 0 such that, for every 0, p(σ, ) > 1 ε k. It follows that p( k i=1 σ i) > 1 ε. Therefore k i=1 σ i T as. Suppose that τ T as,thatis,τ is true i every model of T as. By Compactess (see Corollary/Exercise 9.4) there is a fiite subset {σ 1,σ 2,...,σ k } of T as such that τ is true i all models of {σ 1,σ 2,...,σ k }, that is, i all models of k i=1 σ i. So p(τ,) p( k i=1 σ i,) ad the latter 1as. Hece p(τ,) 1as ad so τ T as, as required. Exercise 8.20 Give a example to show that eve for a laguage with just oe fuctio symbol the 0/1-law will fail. Solutio: We ca try to approach this by thikig up a property which looks like it stads a chace of beig true i (say) about half of all L-structures. The try to prove that this is the case by computig (or estimatig) the proportio of structures o the set {1,...,} of size i which it is true. The umber of L-structures o {1, 2,...,}, that is, fuctios from that set to itself, is. Obvious properties, like beig a bijectio, or omittig just oe elemet from the! 1 rage, do t work (the relevat proportios, ad, ted to zero as teds to ifiity). So we might try to come up with some property that kid of alterates with, or ca roughly be paired with, its opposite. Ad where the umbers ca be computed (at least, estimated) reasoably easily. Ad, if that fails, just try a property where it looks as if the relevat proportios ca be computed/estimated. Which is how I came to try the property of havig at least oe fixed poit (o doubt other properties will work). If we have a set with elemets, the the umber of fuctios from that set to itself with a give fixed poit is ( 1) 1 (sice such a fuctio is essetially a arbitrary fuctio o the remaiig 1 elemets). There are possible fixed poits, so that gives ( 1) 1 such fuctios, except that we have double-couted the fuctios with two or more fixed poits. But, by the same sort of cout, there are o more tha 1 2 ( 1)( 2) 2 of these, so there are at least ( 1) 1 1 2( 1)( 2) 2 fuctios with at least oe fixed poit. Let s ow estimate what fractio this is of all the fuctios from a -elemet set to itself, as. The fractio is: ( 1) 1 1 2( 1)( 2) 2 = ( 1 1 ) 1 1 2 ( 1 1 )( 2 ) 2 1 = ( 1 ) ( 2 ) 1 = ( 1 ) 1 ( 1 ) 1 1 2 ( 2 ). 2 1 1 ( x ) Now take the limit as, usig the formula lim 1+ = e x,to obtai e 1 1 2 e 2, which is about 0.3. This shows that the exact value of the limit of the fractios is ot 0 ad, sice it is also bouded above by e 1 < 1, or is it 1. Therefore the 0/1 law does ot hold for this laguage (observe that the property of havig at least oe fixed poit is defiable, by the L-formula xf(x) =x). 11
4 Additioal Exercises 1. For each of T dlo ad the (theory of the) Radom Graph, determie all (quatifier-free) 1-, 2- ad 3- types. Solutio: The first thig to otice is that ay type ca be realised i a coutable model (by the Dowwards Löweheim-Skolem Theorem). I the case of T dlo that model is, by 7.3, a copy of (Q; ). Furthermore, by 7.1/7.8, if two fiite tuples from (Q; ) have the same quatifier-free type, that is, the same order-type, the there is a automorphism of (Q; ) takig oe to the other, ad hece (by 4.14, or 12.14) they have the same type. So it is eough to cout the possible order-types of -tuples for =1, 2, 3. By 8.4 ad 8.13 the same applies to the Radom Graph. Start with (Q; ). If = 1 the there is just oe possibility (reflectig that, give ay two elemets, there is a automorphism takig oe to the other). If = 2 the the possibilities are: x 1 = x 2, x 1 <x 2, x 1 >x 2, so there are three 2-types. If = 3 the either x 1 = x 2 = x 3,orx 1 = x 2 with either x 3 <x 1 or x 3 >x 1, or the similar cases with x 1 = x 3 ad with x 2 = x 3 - that makes 7 3-types so far - or all three are distict, ad oe ca see that there are 6 ways of orderig these. So we get 13 3-types altogether. I the case of the Radom Graph the cout goes as follows. If = 1 the there is just oe possibility. If = 2 the the possibilities are: x 1 = x 2, x 1 ad x 2 are coected, x 1 ad x 2 are ot coected, so there are three 2-types. If = 3 the either x 1 = x 2 = x 3,orx 1 = x 2 with either x 3 ad x 1 coected or ot coected, or the similar cases with x 1 = x 3 ad with x 2 = x 3 - that makes 7 3-types so far (just as i the T dlo case). It remais to cout the possibilities whe all three elemets are distict. The we may have: o coectios; oe coectio (three possibilities: x 1 coected to x 2 ; x 1 coected to x 3 ; x 2 coected to x 3 ); two coectios (so oe o-coectio - agai three possibilies); all are coected. That gives 1 + 3 + 3 + 1 = 8 more 3-types, hece 15 3-types altogether. 2.. Give a iteger let K deote the complete graph o vertices (i.e. every vertex is joied to every other vertex). Prove that the theory of K has elimiatio of quatifiers. Solutio: As at the ed of Sectio 7 we must show that the type of a tuple of elemets depeds oly o the isomorphism type of the substructure (that is, subgraph) iduced o that tuple. Suppose that (a 1,...,a m )ad(b 1,...,b m ) are tuples such that, if a i = a j,theb i = b j ad also such that the iduced subgraph o the first tuple is isomorphic, uder the map which takes a i to b i,to the iduced subgraph o the secod tuple - hece, ote, each is isomorphic to K k if k is the umber of distict elemets i the tuples. The ay permutatio of K which is a extesio of the partial isomorphism which takes a i to b i will (fairly clearly) be a automorphism of K ; hece these tuples have the same type, as required. 3. Describe a example of a fiite graph whose theory does ot have elimiatio of quatifiers. 12
Solutio: Cosider the graph o {a, b, c} which has a ad b coected but c coected to either. The a ad c satisfy the same quatifier-free formulas (sice the iduced graphs o each of {a} ad {c} are isomorphic, so the apply 4.16(iii)). But the formula yr(x, y) isithetypeofa (take y = b) but ot i the type of c, so that formula caot be equivalet to a quatifier-free oe. 13