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Math 144: Course Notes MATH 144: COURSE NOTES WILL BONEY Contents 4. February 1 1 5. February 3 6 6. February 5 10 References 13 4. February 1 We're going to momentarily cover some stuff not as presented in [TZ]. This is essentially [M, Section 1.3]. 4.1. Definability. What do we do with model theory? There are many answers to this question, some of which we'll get to in this class. At a very basic level, we can take a structure that we want to study R and view it as a model theoretic structure. Then we can use logic to pick out the nice subsets of it, called definable. Definition 4.1. Fix an L-structure M. (1) Given φ(x; y) and m 2 jMj, we write φ(M; m) := fa 2 jMj : M φ(a; m)g We might write φ(M n; m) for this if we want to emphasize that x is of length n. (2) We say that X ⊂ jMjn is definable iff there is some φ(x; y) and m 2 jMj such that X = φ(M; m). In this case, we say φ(x; m) defines X. To specify the parameters, we say that X is A-definable or definable over A if there is a definition with parameters from A. The special case of ;-definable or 0-definable means that there are no parameters. (3) We say that a function or relation is definable iff it is definable when con- sidered as a subset of the universe. Date: February 2, 2016. 1 2 WILL BONEY Example 4.2. (1) Let Z = fZ; 0; +; <g. Then we can define N by φ(x) = \x = 0 _ x > 0". 1 (2) Let R = fR; 0; 1; +; ∗; −; · g be the field of real numbers. The ordering on R is definable by φ(x; y) ≡ \9z(x + z2 = y)00. (3) Let F be a field and set M = fF [X]; +; −; ×; 0; 1g be the ring of polynomials in one variable over F . Then F is definable in M as the set of units. In particular, F = φ(M), where φ(x) ≡ \9y(xy = 1)00. Pm i (4) Let R be a ring and p(X) = i=0 aiX 2 R[X] be a polynomial over R. The zeros of this polynomial are definable over (any set containing) fa0; : : : ; amg. Set m m−1 φ(x; y0; : : : ; ym) = \ymx + ym−1x + ··· + y0 = 0" where xm is an abbreviation for x × · · · × x m times. Then the zeros are exactly φ(R; a). (5) Let Q = fQ; +; −; ×; 0; 1g. Set • φ(x; y; z) ≡ \9a; b; c(xyz2 + 2 = a2 + xy2 − yc2)" • (x) ≡ \8y; z ([φ(y; z; 0) ^ (8w(φ(y; z; w) ! φ(y; z; w + 1)))] ! φ(y; z; x)) " Julia Robinson (see [FlWa91]) showed that (x) defines the integers in Q. One theme of definability is that proving wether or not a set is definable actually requires some knowledge about the structure you're working in. See the homework and imagine that I had said nothing about Lagrange's Theorem. Definable sets are better behaved than sets in general, but still have some very nice closure properties. Here's a purely semantic way about thinking of definable sets. Proposition 4.3 ( [M].1.3.4). Let M be an L-structure. Suppose that Dn is a n subset of jMj and fDn j n 2 Ng is the smallest collection of such subsets such that n • M 2 Dn; M M • for each n-ary f 2 L, the graph of f := f(m; m) j f (m) = mg 2 Dn+1; M • for each n-ary R 2 L, R = fm j M R(m)g 2 Dn; n • for each i; j ≤ n, f(x1; : : : ; xn) 2 jMj : xi = xjg 2 Dn; • if X 2 Dn, then M × X 2 Dn+1; • each Dn is closed under complement, union, and intersection; • if X 2 Dn+1 and π is the projection map (m1; : : : ; mn+1) 7! (m1; : : : ; mn), then π\X 2 Dn; and m n • if X 2 Dn+m and b 2 jMj , then fa 2 jMj :(a; b) 2 Xg 2 Dn, then fDn j n 2 Ng are exactly the definable sets of M. The proof of this is straightforward and I'm not too interested in giving it. Essen- tially, the first two clauses correspond to the basic relations; complements, etc. cor- respond to negations, etc.; and projection corresponds to existential quantification. MATH 144: COURSE NOTES 3 Then a little more work must be done to account for composition/substitution and reordering of variables. Unfortunately (or maybe fortunately), not all subsets are definable. Proposition 4.4. Every infinite structure M in a countable language has unde- finable subsets. Actually, nothing about this requires countability, just that kMk ≥ jLM j. <! Proof: There are @0 many formulas and kMk = kMk many tuples. Thus there are @0 × kMk = kMk many possible definitions for sets, so this is an upper bound for the number of definable sets (turns out we have exactly this many). On the other hand, there are 2kMk > kMk many subsets of M. y Okay, but that's not constructive. In particular, it doesn't help us determine if a particular subset is definable. The following is useful for this. Proposition 4.5 ( [M].1.3.5). Let M be an L-structure. If X ⊂ jMjn is A- definable, then every automorphism of M that fixes A pointwise also fixes X set- wise. ∼ Proof: Let f 2 AutAM := ff : M = M j 8a 2 A; f(a) = a. We want to show that such that X = f 00X := ff(x) j x 2 Xg. By A-definability, there is a 2 A and φ(x; y) such that X = φ(M; a). Let m 2 X; then M φ(m; a). Hitting this − with f gives M φ(f(m); a), so f(m) 2 X. Note that f 1 2 AutAM as well, so f −1(m) 2 X. Thus, X = f 00X. y We can use this to show that the reals are not definable in the complexes (as a field). Proposition 4.6 ( [M].1.3.6). R is not definable in fC; +; ×; 0; 1g. It's necessary to indicate the language because it obviously is definable if we add a unary predicate that consists of the real numbers. Proof: Suppose that R were definable. Then it would be definable over some 1 finite A ⊂ C. We can find r; s 2 C that are algebraically independent over A with the extra property that r 2 R and s 62 R. From Galois theory, we know that there is f 2 AutAC that sends r to s. So we have found an automorphism of C that fixes A pointwise, but doesn't fix R. This contradicts Proposition 4.5. y A converse to Proposition 4.5 would be nice. However, that's impossible in general. If M is rigid (no automorphisms), then that would imply that every set is definable which we know is not the case. When do we have automorphisms? Well, C has lots of automorphisms. This turns out to be an example of a model theoretic 1This means that they are not roots of any polynomial with coefficients in the field generated by A. 4 WILL BONEY property of models called \saturated." When we construct saturated models, they will have lots of automorphisms and that will give us a partial converse. Proposition 4.7 ( [M].4.3.25). Let M be saturated and A ⊂ jMj with jAj < kMk. Suppose X ⊂ jMjn is M-definable. Then X is A-definable iff every automorphism of M that fixes A pointwise also fixes X setwise. 4.2. Interpretation. The following is a motivating example. Example 4.8. C is bigger than R. However, if we have R (as a field), we can reason about C (as a field) in the way you are all familiar with. Set X = f(a; b) j a; b 2 Rg φ+(a; b; c; d; e; f) ≡ \a + c = e ^ b + d = f" φ×(a; b; c; d; e; f) ≡ \ac − bd = e ^ bc + ad = f" ∼ Both are definable function from X×X to X and it is easy to see that (X; φ+; φ×) = (C; +; ×). This motivates the notion of an interpretation. Definition 4.9. Let M be a τ-structure and N be a τ 0 structure. M is defin- ably interperable in N iff we can find a definable X ⊂ jNjn, a definable function 2 φF (xF ; aF ) on X with aF 2 N of the same arity as F for each function F 2 τ, and definable relations φR(yR; aR) on X with aR 2 N of the same arity as R for each relation R 2 τ such that ∼ M = fX; φF (xF ; aF ); φR(yR; aR) j F; R 2 τg Let's look at a more interesting example. We will outline how to interpret any order in a graph; the full details are in [M, Section 1.3]. We want to describe a map from orders I to graphs GI that allows one to recover the order from the graph in a uniform way. Define the vertices of GI as follows: a a a • For each a 2 I, add vertices a; x1; x2; x3. a;b a;b a;b • For each a <I b, add vertices y1 ; y2 ; y3 Define the edges of GI as follows: • For each a 2 I, a a a { x1; x2; x3 are all connected a { a and x1 are connected • For each a <I b, a;b a;b a;b { y1 is connected to a; y2 ; y3 a;b { y2 is also connected to b.
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