The Compactness Theorem Mike Prest Department of Mathematics Alan Turing Building University of Manchester Manchester M13 9PL UK [email protected] July 19, 2017 July 19, 2017 1 / 21 Is 0:9 = 1? meaning, is 0:99999 ··· = 1? Theorem (Compactness Theorem v.1) If you want something and there's no reason you can't have it, then you can get it. Set = 1 - 0:9. + The conditions on , for it to be nonzero, are: f < 1=n : n 2 Z g [ f > 0g. July 19, 2017 2 / 21 There is a non-standard version, R∗, of the reals, which has an infinitesimal (a solution, , to all those conditions). In R∗, we will have 0:9 < 1. R∗ is an elementary extension of R (said otherwise, R is an elementary substructure of R∗) meaning that, whenever '(x) is a formula (in n free variables and with parameters from R) then the solution set, '(R), of ' in R is the intersection of Rn with the solution set '(R∗) in R∗. July 19, 2017 3 / 21 We can \construct" R∗ using ultraproducts. As a simpler case we'll first construct an ultraproduct of finite fields. First we should define infinite products. July 19, 2017 4 / 21 Suppose that M1; M2;:::; Mi ;::: are structures all of the same kind (all groups, or rings, or partially ordered sets, or...). Their product is, as a set, the product Xi=1 Mi = M1 × M2 × · · · × Mi × ::: 1 of their underlying sets. The elements of this set are the sequences + a = (ai ) = (ai )i = (ai )i2Z = (a1; a2;:::; ai ;::: ) with ai 2 Mi for every i. The structure is defined on this set pointwise. For example, applying this with the component structures being the integers mod p for the various primes p, we get the structure Xp Zp where p ranges over the primes. The structure on this is the structure of a ring: there's an addition and a multiplication, both defined coordinatewise, an identity 1 = (1p)p for the multiplication and an identity 0 = (0p)p for the addition, where 1p denotes the image of 1 2 Z in Zp and similarly for 0. But the product, though a ring, is not a field. The aim is to produce a structure whose properties are some kind of \average" of the properties of the component structures. Since all those are fields, we should aim to produce a field. July 19, 2017 5 / 21 We're going to identify a = (ai )i with b = (bi )i if these agree on a \large" set of coordinates. \Large"?? Certainly I itself should be a large subset (equal elements should be identified) and the empty set ; should not be large (collapsing all elements together would not give an interesting result). If J ⊆ I is large and J ⊆ K ⊆ I then surely K should also be large. If we're going to identify a and b and also identify b and c then we're going to have to identify a and c (“identification” will be an equivalence relation). Setting J = fi 2 I : ai = bi g and K = fi 2 I : bi = ci g to be the sets where these pairs of elements agree, then all we can really say about the set of coordinates where ai = ci is that it contains J \ K; so it looks as if we should require this set to be large. Let's extract those conditions. July 19, 2017 6 / 21 Definition If I is a set then a filter on I is a collection F of subsets of I such that: • I 2 F; •; 2= F; • if J ⊆ K ⊆ I and J 2 F then K 2 F; • if J; K 2 F, then J \ K 2 F. Given a filter F on I , we define an equivalence relation ∼ on the product Xi2I Mi by (ai )i ∼ (bi )i iff fi 2 I : ai = bi g 2 F. Denote by Xi2I Mi =F the set of equivalence classes, writing a= ∼ for the equivalence class of any element a 2 Xi2I Mi . We can then turn Xi2I Mi =F into a structure of the same kind as the Mi , defining operations and relations pointwise. This structure is called the reduced product of the Mi (with respect to the filter F). If all the structures Mi are copies of the same structure M then we use the notation MI =F and refer to this as a reduced power of M. July 19, 2017 7 / 21 Let's do this with the field example above, taking F to be the set of cofinite + + subsets of Z (those with a finite complement in Z ). We can then define the algebraic operations by setting (ap)p= ∼ + (bp)p= ∼ = (ap + bp)p= ∼ and (ap)p= ∼ × (bp)p= ∼ = (ap × bp)p= ∼ . 0 It has to be checked that this is well-defined (e.g. that if (ap)p ∼ (ap)p and 0 0 0 (bp)p ∼ (bp)p then (ap + bp)p ∼ (ap + bp)p) but the conditions in the definition of a filter include what we need to do this. We can then check that (0p)p= ∼ is the zero for addition and (1p)p= ∼ is the identity for multiplication and, indeed, that all the axioms for a commutative ring are satisfied by Xp Zp=F. However, it's still not a field. July 19, 2017 8 / 21 Definition An ultrafilter U on a set I is a filter on I which satisfies the further equivalent conditions: • for each J ⊆ I either J 2 U or I n J 2 U; • if J [ K 2 U then either J 2 U or K 2 U; •U is a maximal filter (meaning that any collection of subsets of I which properly includes all the sets in U cannot be a filter). Let's continue our example using an ultrafilter U and check that the quotient ring F = Xp Zp=U, is a field: if a= ∼ = (ap)p= ∼ 6= 0 then Z = fp : ap = 0g 2= U. Since we have an ultrafilter, it follows that I n Z = fp : ap 6= 0g 2 U. But whenever ap 6= 0, ap has an inverse, bp say. Set b = (bp)p (where, if p 2 Z, set bp = 0 say). Then ab= ∼ = 1 so a= ∼ has a multiplicative inverse, b= ∼, as required. July 19, 2017 9 / 21 We will consider only non-principal ultrafilters. But - are there any? If I is any infinite set then the collection of all cofinite sets is a filter, sometimes called the Fr´echet filter F0. If U is any ultrafilter containing F0 then (quick exercise) U cannot be principal. We do need to call on Zorn's Lemma to give the existence of a maximal=ultra filter containing any given filter. July 19, 2017 10 / 21 In the special case where all the `component' structures are the same, M, say, there is a natural map from M to any reduced product, M∗ = MI =F, given by taking a 2 M to (a)i = ∼ (the equivalence class of the constant sequence (a)i ). It's easy to check that this is an embedding of M into M∗, termed the diagonal embedding. So M∗ has a copy of the original structure sitting inside it. Nothing like this is the case when the components are all different - we saw an ultraproduct of finite fields giving a field of characteristic 0 (which cannot, therefore, contain any finite field). July 19, 2017 11 / 21 Theorem (Los' Theorem v.1) If all the component structures (or even just a \large" set of them) have a certain property, then their ultraproduct has that property. Let's come back to infinitesimals. We'll take our index set I to be the set of positive integers. For each n 2 I , take the structure Mn to be the reals R. Choose (rather, apply Zorn's Lemma to get) a non-principal ultrafilter U and form the corresponding ultrapower, R∗. This contains the diagonally-embedded copy of the reals. 1 ∗ Consider the element = n n = ∼ 2 R . Each component is > 0 so, from the way we define the ordering in the ultraproduct, > 0. Also, given a positive integer n, for all but finitely many k, the kth component of is < 1=n; so, again by the definition of the relation in the ultraproduct, < 1=n. Thus we have our infinitesimal, , but we had to move to some \non-standard" version R∗ of the reals to get it. July 19, 2017 12 / 21 Structures: functions, constants, relations Equations: between terms involving variables July 19, 2017 13 / 21 Definition 0 Let M be a structure. Let x1;:::; xn be variables (\unknowns") and let t; t be two terms built up from these variables, using the algebraic operations and also allowing the constants, if there are any, to appear. We write t(x) to display the variables. We refer to t = t 0 as an equation and we define its solution set to be fa 2 Mn : t(a) = t 0(a)g. Definition Suppose first that M is a purely algebraic structure (meaning the \structure" is given by operations and constants - no relations). The definable subsets of (the various finite powers of) M are the sets obtained as follows: • the solution set of every equation t = t 0 is a definable subset; • the complement, Mn n D of any definable subset D of Mn is definable; • the intersection of any two definable subsets of Mn is definable (therefore, in view of the previous clause, their union also is definable); • if D is a definable subset of Mn and i is any of f1;:::; ng then the image of D under projection along the ith axis, that is f(a1;:::; ai-1; ai+1;:::; an): 9a 2 M with (a1;:::; ai-1; a; ai+1;:::; an) 2 Dg, is a definable subset of Mn-1.