Studying the role of induction axioms in reverse mathematics

Transcription

1 Studying the role of induction axioms in reverse mathematics Paul Shafer Universiteit Gent SLS 2014 February 21, 2014 Paul Shafer UGent reverse mathematics and induction February 21, / 36

2 Contents 1 Introduction 2 The pigeonhole principle and induction 3 Π 1 1 -conservativity 4 Back to pigeonhole principles 5 Diagonally non-recursive functions and graph colorability Paul Shafer UGent reverse mathematics and induction February 21, / 36

3 Let s begin with an old friend The system RCA 0 is axiomatized by the axioms of a discretely ordered commutative semi-ring with 1 (or PA if you like); the 0 1 comprehension scheme: n(ϕ(n) ψ(n)) X n(n X ϕ(n)), where ϕ is Σ 0 1 and ψ is Π0 1 ; and the Σ 0 1 induction scheme: where ϕ is Σ 0 1. (ϕ(0) n(ϕ(n) ϕ(n + 1))) nϕ(n), Idea: To get new sets, you have to compute them from old sets. Paul Shafer UGent reverse mathematics and induction February 21, / 36

4 Boy, lots of theorems say that certain kinds of sets exist! RCA 0 gives you a little bit of comprehension and a little bit of induction. In reverse mathematics, we typically analyze which theorems prove which other theorems over RCA 0. Often a theorem looks like a set-existence principle: for every this kind of set, there is a that kind of set. For example: For every continuous function [0, 1] R there is a maximum. For every commutative ring there is a prime ideal. For every coloring of pairs in two colors there is a homogeneous set. For every infinite subtree of 2 <N there is an infinite path. Paul Shafer UGent reverse mathematics and induction February 21, / 36

5 A closer look at induction So reverse mathematics is often concerned with set-existence principles. Almost as often, induction merits no special consideration. (But it can get tricky to make arguments work using only Σ 0 1 -induction.) Today, we look at a few situations in which induction plays a more pronounced role, such as with pigeonhole principles, conservativity results, and diagonally non-recursive functions. It will be fun and interesting. Paul Shafer UGent reverse mathematics and induction February 21, / 36

6 Induction and collection The induction axiom for ϕ is (the universal closure of) (ϕ(0) n(ϕ(n) ϕ(n + 1))) nϕ(n). The collection (or bounding) axiom for ϕ is (the universal closure of) ( n < t)( m)ϕ(n, m) ( b)( n < t)( m < b)ϕ(n, m) IΣ 0 n (IΠ 0 n) is the collection of induction axioms where the ϕ is Σ 0 n (Π 0 n). Note RCA 0 includes IΣ 0 1. BΣ 0 n (BΠ 0 n) is the collection of bounding axioms where the ϕ is Σ 0 n (Π 0 n). Paul Shafer UGent reverse mathematics and induction February 21, / 36

7 Induction and collection Even though there is no such thing as a 0 n formula, we can still make sense of I 0 n. I 0 n is the collection of universal closures of formulas of the form n(ϕ(n) ψ(n)) ([ϕ(0) n(ϕ(n) ϕ(n + 1))] nϕ(n)), where ϕ is Σ 0 n and ψ is Π 0 n. Paul Shafer UGent reverse mathematics and induction February 21, / 36

8 A summary of basic equivalences Let n 1. Over RCA 0 : IΣ 0 n IΠ 0 n; BΣ 0 n+1 BΠ0 n; IΣ 0 n+1 BΣ0 n+1 IΣ0 n (Kirby & Paris); if n 2, then I 0 n BΣ 0 n (Slaman). Two points: (1) RCA 0 proves IΠ 0 1. (2) The bounding axioms are equivalent to induction axioms. Paul Shafer UGent reverse mathematics and induction February 21, / 36

9 Contents 1 Introduction 2 The pigeonhole principle and induction 3 Π 1 1 -conservativity 4 Back to pigeonhole principles 5 Diagonally non-recursive functions and graph colorability Paul Shafer UGent reverse mathematics and induction February 21, / 36

10 Equivalence between a theorem and an induction scheme RT 1 <N is the following statement (it s the infinite pigeonhole principle): For every k > 0 and every f : N k, there is an infinite H N and a c < k such that ( n H)(f(n) = c). Theorem (Hirst) BΣ 0 2 and RT1 <N are equivalent over RCA 0. Remember that BΣ 0 2 BΠ0 1 over RCA 0 (from the previous slide). So Hirst s theorem also means that BΠ 0 1 and RT1 <N RCA 0. are equivalent over Paul Shafer UGent reverse mathematics and induction February 21, / 36

11 RCA 0 + BΠ 0 1 RT 1 <N Let f : N k. Suppose for a contradiction that no color c < k appears infinitely often: ( c < k)( n)( m)(m > n f(m) c). Then by BΠ 0 1 there is a b such that ( c < k)( n < b)( m)(m > n f(m) c). This means that f(b + 1) k, a contradiction. So in fact some color c < k appears infinitely often. Let H = {n : f(n) = c}. Paul Shafer UGent reverse mathematics and induction February 21, / 36

12 RCA 0 + RT 1 <N BΠ 0 1 Suppose ( n < t)( m)( z)ϕ(n, m, z) (where ϕ has only bounded quantifiers). Define f : N N by { (µb < l)( n < t)( m < b)( z < l)ϕ(n, m, z) f(l) = l if a b exists otherwise f(l) is the least b that witnesses bounding when only looking up to l. Paul Shafer UGent reverse mathematics and induction February 21, / 36

13 RCA 0 + RT 1 <N BΠ 0 1 Reminder: f(l) = { (µb < l)( n < t)( m < b)( z < l)ϕ(n, m, z) l if a b exists otherwise If ran(f) is bounded, then RT 1 <N applies, and there is a b and an infinite H such that ( l H)(f(l) = b). In this case, one proves that b is the desired bound: ( n < t)( m < b)( z)ϕ(n, m, z) Otherwise ran(f) is unbounded. Define a sequence (l i : i N) so that l i < l i+1 and f(l i ) < f(l i+1 ). Now define g : N t by g(i) = (µn < t)( m < f(l i ) 1)( z < l i )( ϕ(n, m, z)). Paul Shafer UGent reverse mathematics and induction February 21, / 36

14 RCA 0 + RT 1 <N BΠ 0 1 Reminder: g(i) = (µn < t)( m < f(l i ) 1)( z < l i )( ϕ(n, m, z)). By RT 1 <N applied to g, there is an n < t and an infinite H such that ( i H)(g(i) = n). Let m be such that zϕ(n, m, z). Let i H be such that f(l i ) 1 > m. Then g(i) = n, so z( ϕ(n, m, z)). Contradiction! Paul Shafer UGent reverse mathematics and induction February 21, / 36

15 Contents 1 Introduction 2 The pigeonhole principle and induction 3 Π 1 1 -conservativity 4 Back to pigeonhole principles 5 Diagonally non-recursive functions and graph colorability Paul Shafer UGent reverse mathematics and induction February 21, / 36

16 Π 1 1-conservativity Definition (Π 1 1-conservativity) Let S and T be theories with S T. Then T is Π 1 1-conservative over S if ϕ T ϕ S whenever ϕ is a Π 1 1 sentence. So although T may be stronger than S, this additional strength is not witnessed by a Π 1 1 sentence. (Of course you can study conservativity for other formula classes too.) Classic examples: WKL 0 is Π 1 1 -conservative over RCA 0 (Harrington). WKL 0 + BΣ 0 2 is Π1 1 -conservative over RCA 0 + BΣ 0 2 (Hájek). Paul Shafer UGent reverse mathematics and induction February 21, / 36

17 Why study conservativity? Conservativity gives a way to express that T is stronger than S but not too much stronger than S. Conservativity is useful for studying the first-order consequences of a theory. For example, RCA 0 and WKL 0 have the same first-order part because WKL 0 is Π 1 1 -conservative over RCA 0. Paul Shafer UGent reverse mathematics and induction February 21, / 36

18 How do you prove conservativity results? Proving that T is Π 1 1-conservative over S is about proving that every countable model of S can be extended to a model of T with the same first-order part. For example: Theorem (Harrington) Every countable model of RCA 0 is a submodel of a countable model of WKL 0 with the same first-order part. Paul Shafer UGent reverse mathematics and induction February 21, / 36

19 How do you prove conservativity results? Suppose RCA 0 Xϕ(X), where ϕ is arithmetic. Then there is a countable model M = (N, S) such that M RCA 0 and M X( ϕ(x)). Let X S witness M ϕ(x). M is a submodel of a countable model N = (N, T ) of WKL 0, where S T. Thus X T and so N ϕ(x). This is because ϕ is arithmetic, so the truth of ϕ(x) depends only on N and X. So N WKL 0 and N X( ϕ(x)). So WKL 0 Xϕ(X). Paul Shafer UGent reverse mathematics and induction February 21, / 36

20 Extending models while preserving induction Suppose ϕ and ψ are two theorems, and we want to prove that RCA 0 + ϕ ψ. So we have to build a model of RCA 0 + ϕ that is not a model of ψ. In many cases, this can be done by working over ω and proving that you can add sets witnessing ϕ that do not compute sets witnessing ψ. Induction is never a problem. With ω, you have as much induction as you could ever want, and you can never add a set that breaks IΣ 0 1. When extending models, the situation is different. You are given a model (N, S) of RCA 0, and all you know is that IΣ 0 1 holds relative to the X S. When adding sets to S, you need to make sure that IΣ 0 1 them. holds relative to Paul Shafer UGent reverse mathematics and induction February 21, / 36

21 Weak Ramsey principles Definition (RT 2 2, ADS, and CAC) RT 2 2: For every f : [N] 2 2, there is an infinite H such that f([h] 2 ) = 1. CAC: In every partial order, there is either an infinite chain or an infinite antichain. ADS: In every linear order, there is either an infinite increasing sequence or an infinite decreasing sequence. RT 2 2 is strictly stronger than CAC over RCA 0 (Hirschfeldt & Shore). CAC is strictly stronger than ADS over RCA 0 (Lerman, Solomon, Towsner). RT 2 2 proves BΣ 0 2 over RCA 0 (Hirst). CAC and ADS both prove BΣ 0 2 over RCA 0 (Chong, Lempp, Yang). Paul Shafer UGent reverse mathematics and induction February 21, / 36

22 What about IΣ 0 2? Question Does RT 2 2 imply IΣ 0 2 over RCA 0? Does CAC? Does ADS? Answer: None of these principles implies IΣ 0 2 over RCA 0 (Chong, Slaman, Yang). Theorem (Chong, Slaman, Yang) Both ADS and CAC are Π 1 1 -conservative over RCA 0 + BΣ 0 2. It follows that neither ADS nor CAC proves IΣ 0 2 over RCA 0 because IΣ 0 2 consists of arithmetical axioms, and RCA 0 + BΣ 0 2 IΣ0 2. Paul Shafer UGent reverse mathematics and induction February 21, / 36

23 Contents 1 Introduction 2 The pigeonhole principle and induction 3 Π 1 1 -conservativity 4 Back to pigeonhole principles 5 Diagonally non-recursive functions and graph colorability Paul Shafer UGent reverse mathematics and induction February 21, / 36

24 The pigeonhole principle for trees Definition (TT 1 ) TT 1 : For every k > 0 and every f : 2 <N k, there is an infinite T 2 <N that is order-isomorphic to 2 <N and a c < k such that ( σ T )(f(σ) = c). Theorem RCA 0 + IΣ 0 2 TT1 (Chubb, Hirst, McNicholl). RCA 0 + TT 1 BΣ 0 2 (Chubb, Hirst, McNicholl). RCA 0 + BΣ 0 2 TT1 (Corduan, Groszek, Mileti). Theorem (Breaking news! Chong and Li) RCA 0 + TT 1 IΣ 0 2 Paul Shafer UGent reverse mathematics and induction February 21, / 36

25 RCA 0 + BΣ 0 2 TT 1 Theorem (Corduan, Groszek, Mileti) If S is an extension of RCA 0 by Π 1 1 axioms, then S TT1 if and only if S IΣ 0 2. Thus RCA 0 + BΣ 0 2 TT1 because RCA 0 + BΣ 0 2 RCA 0 + IΣ 0 2. is strictly weaker than The idea here is to prove that if M RCA 0 but M IΣ 0 2, then there is a coloring f : 2 <N k such that no T T f is a monochromatic subtree of 2 <N isomorphic to 2 <N. Thus if M RCA 0 and M IΣ 0 2, then there is a submodel N of M with the same first-order part that satisfies all the Π 1 1 sentences true in M but not TT 1. Paul Shafer UGent reverse mathematics and induction February 21, / 36

26 Contents 1 Introduction 2 The pigeonhole principle and induction 3 Π 1 1 -conservativity 4 Back to pigeonhole principles 5 Diagonally non-recursive functions and graph colorability Paul Shafer UGent reverse mathematics and induction February 21, / 36

27 Diagonally non-recursive functions Definition Let f, g : N N, and let k 2. f is diagonally non-recursive relative to g (i.e., DNR(g)) if e(f(e) Φ g e(e)). f is k-bounded diagonally non-recursive relative to g (i.e., DNR(k, g)) if f is DNR(g) and ran(f) {0, 1,..., k 1}. (In the case g = 0, we just write DNR and DNR(k).) Notice that every DNR(k) function is also a DNR(k + 1) function. Does every DNR(k + 1) function compute a DNR(k) function? Paul Shafer UGent reverse mathematics and induction February 21, / 36

28 Classical theorems concerning DNR(k) functions Theorem (Jockusch attributes to Friedberg) For every k 2, every DNR(k) function computes some DNR(2) function. Theorem (Jockusch) The reduction from DNR(k) to DNR(2) is not uniform. That is, for each k 2, there is no Turing functional Φ such that ( f DNR(k + 1))(Φ f DNR(k)). Theorem (Jockusch & Soare) Every DNR(2) function computes an infinite path through every infinite, computable tree T 2 <N. Paul Shafer UGent reverse mathematics and induction February 21, / 36

29 Reverse math and DNR functions Here we overload the DNR notation: DNR(g) also denotes the statement there is an f that is DNR(g). DNR(k, g) also denotes the statement there is an f that is DNR(k, g). Theorem RCA 0 WKL g(dnr(2, g)). For each standard k 2, RCA 0 WKL g(dnr(k, g)). Theorem (Ambos-Spies, Kjos-Hanssen, Lempp, Slaman) g(dnr(g)) is strictly weaker than WWKL over RCA 0. Paul Shafer UGent reverse mathematics and induction February 21, / 36

30 k g(dnr(k, g)) versus WKL What if you know that there are DNR(k, g) functions for some k 2, but you do not know which k? This is essentially a question of Simpson, who asked: Question (Simpson, 2001) Is k g(dnr(k, g)) equivalent to WKL over RCA 0? Of course, we could ask the same question about the (formally weaker) statement g k(dnr(k, g)). Paul Shafer UGent reverse mathematics and induction February 21, / 36

31 Connection to graph colorability By compactness, every locally l-colorable graph is l-colorable. There has been considerable work analyzing the computability-theoretic content of this fact, culminating (for our purposes) in a theorem of Schmerl: Theorem (Schmerl) Fix standard l and k such that 2 l k. Then the following are equivalent over RCA 0 : (i) WKL (ii) Every locally l-colorable graph is k-colorable. The content of this theorem is that the ability to color a graph with a fixed sub-optimal number of colors is as strong as the ability to color the graph with the optimal number of colors. Paul Shafer UGent reverse mathematics and induction February 21, / 36

32 Connection to graph colorability Denote by COL(l, k, G) the statement if G is a locally l-colorable graph, then G is k-colorable. Question (essentially of Simpson) Are either of the following statements equivalent to WKL over RCA 0? l k G(COL(l, k, G)) l G k(col(l, k, G)) Paul Shafer UGent reverse mathematics and induction February 21, / 36

33 DNR(k) and induction Recall: every DNR(k) function computes a DNR(2) function. With a little work, the proof can be implemented using IΣ 0 2. Hence the following are equivalent over RCA 0 + IΣ 0 2 : (i) WKL (ii) k g(dnr(k, g)) (iii) g k(dnr(k, g)) Dorais, Hirst, and I show that the three equivalences fail if you only assume RCA 0 + BΣ 0 2. Paul Shafer UGent reverse mathematics and induction February 21, / 36

34 The strength of DNR(k) functions We show that, over RCA 0 + BΣ 0 2, WKL k g(dnr(k, g)) g k(dnr(k, g)), and WKL l k G(COL(l, k, G)), and l G k(col(l, k, G)) k g(dnr(k, g)) The results rewritten in an authoritative purple box: Theorem (Dorais, Hirst, S) (i) RCA 0 + BΣ k g(dnr(k, g)) WKL (ii) RCA 0 + BΣ g k(dnr(k, g)) k g(dnr(k, g)) (iii) RCA 0 + BΣ l k G(COL(l, k, G)) WKL (iv) RCA 0 + BΣ l G k(col(l, k, G)) k g(dnr(k, g)) Paul Shafer UGent reverse mathematics and induction February 21, / 36

35 Remaining questions Are k g(dnr(k, g)) and k G(COL(k, G)) equivalent over RCA 0? Are g k(dnr(k, g)) and l G k(col(l, k, G)) equivalent over RCA 0? Does l k G(COL(l, k, G)) (or l G k(col(l, k, G))) imply WKL over RCA 0 + IΣ 0 2? Is l G k(col(l, k, G)) strictly weaker than l k G(COL(l, k, G)) over RCA 0 (or over RCA 0 + BΣ 0 2 )? Best question: Does k g(dnr(k, g)) imply WWKL over RCA 0 (or over RCA 0 + BΣ 0 2 )? Paul Shafer UGent reverse mathematics and induction February 21, / 36

36 Thank you! Thank you for coming to my talk! Do you have a question about it? Paul Shafer UGent reverse mathematics and induction February 21, / 36