Studying the role of induction axioms in reverse mathematics Paul - PowerPoint PPT Presentation

Studying the role of induction axioms in reverse mathematics Paul Shafer Universiteit Gent Paul.Shafer@UGent.be http://cage.ugent.be/~pshafer/ SLS 2014 February 21, 2014 Paul Shafer – UGent reverse mathematics and induction February 21, 2014 1 / 36

Contents Introduction 1 The pigeonhole principle and induction 2 Π 1 1 -conservativity 3 Back to pigeonhole principles 4 Diagonally non-recursive functions and graph colorability 5 Paul Shafer – UGent reverse mathematics and induction February 21, 2014 2 / 36

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 Π 0 1 ; and • the Σ 0 1 induction scheme: ( ϕ (0) ∧ ∀ n ( ϕ ( n ) → ϕ ( n + 1))) → ∀ nϕ ( n ) , where ϕ is Σ 0 1 . Idea: To get new sets, you have to compute them from old sets. Paul Shafer – UGent reverse mathematics and induction February 21, 2014 3 / 36

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, 2014 4 / 36

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, 2014 5 / 36

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 ) I Σ 0 n (I Π 0 n ) is the collection of induction axioms where the ϕ is Σ 0 n ( Π 0 n ). Note RCA 0 includes I Σ 0 1 . B Σ 0 n (B Π 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, 2014 6 / 36

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, 2014 7 / 36

A summary of basic equivalences Let n ≥ 1 . Over RCA 0 : • I Σ 0 n ↔ I Π 0 n ; • B Σ 0 n +1 ↔ B Π 0 n ; • I Σ 0 n +1 ⇒ BΣ 0 n +1 ⇒ I Σ 0 n (Kirby & Paris); • if n ≥ 2 , then I ∆ 0 n ↔ B Σ 0 n (Slaman). Two points: (1) RCA 0 proves I Π 0 1 . (2) The bounding axioms are equivalent to induction axioms. Paul Shafer – UGent reverse mathematics and induction February 21, 2014 8 / 36

Contents Introduction 1 The pigeonhole principle and induction 2 Π 1 1 -conservativity 3 Back to pigeonhole principles 4 Diagonally non-recursive functions and graph colorability 5 Paul Shafer – UGent reverse mathematics and induction February 21, 2014 9 / 36

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) B Σ 0 2 and RT 1 < N are equivalent over RCA 0 . Remember that B Σ 0 2 ↔ B Π 0 1 over RCA 0 (from the previous slide). So Hirst’s theorem also means that B Π 0 1 and RT 1 < N are equivalent over RCA 0 . Paul Shafer – UGent reverse mathematics and induction February 21, 2014 10 / 36

1 ⊢ RT 1 RCA 0 + B Π 0 < 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 B Π 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, 2014 11 / 36

RCA 0 + RT 1 < N ⊢ B Π 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 ) if a b exists f ( ℓ ) = otherwise ℓ f ( ℓ ) is the least b that witnesses bounding when only looking up to ℓ . Paul Shafer – UGent reverse mathematics and induction February 21, 2014 12 / 36

RCA 0 + RT 1 < N ⊢ B Π 0 1 Reminder: � ( µb < ℓ )( ∀ n < t )( ∃ m < b )( ∀ z < ℓ ) ϕ ( n, m, z ) if a b exists f ( ℓ ) = otherwise ℓ 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, 2014 13 / 36

RCA 0 + RT 1 < N ⊢ B Π 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, 2014 14 / 36

Contents Introduction 1 The pigeonhole principle and induction 2 Π 1 1 -conservativity 3 Back to pigeonhole principles 4 Diagonally non-recursive functions and graph colorability 5 Paul Shafer – UGent reverse mathematics and induction February 21, 2014 15 / 36

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 + B Σ 0 2 is Π 1 1 -conservative over RCA 0 + B Σ 0 2 (H´ ajek). Paul Shafer – UGent reverse mathematics and induction February 21, 2014 16 / 36

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, 2014 17 / 36

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, 2014 18 / 36

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, 2014 19 / 36