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

1

Introduction

2

The pigeonhole principle and induction

3

Π1

1-conservativity 4

Back to pigeonhole principles

5

Diagonally non-recursive functions and graph colorability

Paul Shafer – UGent reverse mathematics and induction February 21, 2014 2 / 36

Let’s begin with an old friend

The system RCA0 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 Σ0

1 and ψ is Π0 1; and

• the Σ0

1 induction scheme:

(ϕ(0) ∧ ∀n(ϕ(n) → ϕ(n + 1))) → ∀nϕ(n), where ϕ is Σ0

1.

Idea: To get new sets, you have to compute them from old sets.

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Boy, lots of theorems say that certain kinds of sets exist!

RCA0 gives you

• a little bit of comprehension and
• a little bit of induction.

In reverse mathematics, we typically analyze which theorems prove which

• ther theorems over RCA0.

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.

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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 Σ0

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) IΣ0

n (IΠ0 n) is the collection of induction axioms where the ϕ is Σ0 n (Π0 n).

Note RCA0 includes IΣ0

1.

BΣ0

n (BΠ0 n) is the collection of bounding axioms where the ϕ is Σ0 n (Π0 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 Σ0

n and ψ is Π0 n.

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A summary of basic equivalences

Let n ≥ 1. Over RCA0:

• IΣ0

n ↔ IΠ0 n;

• BΣ0

n+1 ↔ BΠ0 n;

• IΣ0

n+1 ⇒ BΣ0 n+1 ⇒ IΣ0 n (Kirby & Paris);

• if n ≥ 2, then I∆0

n ↔ BΣ0 n (Slaman).

Two points: (1) RCA0 proves IΠ0

1.

(2) The bounding axioms are equivalent to induction axioms.

Paul Shafer – UGent reverse mathematics and induction February 21, 2014 8 / 36

Contents

1

Introduction

2

The pigeonhole principle and induction

3

Π1

1-conservativity 4

Back to pigeonhole principles

5

Diagonally non-recursive functions and graph colorability

Paul Shafer – UGent reverse mathematics and induction February 21, 2014 9 / 36

Equivalence between a theorem and an induction scheme

RT1

<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)

BΣ0

2 and RT1 <N are equivalent over RCA0.

Remember that BΣ0

2 ↔ BΠ0 1 over RCA0 (from the previous slide).

So Hirst’s theorem also means that BΠ0

1 and RT1 <N are equivalent over

RCA0.

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RCA0 + BΠ0

1 ⊢ RT1 <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 BΠ0

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}.

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RCA0 + RT1

<N ⊢ BΠ0 1

Suppose (∀n < t)(∃m)(∀z)ϕ(n, m, z) (where ϕ has only bounded quantifiers). Define f : N → N by f(ℓ) =

• (µb < ℓ)(∀n < t)(∃m < b)(∀z < ℓ)ϕ(n, m, z)

if a b exists ℓ

• therwise

f(ℓ) is the least b that witnesses bounding when only looking up to ℓ.

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RCA0 + RT1

<N ⊢ BΠ0 1

Reminder: f(ℓ) =

• (µb < ℓ)(∀n < t)(∃m < b)(∀z < ℓ)ϕ(n, m, z)

if a b exists ℓ

• therwise

If ran(f) is bounded, then RT1

<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 (ℓi : i ∈ N) so that ℓi < ℓi+1 and f(ℓi) < f(ℓi+1). Now define g: N → t by g(i) = (µn < t)(∀m < f(ℓi) − 1)(∃z < ℓi)(¬ϕ(n, m, z)).

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RCA0 + RT1

<N ⊢ BΠ0 1

Reminder: g(i) = (µn < t)(∀m < f(ℓi) − 1)(∃z < ℓi)(¬ϕ(n, m, z)). By RT1

<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(ℓi) − 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

1

Introduction

2

The pigeonhole principle and induction

3

Π1

1-conservativity 4

Back to pigeonhole principles

5

Diagonally non-recursive functions and graph colorability

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Π1

1-conservativity

Definition (Π1

1-conservativity)

Let S and T be theories with S ⊆ T. Then T is Π1

1-conservative over S if

ϕ ∈ T ↔ ϕ ∈ S whenever ϕ is a Π1

1 sentence.

So although T may be stronger than S, this additional strength is not witnessed by a Π1

1 sentence.

(Of course you can study conservativity for other formula classes too.) Classic examples:

• WKL0 is Π1

1-conservative over RCA0 (Harrington).

• WKL0 + BΣ0

2 is Π1 1-conservative over RCA0 + BΣ0 2 (H´

ajek).

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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, RCA0 and WKL0 have the same first-order part because WKL0 is Π1

1-conservative over RCA0.

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How do you prove conservativity results?

Proving that T is Π1

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 RCA0 is a submodel of a countable model of WKL0 with the same first-order part.

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How do you prove conservativity results?

Suppose RCA0 ∀Xϕ(X), where ϕ is arithmetic. Then there is a countable model M = (N, S) such that M RCA0 and M ∃X(¬ϕ(X)). Let X ∈ S witness M ¬ϕ(X). M is a submodel of a countable model N = (N, T ) of WKL0, 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 WKL0 and N ∃X(¬ϕ(X)). So WKL0 ∀Xϕ(X).

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Extending models while preserving induction

Suppose ϕ and ψ are two theorems, and we want to prove that RCA0 + ϕ ψ. So we have to build a model of RCA0 + 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 IΣ0

1.

When extending models, the situation is different. You are given a model (N, S) of RCA0, and all you know is that IΣ0

1 holds relative to the X ∈ S.

When adding sets to S, you need to make sure that IΣ0

1 holds relative to

them.

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Weak Ramsey principles

Definition (RT2

2, ADS, and CAC)

• RT2

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. RT2

2 is strictly stronger than CAC over RCA0 (Hirschfeldt & Shore).

CAC is strictly stronger than ADS over RCA0 (Lerman, Solomon, Towsner). RT2

2 proves BΣ0 2 over RCA0 (Hirst).

CAC and ADS both prove BΣ0

2 over RCA0 (Chong, Lempp, Yang).

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What about IΣ0

2?

Question

Does RT2

2 imply IΣ0 2 over RCA0? Does CAC? Does ADS?

Answer: None of these principles implies IΣ0

2 over RCA0 (Chong, Slaman,

Yang).

Theorem (Chong, Slaman, Yang)

Both ADS and CAC are Π1

1-conservative over RCA0 + BΣ0 2.

It follows that neither ADS nor CAC proves IΣ0

2 over RCA0 because IΣ0 2

consists of arithmetical axioms, and RCA0 + BΣ0

2 IΣ0 2.

Paul Shafer – UGent reverse mathematics and induction February 21, 2014 22 / 36

Contents

1

Introduction

2

The pigeonhole principle and induction

3

Π1

1-conservativity 4

Back to pigeonhole principles

5

Diagonally non-recursive functions and graph colorability

Paul Shafer – UGent reverse mathematics and induction February 21, 2014 23 / 36

The pigeonhole principle for trees

Definition (TT1)

TT1: 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

• RCA0 + IΣ0

2 ⊢ TT1 (Chubb, Hirst, McNicholl).

• RCA0 + TT1 ⊢ BΣ0

2 (Chubb, Hirst, McNicholl).

• RCA0 + BΣ0

2 TT1 (Corduan, Groszek, Mileti).

Theorem (Breaking news! Chong and Li)

RCA0 + TT1 ⊢ IΣ0

2

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RCA0 + BΣ0

2 TT1

Theorem (Corduan, Groszek, Mileti)

If S is an extension of RCA0 by Π1

1 axioms, then S ⊢ TT1 if and only if

S ⊢ IΣ0

2.

Thus RCA0 + BΣ0

2 TT1 because RCA0 + BΣ0 2 is strictly weaker than

RCA0 + IΣ0

2.

The idea here is to prove that if M RCA0 but M IΣ0

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 RCA0 and M IΣ0

2, then there is a submodel N of M with

the same first-order part that satisfies all the Π1

1 sentences true in M but

not TT1.

Paul Shafer – UGent reverse mathematics and induction February 21, 2014 25 / 36

Contents

1

Introduction

2

The pigeonhole principle and induction

3

Π1

1-conservativity 4

Back to pigeonhole principles

5

Diagonally non-recursive functions and graph colorability

Paul Shafer – UGent reverse mathematics and induction February 21, 2014 26 / 36

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) = Φg

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, 2014 27 / 36

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))(Φf ∈ DNR(k)).

Theorem (Jockusch & Soare)

Every DNR(2) function computes an infinite path through every infinite, computable tree T ⊆ 2<N.

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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

• RCA0 ⊢ WKL ↔ ∀g(DNR(2, g)).
• For each standard k ≥ 2, RCA0 ⊢ WKL ↔ ∀g(DNR(k, g)).

Theorem (Ambos-Spies, Kjos-Hanssen, Lempp, Slaman)

∀g(DNR(g)) is strictly weaker than WWKL over RCA0.

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∃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 RCA0? Of course, we could ask the same question about the (formally weaker) statement ∀g∃k(DNR(k, g)).

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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 RCA0: (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.

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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 RCA0?

• ∀ℓ∃k∀G(COL(ℓ, k, G))
• ∀ℓ∀G∃k(COL(ℓ, k, G))

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DNR(k) and induction

Recall: every DNR(k) function computes a DNR(2) function. With a little work, the proof can be implemented using IΣ0

2.

Hence the following are equivalent over RCA0 + IΣ0

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 RCA0 + BΣ0

2.

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The strength of DNR(k) functions

We show that, over RCA0 + BΣ0

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) RCA0 + BΣ0

2 + ∃k∀g(DNR(k, g)) WKL

(ii) RCA0 + BΣ0

2 + ∀g∃k(DNR(k, g)) ∃k∀g(DNR(k, g))

(iii) RCA0 + BΣ0

2 + ∀ℓ∃k∀G(COL(ℓ, k, G)) WKL

(iv) RCA0 + BΣ0

2 + ∀ℓ∀G∃k(COL(ℓ, k, G)) ∃k∀g(DNR(k, g))

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Remaining questions

Are ∃k∀g(DNR(k, g)) and ∃k∀G(COL(k, G)) equivalent over RCA0? Are ∀g∃k(DNR(k, g)) and ∀ℓ∀G∃k(COL(ℓ, k, G)) equivalent over RCA0? Does ∀ℓ∃k∀G(COL(ℓ, k, G)) (or ∀ℓ∀G∃k(COL(ℓ, k, G))) imply WKL over RCA0 + IΣ0

2?

Is ∀ℓ∀G∃k(COL(ℓ, k, G)) strictly weaker than ∀ℓ∃k∀G(COL(ℓ, k, G)) over RCA0 (or over RCA0 + BΣ0

2)?

Best question: Does ∃k∀g(DNR(k, g)) imply WWKL over RCA0 (or over RCA0 + BΣ0

2)?

Paul Shafer – UGent reverse mathematics and induction February 21, 2014 35 / 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, 2014 36 / 36