Forcing Arguments in Second-Order Arithmetic Kenji Omoto Tohoku University
2012/2/20
Reverse Mathematics I
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Setting: Second order arithmetic.
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Main Question: What axioms are necessary to prove the theorems of Mathematics?
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Axiom systems(Big 5): RCA0 ≡ ∆11 -Comprehension + Σ01 -Induction + Semiring ax WKL0 ≡ RCA0 + Weak K¨onig’s lemma ACA0 ≡ RCA0 + Arithmetic Comprehension ATR0 1 Π1 -CA0
≡ ACA0 + Arithmetic Transfinite recursion ≡ ACA0 + Π11 -Comprehension
Theories of Second-Order Arithmetic I ▶
∪ A set is called arithmetical ( n<ω Σ0n (X )) if it is defined by a formula consisting of only number quantifiers(no set quantifiers).
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ACA0 is a theory consists of RCA0 and arithmetical comprehension.
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ACA0 is a Π11 -conservative extension of PA.
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ACA0 satisfies good theorems about arithmetical properties.
Fact (examples of properties of ACA0 ) 1. ARITH(X ) ⊨ ACA0 for any X ⊆ ω. Where ARITH(X ) is a class of all sets which are arithmetical in X. 2. Every ω model M of ACA0 is arithmetically closed. That is, for any X ∈ M, ARITH(X ) ⊆ M.
Theories of Second-Order Arithmetic II RCA0 WKL0
∆01
Comprehension Weak K¨onig’s Lemma
∆01 Recursive, Computable ∪
Σ0n Arithmetical ∪ ∆11 , µ<ωCK Σ0µ n<ω
ACA0
Arithmetical Comprehension
? ATR0 Π11 -CA0
What axiom is placed here? Arithmetical Transfinite Recursion Π11 Comprehension
1
Hyperarithmetical Π11
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A set defined by transfinite induction of arithmetical definition. is called hyperarithmetical set.
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A ∪ class of0 sets which hyperarithmetical in1 X is denoted by µ<ω X Σµ . In fact, It is equivalent to ∆1 (X )
Theories of Second-Order Arithmetic III
∆01 Comprehension Weak K¨onig’s lemma
RCA0 WKL0
∆01 Recursive, Computable ∪
Σ0n Arithmetical ∪ ∆11 , µ<ωCK Σ0µ n<ω
ACA0
Arithmetical Comprehension
? ATR0 Π11 -CA0
What axiom is placed here? Arithmetical Transfinite Recursion Π11 Comprehension
1
Hyperarithmetical Π11
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Question: What theory of second-order arithmetic corresponds to Hyperarithmetical?
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∆11 -CA0 ?
Theory of Hyperarithemetic Analysis I
Definition A theory T is called theory of hyperarithmetic analysis if it satisfies following: 1. HYP(X ) ⊨ T for any X ⊆ ω. Where HYP(X ) is a class of all sets which are hyperarithmetical in X. 2. Every ω model of T is hyperarithmetically closed. ▶
∆11 -CA0 is a theory of hyperarithmetic analysis.
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However, beside ∆11 -CA0 , there exists other various theories of hyperarithmetic.
Theory of Hyperarithemetic Analysis II
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the following are theories of hyperarithmetic analysis. Σ11 -AC0 ≡ ACA0 + Σ11 -Axiom of Choice Π11 -SEP0 ≡ ACA0 + Π11 -Separation ∆11 -CA0 ≡ ACA0 + ∆11 -Comprehension
weak-Σ11 -AC0 ≡ ACA0 + weak-Σ11 -Axiom of Choice INDEC ≡ ACA0 + Jullien’s Indecomposability Theorem ABW0 ≡ ACA0 + Arithmetical Bolzano-Weierstraß’s Theorem JI ≡ ACA0 + (∀X )(∀α){[(∀β < α)X (β) exists] → X (α) exists}
Theory of Hyperarithemetic Analysis III CDG-CA ≡ ACA0 + Given a sequence {Tn | n ∈ N} of completely determined trees, there exists a set X s.t. n ∈ X ⇔ I has a winning strategy in Tn CDG-AC ≡ ACA0 + Given a sequence {Tn | n ∈ N} of completely determined trees, Σn∈N Tn is also completely determined. DG-CA ≡ ACA0 + Given a sequence {Tn | n ∈ N} of determined trees, there exists a set X s.t. n ∈ X ⇔ I has a winning strategy in Tn DG-AC ≡ ACA0 + Given a sequence {Tn | n ∈ N} of determined trees, Σn∈N Tn is also completely determined. etc...
Theory of Hyperarithemetic Analysis IV → is proved over RCA0
Σ11 -AC0
⇒ is proved over RCA0 + Σ11 -IND ?
DG-AC H
×
̸→ is proved by tagged tree forcing
Π11 -SEP0 J
ABW0 o
× × ×
∆11 -CA0 o J
'
×
/ DG-CA
INDEC H ×
"
weak-Σ11 -AC0 K
o × CDG-CA H ×
JI
/ CDG-AC
Definition of Tagged Tree I We shall see how Π11 -SEP0 ̸→ Σ11 -AC0 is proved by Montalb´an.
Definition Σ11 -AC0 is ACA0 plus following scheme: (∀n)(∃X )ψ(n, X ) ⇒ (∃Z )(∀n)ψ(n, Zn ), where ψ(n, X ) is Σ11 .
Definition Π11 -SEP0 is ACA0 plus following scheme: ¬(∃n)[¬ψ(n) ∧ ¬ϕ(n)] ⇒ (∃Z )(∀n)[(¬ψ(n) → n ∈ Z ) ∧ (¬ϕ(n) → n ∈ / Z )], where ψ(n) and ϕ(n) are Σ11 formula.
Definition of Tagged Tree II Definition p = (T p , f p , hp ) ∈ P if and only if ▶ T p ⊆ ω <ω is a finite tree. ▶
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f p is a non-empty finite partial function such that dom(f p ) ⊆ ω, ran(f p ) ⊆ T p . ▶
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T p grows to a infinite tree T G .
Every f p (i) grows to a path f G (i) of T G .
hp : T p → ω1CK ∪ {∞} s.t. ▶
(∀s, t ∈ T p )[s ⊊ t → hp (s) > hp (t)] (∀s ∈ T p )[(∃i)[s ⊆ f p (i)] → hp (s) = ∞]
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Every hp grows to a rank function hG of T G
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Definition of Tagged Tree III
Definition For any P-generic G, we define objects as follows: ▶
T G := ∪p∈G T p
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αiG = f G (i) := ∪p∈G ,i∈dom(f p ) f p (i)
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hG := ∪p∈G hp
▶ ▶
⊕ MFG := {X | (∃µ < ω1CK )[X is Σ0µ (T G ⊕ i∈F αG (i))]} ∪ G := G G G M∞ F ⊆fin ω MF = HYP({T } ∪ {α (i) | i ∈ F })
where, MFG ∩ [T G ] = {αiG | i ∈ F } holds.
G M∞ ⊭ Σ11 -AC0
Theorem G ⊭ Σ1 -AC M∞ 0 1 ▶
If there finitely arbitrary many sets satisfy a Σ11 formula, Σ11 -AC0 certifies that it is possible to take infinitely many sets satisfying the formula.
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”X is an infinite path of T” is also Σ11 formula with argument X.
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If F is large enough, MFG contains many paths, therefore it is G possible to take they from M∞
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However, every MFG contains at most finitely many paths. Therefore it is impossible to take infinitely many paths from G. M∞
Forcing relation I
Definition For p ∈ P and a formula ψ, p ⊩ ψ is defined as follows: ▶
p ⊩ σ ∈ T ⇔def σ ∈ T p ∨ (1 ≤ hp (σ − ) ∧ σ − ∈ T p )
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p ⊩ (¯ n, m) ¯ ∈ αi ⇔def i ∈ dom(f p ) ∧ f p (i)(n) = m
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p ⊩ n ∈ Sν,F ,e ⇔def p ⊩ (∃x)θ(e, x, n, Hν,F ) (where (∃x)θ(e, x, y , X ) is a universal Σ01 formula for some θ Σ00 formula.)
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p ⊩ (e, n, ν) ∈ Hµ,F ⇔def ν < µ ∧ p ⊩ n ∈ Sν,F ,e
Forcing relation II Definition ▶
p ⊩ ∀XFµ ϕ(XFµ ) ⇔def (∀ν < µ)(∀e ∈ ω)p ⊩ ϕ(Sν,F ,e )
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p ⊩ ∀X µ ϕ(X µ ) ⇔def (∀ν < µ)(∀e ∈ ω)(∀F ⊆fin ω)p ⊩ ϕ(Sν,F ,e )
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p ⊩ ∀XF ϕ(XF ) ⇔def (∀ν < ω1CK )(∀e ∈ ω)p ⊩ ϕ(Sν,F ,e )
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p ⊩ ∀X ϕ(X ) ⇔def (∀ν < ω1CK )(∀e ∈ ω)(∀F ⊆fin ω)p ⊩ ϕ(Sν,F ,e )
where a subscript F of the variable X means that values of X are in MFG , a superscript µ of the variable X means that values of X are Σ0µ set.
Retagging I
Definition
Ret(µ, F , p, p ∗ ) (p is a µ-F -absolute-retagging of p ∗ ) if and only if ∗
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Tp = Tp
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(∀i ∈ F )[f p (i) = f p (i)]
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(∀s ∈ T p )[hp (s) ≥ µ → hp (s) ≥ µ]
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(∀s ∈ T p )[hp (s) < µ → hp (s) = hp (s)]
∗
∗
Lemma
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Ret(ω · µ, F , p, p ∗ ) ∧ q ≤ p ∧ ν < µ ⇒ (∃q ∗ ≤ p ∗ )Ret(ω · ν, F , q, q ∗ )
Retagging II Definition A formula ψ is F -restricted if every subscript of variable of ψ is a subset of F .
Lemma Let a formula ψ be such that F-restricted and rk(ψ) < ω1CK . Let p, p ∗ ∈ P. Then Ret(ω · rk(ψ), F , p, p ∗ ) ⇒ (p ⊩ ψ ⇒ p ∗ ⊩ ψ) p⊩ψ
Ret
p∗
q⊩ψ
Ret
q∗
Retagging III
Lemma For a formula ψ with rank µ, 0(µ) is able to decide whether p ⊩ ψ uniformly in p, ψ, and µ.
Σ-over-L∞ Definition ▶
A formula ψ is F -restricted if every subscript of variable of ψ is a subset of F .
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A formula ψ is ranked if every variable of ψ is ranked.
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A formula is Σ-over-LF if it is built up from ranked and F -restricted formulas using ∧, (∀n), (∃X ).
Definition For any formula ψ and µ < ω1CK , ψ µ is a result of replacing of (∃X ) by (∃X µ ).
Lemma Let F ⊆fin ω, ψ be Σ-over-LF sentence, and µ < ω1CK . Then, Ret(ω · rk(ψ) + ω 2 + ω, F , p, p ∗ ) ∧ dom(f p ) ⊆ F
⇒ (p ⊩ ¬ψ µ ⇒ p ∗ ⊩ ¬ψ µ )
G M∞ ⊨ Π11 -SEP0 I
Theorem G ⊨ Π1 -SEP M∞ 0 1
Proof ▶
Let ϕ(n) and ψ(n) be a Σ-over-LF formulas(Σ11 formulas) with a free variable n such that G M∞ ⊨ (∀n)[ψ(n) ∨ ϕ(n)]
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G will be constructed such that D ∈ M∞ G M∞ ⊨ (∀n)[(¬ϕ(n) → n ∈ D) ∧ (n ∈ D → ψ(n))]
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Let p ∈ G be such that p ⊩ (∀n)[ψ(n) ∨ ϕ(n)].
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Let µ < ω1CK be such that p ⊩ (∀n)[ψ µ (n) ∨ ϕµ (n)].
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Let ν < ω1CK be large enough for p ∈ Pν ∧ (∀n ∈ ω)[rk(ψ µ (¯ n) ∨ ϕµ (¯ n)) < ν]
G M∞ ⊨ Π11 -SEP0 II
▶ ▶
D is defined as follows. d ∈ D if and only if there exists q ∈ P such that 1. 2. 3. 4. 5. 6.
q ⊩ ¬ϕν (d) Tq ⊆ TG (∀i ∈ F )[f q (i) = αiG ∩ T q ] ran(hq (s)) ⊆ (ων + ω 2 + 2ω) ∪ {∞} hG (s) < ων + ω 2 + 2ω → g (s) = hG (s) hG (s) ≥ ων + ω 2 + 2ω → g (s) ≥ ν
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D is hyperarithmetical in T ⊕
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G. Therefore D ∈ M∞
⊕
i∈F
αiG .
(¬ϕ(n) → n ∈ D) and (n ∈ D → ψ(n)) will be proved in next page.
G M∞ ⊨ Π11 -SEP0 III
¬ϕ(n) → n ∈ D. ▶
G ⊨ ¬ϕµ (n) holds. Assume that M∞
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Let q ∗ ∈ G be such that q ∗ ≤ p ∧ q ∗ ⊩ ¬ϕµ (n).
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q ∗ satisfies almost every condition of D.
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However, it is not necessarily true that ∗ ran(hq (s)) ⊆ (ων + ω 2 + 2ω) ∪ {∞}. Let retagging q of q ∗ as follows. Then q is a witness of d ∈ D.
▶
∗
1. T q = T q ∗ 2. f q = f q { 3. hq (s) =
∗
hq (s) ∞
∗
(if hq (s) < ων + ω 2 + 2ω) (otherwise)
G M∞ ⊨ Π11 -SEP0 IV
¬ψ(n) → n ∈ /D ▶
Assume that ¬ψ(n) ∧ n ∈ D and we prove it by contradiction.
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G ⊨ ¬ψ µ (n), there exists r ≤ p s.t. r ⊩ ¬ψ µ (n). Since M∞
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G ⊨ n ∈ D, there exists q ≤ p such that q ⊩ ¬ϕµ (n). Since M∞
p ⊩ ψ µ (n) ∨ ϕµ (n) u
)
q ⊩ ¬ϕµ (n)
r ⊩ ¬ψ µ (n)
q ∗ ⊩ ¬ϕµ (n)
Ret
s∗
Ret
r ∗ ⊩ ¬ψ µ (n)
Let q ∗ ≤ q, r ∗ ≤ r , s ∗ ≤ p be such that Ret(ων + ω 2 + ω, F , q ∗ , s ∗ ) ∧ Ret(ων + ω 2 + ω, F , r ∗ , s ∗ ). Then s ∗ ⊩ ¬ϕµ (n) ∧ ¬ψ µ (n) holds. It is contradiction. Q.E.D.
Bibliography ▶
Antonio Montalb´an “Indecomposable linear ordering and hyperarithmetic analysis” Journal of Mathematical Logic, Vol.6, No.1, 89-120 2006.
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Antonio Montalb´an “On the Π11 -separation principle” Mathematical Logic Quarterly 2008.
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Itay Neeman “The strength of Jullien’s indecomposability theorem” Journal of Mathematical Logic 2008.
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Chris J. Conidis “Comparing theorems of hyperarithmetic analysis with the arithmetical Bolzano-Weierstrass theorem” to appear.
Thank you for your attention.
