Download Equivalents of the (Weak) Fan Theorem

Equivalents of the (Weak) Fan Theorem
Constructive Reverse Mathematics
Supervisor: Wim Veldman
Iris Loeb
Radboud Universiteit Nijmegen
Reverse Mathematics
* Different kinds of reverse mathematics:
– classical reverse mathematics
– informal constructive reverse mathematics
– formal constructive reverse mathematics
* Similarity: proving that a mathematical theorem is equivalent
to some axiom
Classical Reverse Mathematics
H.Friedman, Simpson
* Classify mathematical theorems:
Which set existence axioms are needed to prove the theorems
of ordinary mathematics?
* Prove conservativity of the extensions
Informal Constructive Reverse Mathematics
Ishihara, Mandelkern, Schuster
* Classify intuitionistic, recursive constructive and
classical theorems:
Which logical principles are needed to prove these theorems?
Formal Constructive Reverse Mathematics (I)
Kohlenbach, Toftdal
* Classify recursive constructive and classical
theorems in a formal, refined way:
Which logical principles are needed to prove these theorems?
Formal Constructive Reverse Mathematics (II)
van den Brink, Ishihara, Loeb, Veldman
* Classify intuitionistic, recursive constructive and
classical theorems:
Which logical principles and function existence axioms are
needed to prove these theorems?
Language of intuitionistic second order arithmetic:
* first order variables: i, k, m, n, . . .
* second order variables: α, β, γ, . . .
* logical constants: ∧, ¬, ∨, →, ∀, ∃
* mathematical constants: =, +, ·, j, j1, j2, 0, S
* Church’s lambda: λ
Axioms of WKV:
* Equivalence axioms
* Basic axioms
* Pairing axioms
* Function axiom
* Conversion axiom scheme
* Axiom scheme of induction
* Axiom scheme of primitive recursion
* Axiom scheme of unique choice
Fan
β(hi) = 0
∀m[β(m) = 0 ↔ ∃n[β(m ∗ hni) = 0]]
∀m[β(m) = 0 → ∃k∀n[β(m ∗ hni) = 0 → n < k]]
Bar
∀α ∈ β∃n[γ(αn) = 1]
Bar
∀α ∈ β∃n[γ(αn) = 1]
Bar
∀α ∈ β∃n[γ(αn) = 1]
Bar
∀α ∈ β∃n[γ(αn) = 1]
Bar
∀α ∈ β∃n[γ(αn) = 1]
Bar
∀α ∈ β∃n[γ(αn) = 1]
Fan theorem
∀α ∈ β∃n[γ(αn) = 1] →
∃m∀α ∈ β∃n[n ≤ m ∧ γ(αn) = 1]
The interval [0, 1]
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1 3
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1 5
9 11
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3 7
13 15
7
[0, 8 ] [ 16 , 16 ] [ 8 , 4 ] [ 16 , 16 ] [ 4 , 8 ] [ 16 , 16 ] [ 8 , 2 ] [ 16 , 16 ] [ 2 , 8 ] [ 16 , 16 ] [ 8 , 4 ] [ 16 , 16 ] [ 4 , 8 ] [ 16 , 16 ] [ 8 , 1]
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// // // // // // // [0, 41 ]
[ 18 , 38 ]
[ 41 , 12 ]
[ 83 , 58 ]
[ 12 , 43 ]
[ 58 , 78 ]
[ 43 , 1]
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[0, 1]
Definition Heine-Borel
* Every open cover of the interval [0, 1] has a finite sub-cover.
* Let (α)i be a sequence of real numbers (the lower bounds).
Let (β)i be a sequence of real numbers (the upper bounds).
If for all x ∈ [0, 1] there exists an i such that αi < x < β i,
then there exists an m such that for all x ∈ [0, 1] there exists
an i ≤ m such that αi < x < β i.
Heine-Borel vs. Fan Theorem
∀x ∈ [0, 1]∃i[(α)i < x < (β)i] → ∃m∀x ∈ [0, 1]∃i[i ≤ m ∧ (α)i < x < (β)i]
∀α ∈ β∃n[γ(αn) = 1] → ∃m∀α ∈ β∃n[n ≤ m ∧ γ(αn) = 1]
Idea: identify a real number in [0, 1] with a path
through a fan.
Problem: not a 1-1 correspondence.
Fan Theorem implies Heine-Borel
Suppose for all x ∈ [0, 1] there exists an i such that (α)i < x <
(β)i. Then for all x there exist i,n such that the nth approximation of x is between the nth approximation of (α)i and the nth
approximation of (β)i:
(α)i(n)00 < x(n)0 ≤ x(n)00 < (β)i(n)0
Then (by making n bigger): for all x there exist i,n such that
i < n and the nth approximation of x is between the nth approximation of (α)i and the nth approximation of (β)i:
(α)i(n)00 < x(n)0 ≤ x(n)00 < (β)i(n)0(decidable)
So (Fan Theorem): There exists an m such that for all x ∈ [0, 1]
there are n, i s.t. n ≤ m and i < n and
(α)i(n)00 < x(n)0 ≤ x(n)00 < (β)i(n)0
Heine-Borel implies Fan Theorem
Find a subset of [0,1] that does have a 1-1
correspondence with a fan: Cantor’s Discontinuum
Cantor’s Discontinuum
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Definition Compactness Theorem
* A valuation is a function from the basic formulas to {0, 1}.
* A decidable set Γ of formulas is called lying when for every
valuation v there exists an X in Γ such that v ∗(X) = 0.
* Compactness Theorem for Propositional Logic
For every decidable set Γ of formulas: if Γ is lying, then there
is a finite lying subset of Γ.
Compactness Theorem implies Fan Theorem
A2
0
1
0
0
1
1
0
0
1
0
0
1
1
1
A1
A0
Compactness Theorem implies Fan Theorem
A2
0
1
0
0
1
1
0
A0 /\ ~A1 −> A0 /\ ~A0
0
1
0
0
1
1
1
A1
A0
Compactness Theorem implies Fan Theorem
A2
0
1
0
0
1
0
1
0
~A0 /\ ~A1 /\ A2 −> A0 /\ ~A0
1
0
0
1
1
1
A1
A0
Compactness Theorem implies Fan Theorem
Suppose γ is a bar. Define
Γ := {X|∃a ∈ {0, 1}∗[γ(a) = 1 ∧ X =
^
(a)i
{Ai
}→A0∧¬A0]}
i≤lg(n)
Suppose v is a valuation. Determine an n s.t. γ(vn) = 1. Then
v ∗(
^
(vn)i
{Ai
}) = 1 ,but v ∗(A0∧¬A0) = 0
i≤n
(vn)
i }→A ∧¬A ) = 0. So Γ is lying.
So v ∗( i≤n{Ai
0
0
Then (Compactness Theorem): there exists a finite lying subset
Γ0 of Γ. Now we can find a bound for the bar.
V
Equivalents of the (Weak) Fan Theorem
From analysis
* Every continuous function from {0, 1}N to N is u. continuous.
* Every continuous function from [0, 1] to R is u. continuous.
* Every continuous function from [0, 1] to R is bounded.
* Every continuous function from [0, 1] to R is Riemann
integrable.
* Heine-Borel theorem.
Equivalents of the (Weak) Fan Theorem
From logic
* Lindenbaum’s Lemma.
* Completeness Theorem for propositional logic.
* Compactness Theorem for propositional logic.