The Almost-Fan Theorem
Wim Veldman
Fifth Workshop on Formal Topology
Mittag-Leffler Institute
June 10, 2015
Finite and almost-finite
For every α in C, Dα := {n|n ∈ N|α(n) = 1}.
A subset A of N is a decidable subset of N: ∃α[A = Dα ].
A ⊆ N is finite: A is a decidable subset of N and
∃n∀m ≥ n[m ∈
/ A].
For every γ in N , Eγ := {n|n ∈ N|∃p[γ(p) = n + 1]}.
A ⊆ N is an enumerable subset of N: ∃γ[A = Eγ ].
Eγ is almost-finite:
∀ζ ∈ N ∃m∃n[m < n ∧ γ ◦ ζ(m) = γ ◦ ζ(n)]
.
Almost-finite × almost-finite
Let s be an element of N.
Then s = hs(0), s(1), . . . s(m − 1)i, and m = length(s).
For all s, t in N: s ∗ t = hs(0), . . . s(m − 1), t(0), . . . , t(n − 1)i.
s ∗ A = {s ∗ t|t ∈ A}.
Let A0 , A1 , A2 , . . . be an infinite sequence of almost-finite
subsets of N such that ∀ζ ∈ [ω]ω ∃n[Aζ(n) = ∅].
S
Then:
hni ∗ An is almost-finite.
n∈N
Spreads, fans and almost-fans
An element β of C is a spread-law:
for all s, β(s) = 0 iff ∃n[β(s ∗ hni) = 0].
Fβ := {α|∀n[β(αn) = 0]}.
A subset F of N is a spread: there exists a spread-law β such
that F = Fβ .
A subset F of N is a fan: there exists a spread-law β such
that F = Fβ and: for each s, {n|β(s ∗ hni) = 0} is finite.
A subset F of N is an almost-fan: there exists a spread-law
β such that F = Fβ and: for each s, {n|β(s ∗ hni) = 0} is
almost-finite.
Fan Theorem and Almost-Fan Theorem
Let B ⊆ N and let F ⊆ N .
B is bar in F: ∀α ∈ F∃n[αn ∈ B].
B is thin: ∀s ∈ B∀s ∈ B[s v t → s = t].
(Almost-)Fan Theorem (Alm)FT:
Let β be a spread-law such that Fβ is an (almost-)fan and let
B be a decidable subset of N that is thin and a bar in F.
Then:
(i) The set of all s such that β(s) = 0 and s ∈ B is
(almost-)finite.
(ii) The set of all s such that β(s) = 0 and ∀t v s[t ∈
/ B] is
(almost-)finite.
Proof by bar induction
P := {s|β(s) 6= 0 ∨ s ∈ B}.
P is a decidable subset of N and a bar in N .
Q := {s| the set {t|s v t ∧ β(t) = 0 ∧ t ∈ B} is
(almost-)finite}.
(Q := {s| the set {t|s v t ∧ β(t) = 0 ∧ ∀u v t[u ∈
/ B]}) is
(almost-)finite}).
P ⊆ Q.
Q is inductive: if, for all n, s ∗ hni ∈ Q, then s ∈ Q.
Conclusion: h i ∈ Q:
the set {t|β(t) = 0 ∧ t ∈ B} is (almost-)finite
(the set {t|β(t) = 0 ∧ ∀u v t[u ∈
/ B]} is (almost-)finite).
AlmFT ⇒ FT
Let β be a fan-law and let B be a decidable subset of N that
is thin and a bar in Fβ .
Assume:
the set {s|β(s) = 0 ∧ ∀t v s[t ∈
/ B]} is almost-finite.
ζ(0) = h i, ζ(n + 1) := the least s such that length(s) = n + 1
and β(s) = 0 and ∀t v s[t ∈
/ B] if such s exists,
ζ(n + 1) := ζ(n) if not.
Find n such that ζ(n) = ζ(n + 1).
Conclude: every s such that β(s) = 0 and length(s) = n + 1
meets B and:
the set {s|β(s) = 0 ∧ ∀t v s[t ∈
/ B]} is finite.
Extension to almost-semi-fans
γ ∈ N is a semi-spread-law if and only if, for each s,
s ∈ Eγ iff ∃n[s ∗ hni ∈ Eγ ].
For each γ, SFγ = {α|∀n[αn ∈ Eγ ]}.
A semi-spread Fγ is an almost-semi-fan if and only if, for
each s, the set {n|s ∗ hni ∈ Eγ } is almost-finite.
AlmFT implies: Let F = SFγ be an almost-semi-fan and let
B be a decidable subset of N that is thin and a bar in F.
Then:
(i) The set {s|s ∈ Eγ ∧ s ∈ B} is almost-finite.
(ii) The set {s|s ∈ Eγ ∧ ∀t v s[t ∈
/ B]} is almost-finite.
Open Induction on C
α < β : ∃n[αn = βn ∧ α(n) < β(n)].
C<α := {β|β ∈ C ∧ β < α}.
Gβ := {α|∃n[β(αn) = 1]}
A subset G of N is open: ∃β ∈ C[G = Gβ ].
G is progressive in C: ∀α ∈ C[C<α ⊆ G → α ∈ G].
OI(C): For all β, if Gβ is progressive in C, then C ⊆ Gβ .
(T. Coquand proved: BID ⇒ OI(C)).
AlmFT ⇒ OI(C)
α < s: ∃n < length(s)[αn = sn ∧ α(n) < s(n)].
Assume: Gβ is progressive in C.
F := {s|s ∈ Bin|∃k∀α < s∃n[β(αn) = 1 ∧ αn < k]}.
F is an enumerable subset of N.
For every s, s ∈ F if and only if s ∗ h0i ∈ F .
F := {α|∀n[αn ∈ F ]}.
F is an almost-semi-fan and ∀α ∈ F∀γ < α[γ ∈ Gβ ].
Conclude: ∀α ∈ F[α ∈ Gβ ] and: Dβ is a bar in F.
Let B be an almost-finite subset of Dβ that is a bar in F.
Assume C ⊆ B, C finite.
One may decide: C is a bar in C or C is not a bar in C.
If not, let α be the <-first element of C that does not meet C ,
Find n such that αn ∈ B. Then αn ∈ B \ C .
Define ζ : N → B. For each n,
if {ζ(0), ζ(1), . . . , ζ(n − 1)} is a bar in C, ζ(n) := ζ(n − 1),
and if not, ζ(n) := µs[s ∈ B ∧ ∀i < n[ζ(i) 6= s]].
Find n such that ζ(n) = ζ(n + 1). Conclude:
{ζ(0), ζ(1), . . . , ζ(n)} is a bar in C and C ⊆ Gβ .
OI(C) ⇒ EnDec?!
Let γ be given. Assume:
∀α ∈ C[(0 ∈
/ Dα ∧ Dα ⊆ Eγ ) → ∃p ∈ Eγ \ Dα ].
QED := 0 ∈ Eγ , that is: ∃i[γ(i) = 1].
G := {α|α ∈ C|∃i[(α(i 0 ) = 0 ∧ γ(i 00 ) = i 0 + 1) ∨ γ(i) = 1]}.
G is progressive in C:
Assume α ∈ C and C<α ⊆ G.
Assume α(k) = 1. Define β such that β(k) = 0 and
∀n 6= k[β(n) = α(n)]. Then: β < α, so β ∈ G. Find i such
that (β(i 0 ) = 0 ∧ γ(i 00 ) = i 0 + 1) ∨ γ(i) = 1.
If i 0 6= k, then α ∈ G. If i 0 = k, k ∈ Eγ .
Conclude: if α(k) = 1, either α ∈ G or k ∈ Eγ .
Define δ: for each k, if α(k) = 1, then δ(k) := µi[(α(i 0 ) =
0 ∧ γ(i 00 ) = i 0 + 1) ∨ γ(i) = k + 1 ∨ γ(i) = 1].
Define α∗ : for each k, α∗ (k) = 1 if and only if α(k) = 1 and
γ δ(k) = k + 1.
Dα∗ ⊆ Eγ . Find p in Eγ \ Dα∗ . Then α∗ (p) = 0, and either:
α(p) = 0 and α ∈ G, or: α(p) = 1 and γ δ(p) 6= p + 1, so
γ δ(p) = 1, and again: α ∈ G.
In any case: α ∈ G.
Using OI(C), we conclude: 1 ∈ G, and: 0 ∈ Eγ .
EnDec?! :
if ∀α ∈ C[(0 ∈
/ Dα ∧ Dα ⊆ Eγ ) → ∃p ∈ Eγ \ Dα ],
then 0 ∈ Eγ .
EnDec?! ⇒ OI([0, 1])
Let H be an open subset of R that is progressive in [0, 1].
S
(qn , rn ).
H=
n∈N
S
H := {q|q ∈ Q ∩ [0, 1]|∃N[ (qn , rn ) covers [0, q]]}.
n≤N
The set H is enumerable.
Let D be a decidable set of rationals such that 0 ∈ D ⊆ H
and 1 ∈
/ D.
Successive bisection:
0 = a0 ≤ a1 ≤ a2 ≤ . . . ≤ . . . x . . . ≤ . . . b2 ≤ b1 ≤ b0 = 1
For each i, ai ∈ D ⊆ H, bi ∈
/ D, therefore [0, x) ⊆ H and
x ∈ H and some bi ∈ H \ D.
Conclude: 1 ∈ H and [0, 1] ⊆ H.
Dedekind’s Theorem, a contraposition
Assume: α = α(0) ≤Q α(1) ≤Q α(2) ≤Q . . ..
Assume: ∀ζ ∈ [ω]ω ∃n[α ◦ ζ(n + 1) −Q α ◦ ζ(n) ≥Q
1
].
2n
Define H := {x|x ∈ [0, ∞)|∃n[x <R α(n)]}.
H is open and progressive in [0, ∞):
Assume: x ≥R 0 and [0, x) ⊆ H.
Define ζ in [ω]ω such that, for each n, α(n) >R x −
Find n such that α ◦ ζ(n + 1) −Q α ◦ ζ(n) >Q
1
.
2n
1
.
2n
Conclude: α(n + 1) >R x and: x ∈ H.
Conclude: 1 ∈ H.
CMC : If α is nondecreasing and positively fails to converge,
then ∀M∃n[α(n) ≥Q M]: α grows beyond all bounds.
An extension
Assume: α is one-to-one and semi-monotone:
for all m, n, p, if m < n and m < p, then:
either: α(m) <Q α(n) and α(m) <Q α(p)
or: α(m) >Q α(n) and α(m) >Q α(p).
Assume also: α positively fails to have a limit point:
∀ζ ∈ [ω]ω ∃n[|α ◦ ζ(n + 1) −Q α ◦ ζ(n)| ≥Q
1
].
2n
QED := ∃n[α(n) >Q 1 ∨ α(n) <Q −1].
(i) ∀m∃n > m[α(n + 1) <Q α(n) ∨ QED].
(ii) ∃ζ ∈ [ω]ω ∀n[α ζ(n) + 1 <Q α ζ(n) ∨ QED].
(iii) ∃ζ ∈ [ω]ω ∀n[α ◦ ζ(n + 1) <Q α ◦ ζ(n) ∨ QED].
(iv) QED, and: α grows beyond all bounds.
Bolzano-Weierstrass?
Assume:
α : N → Q, one-to-one, positively fails to have a limit
point.
QED := ∃n[α(n) >Q 1 ∨ α(n) <Q −1].
Define τ : τ (0) = h i, and, for each n, τ (n + 1) = τ (j) ∗ hni
where j is the largest i ≤ n such that τ (i) ∗ hni is
semi-monotonous.
F := {τ (n) ∗ †k|k, n ∈ N}.
F is enumerable.
For each s in F , #{n|s ∗ hni ∈ F } ≤ length(s) + 2.
F := {α|∀n[αn ∈ F ]}.
Define B := {s|∃i[s(i) = † ∨ s(i) >Q 1 ∨ s(i) <Q −1]}.
B is a bar in F.
Define B 0 := {s|s ∈ B|¬∃t @ s[t ∈ B]}.
B 0 is thin and a bar in F.
B 0 is almost-finite.
Assume C ⊆ B 0 , C finite, and ∀s ∈ C ∃i[s(i) = †].
Then one may find s ∈ B 0 \ C .
Define ζ: for each n,
either: ∀j < n∃i[ ζ(j) (i) = †] and
ζ(n) = µs ∈ B 0 ∀j < n[ζ(j) 6= s],
or: ∃j < n∃i[ ζ(j) (i) >Q 1 ∨ ζ(j) (i) <Q −1] and
ζ(n) = ζ(n − 1).
Find n such that ζ(n) = ζ(n − 1) and conclude: QED.
Conclusion
CBW: Let γ = γ 0 , γ 1 , . . . be an infinite sequence of reals that
positively fails to have a limit point:
∀ζ ∈ [ω]ω ∃n[|γ ζ(n+1) − γ ζ(n) | >R
1
].
2n
Then γ grows beyond all bounds:
∀M∃n[γ n <R −M ∨ γ n >R M].
Approximate Fans
A subset Eγ of N is bounded in number iff, for some k, Eγ
has at most k elements, that is, for all s in [ω]k+2 there
exist i, j ≤ k + 1 such that i < j and γ s(i) = γ s(j) .
A subset F of N is an approximate fan iff there exists a
spread-law β such that F = Fβ and: for each n, the set
{s|β(s) = 0 ∧ length(s) = n} is bounded in number.
A subset F of N is an explicit approximate fan iff there
exist a spread-law β such that F = Fβ and some δ such that,
for each n, the set {s|β(s) = 0 ∧ length(s) = n} has at
most δ(n) elements.
Extending CBW
Using this lemma:
Let F = Fβ be an explicit approximate fan. There exists a
function φ from F into [0, 1] such that
∀m∃n∀α, γ ∈ F[αn 6= γn → |φ|α −R φ|γ| ≥
1
].
2m
one obtains CBW(N ):
Let F = Fβ be an explicit approximate fan.
Let γ = γ 0 , γ 1 , . . . be an infinite sequence in N that
positively fails to have a limit point:
∀ζ ∈ [ω]ω ∃n[γ ζ(n+1) n 6= γ ζ(n) n].
Then γ leaves F: ∃n∃m[β(γ n m) 6= 0].
CBW(N ) ⇒ AFT
Approximate-Fan Theorem AFT: Let F = Fβ be an
explicit approximate fan and let B be a decidable subset of N
that is thin and a bar in F.
Then:
(i) the set {s|β(s) = 0 ∧ s ∈ B} is almost-finite, and
(ii) the set {s|β(s) = 0 ∧ ∀t v s[t ∈
/ B]} is almost-finite.
Assume: ζ(0) < ζ(1) < ζ(2) < . . . and ∀n[β ζ(n) = 0].
QED := ∃n[ζ(n) meets B].
Assume: length ζ(n) ≥ n. Find γ such that, for each n,
if ∀i ≤ n[ζ(n) does not meet B], then ζ(n) @ γ n ∈ F, and,
if ∃i ≤ n[ζ(n) meets B], then γ n = p.
(p = µj[β(hji) 6= 0 ∧ ∀i < n[j 6= γ i (0)]].)
γ positively fails to have a limit point:
Assume η ∈ [ω]ω . Define ε:
if ∀i ≤ n[γ η(i+1) (i + 1) = γ η(i) (i + 1)], then ε(n) = γ η(n) (n),
and,
if not, then ε(n) = the least i such that β(ε(n) ∗ hii) = 0.
Find n such that εn ∈ B.
Conclude: ∃i ≤ n[γ η(i+1) (i + 1) 6= γ η(i) (i + 1)].
Conclude: γ leaves F, and: QED.
AFT extends to approximate semi-fans
A subset F of N is an explicit approximate semi-fan iff
there exist a semi-spread-law γ such that F = SF γ and some
δ such that, for each n, the set {s|s ∈ Eγ ∧ length(s) = n}
has at most δ(n) elements.
AFT extends to explicit approximate semi-fans.
AFT implies OI(C) and CBW.
Ramsey’s Theorem
Dα is k-almost-full: ∀ζ ∈ [ω]ω ∃s ∈ [ω]k [α(ζ ◦ s) = 1].
IRT(k): For all α, β, if Dα , Dβ are k-almost-full, then
Dα ∩ Dβ is k-almost-full.
AFT ⇒ IRT(2):
s in [ω]r is α-2-homeogeneous:
for all m, n, p such that m < n < r and m < p < r :
either: s(m), s(n) ∈ Dα and s(m), s(p) ∈ Dα
or: s(m), s(n) ∈
/ Dα and s(m), s(p) ∈
/ Dα .
Assume Dα , Dβ are 2-almost-full.
QED := ∃s ∈ [ω]2 [s ∈ Dα ∩ Dβ ].
Define τ : τ (0) = h i, and, for each n, τ (n + 1) = τ (j) ∗ hni
where j is the largest i ≤ n such that τ (i) ∗ hni is
α-2-homeogeneous.
Apply RT(1) and AFT: there exist q, r such that t := τ (q)
has length r and there exist m, n, p such that m < n < r and
m < p < r and t(m), t(n) ∈ Dα and t(m), t(p) ∈ Dβ .
Conclude: t(m), t(p) ∈ Dα ∩ Dβ and: QED.
RT(3) ⇒ EnDec?!
Let γ be given such that
∀α ∈ C[(0 ∈
/ Dα ∧ Dα ⊆ Eγ ) → ∃p ∈ Eγ [p ∈
/ Dα ]].
QED := 0 ∈ Eγ , that is: ∃n[γ(n) = 1].
Define α in C: for all s in [ω]3 , α(s) = 1 iff
∃i < s(0)[∀j < s(1)[γ(j) 6= i + 1] ∧ ∃j < s(2)[γ(j) = i + 1]]
or ∃i < s(2)[γ(i) = 1].
Dα is 3-almost-full: let ζ in [ω]ω be given. Define ε in C:
ε(n) = 1 iff 0 < n and either n < ζ(0) and
∃j < ζ(1)[γ(j) = n + 1], or ζ(k) ≤ n < ζ(k + 1) and
∃j < ζ(k + 2)[γ(j) = n + 1].
Then 0 ∈
/ Dε ⊆ Eγ .
Find p in Eγ \ Dε . Find k such that ζ(k) ≤ p < ζ(k + 1).
Find j such that γ(j) = p + 1 and q such that j < ζ(q). Then
ζ(n), ζ(n + 1), ζ(q) ∈ Dα .
Define β in C: for all s in [ω]3 , β(s) = 1 iff
∀i < s(0)[∃j < s(2)[s(j) = i + 1] → ∃j < s(1)[γ(j) = i + 1]].
Dβ is 3-almost-full.
Therefore: Dα ∩ Dβ is 3-almost-full and ∃i[γ(i) = 1] and:
QED.
Using ∀k[RT(k)] one may prove the Paris-Harrington
Theorem.
Unlike FT, AFT is not conservative over Heyting
arithmetic HA.
A survey
BID
AlmFT
AFT ⇔ CBW ⇔ CBW(N ) ⇔ RT(3) ⇔ ∀k[RT(k)]
OI(C) ⇔ EnDec?! ⇔ CMC ⇔ OI([0, 1])
FT ⇔ HB ⇔ Σ01 -Contr(AC)<2
0,0
Markov’s Principle
MP1 : ∀α[¬¬∃n[α(n) = 1] → ∃n[α(n) = 1]]
MP1 ⇒ (∀㬬∃α[Eγ = Dα ] ↔ EnDec?!):
‘→’: Assume 0 ∈
/ Eγ and Dα = Eγ . Then 0 ∈
/ Dα . So 0 ∈ Eγ .
Contradiction. Therefore: ¬¬(0 ∈ Eγ ).
But: ¬¬∃α[Eγ = Dα ]. So ¬¬(0 ∈ Eγ ). So 0 ∈ Eγ .
‘←’: (R. Solovay, J. Moschovakis):
Assume ¬∃α[Eγ = Dα ]. Then
∀α[Dα ⊆ Eγ → ¬¬∃p∃n[γ(p) = n + 1 ∧ α(n) = 0]].
Conclude: N ⊆ Eγ . Contradiction.
From ∀㬬∃α[Eγ = Dα ], one obtains:
For every arithmetical subset P of N:
¬¬∃α∀n[P(n) ↔ α(n) = 1].
and also:
DNS: For every arithmetical subset P of N:
∀n[¬¬P(n)] → ¬¬∀n[P(n)].
Life becomes easier
MP1 ⇔ ∀α[Dα is not-finite → ∀m∃n > m[α(n) 6= 0]].
MP1 ⇒ ∀α[Dα is almost-finite → Dα is not-not-finite].
MP1 ⇔ ∀α[Dα is almost-finite ↔ Dα is not-not-finite].
MP1 + DNS + FT ⇒ AlmFT.
MP1 ⇒
(OI(C) ⇔ AlmFT ⇔ AFT ⇔ DNS ⇔ Σ01 -BID ⇔ . . .)