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Strongly unbounded colorings
(joint work with Assaf Rinot)
Chris Lambie-Hanson
Department of Mathematics and Applied Mathematics
Virginia Commonwealth University
Kobe Set Theory Workshop
on the occasion of Saka´
e Fuchino’s retirement
11 March 2021
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Prologue
Productivity of chain conditions
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Knaster property
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Knaster property
Drawing by
Leon Jesmanowicz
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Knaster property
Drawing by
Leon Jesmanowicz
Throughout the talk, κ denotes a regular
uncountable cardinal.
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Knaster property
Drawing by
Leon Jesmanowicz
Throughout the talk, κ denotes a regular
uncountable cardinal. Recall that a poset
P is κ-Knaster if, whenever, A ∈ [P]κ,
there is B ∈ [A]κ consisting of pairwise
compatible conditions.
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Knaster property
Drawing by
Leon Jesmanowicz
Throughout the talk, κ denotes a regular
uncountable cardinal. Recall that a poset
P is κ-Knaster if, whenever, A ∈ [P]κ,
there is B ∈ [A]κ consisting of pairwise
compatible conditions.
The κ-Knaster property is a strengthening
of the κ-cc.
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Knaster property
Drawing by
Leon Jesmanowicz
Throughout the talk, κ denotes a regular
uncountable cardinal. Recall that a poset
P is κ-Knaster if, whenever, A ∈ [P]κ,
there is B ∈ [A]κ consisting of pairwise
compatible conditions.
The κ-Knaster property is a strengthening
of the κ-cc.
In contrast to the κ-cc, the κ-Knaster
property is always productive: if P and Q
are κ-Knaster, then P × Q are κ-Knaster.
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Infinite productivity
For an infinite cardinal θ, we say that the κ-Knaster property is
θ-productive if, whenever {Pi | i < θ} are all κ-Knaster, the
full-support product i<θ Pi is κ-Knaster.
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Infinite productivity
For an infinite cardinal θ, we say that the κ-Knaster property is
θ-productive if, whenever {Pi | i < θ} are all κ-Knaster, the
full-support product i<θ Pi is κ-Knaster. It’s not hard to show
that, if κ is weakly compact, then the κ-Knaster property is
<κ-productive.
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Infinite productivity
For an infinite cardinal θ, we say that the κ-Knaster property is
θ-productive if, whenever {Pi | i < θ} are all κ-Knaster, the
full-support product i<θ Pi is κ-Knaster. It’s not hard to show
that, if κ is weakly compact, then the κ-Knaster property is
<κ-productive.
Theorem (Cox-L¨
ucke, ’17)
Assuming the consistency of a weakly compact cardinal, there is
consistently an inaccessible cardinal κ that is not weakly compact
for which the κ-Knaster property is <κ-productive.
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Infinite productivity
For an infinite cardinal θ, we say that the κ-Knaster property is
θ-productive if, whenever {Pi | i < θ} are all κ-Knaster, the
full-support product i<θ Pi is κ-Knaster. It’s not hard to show
that, if κ is weakly compact, then the κ-Knaster property is
<κ-productive.
Theorem (Cox-L¨
ucke, ’17)
Assuming the consistency of a weakly compact cardinal, there is
consistently an inaccessible cardinal κ that is not weakly compact
for which the κ-Knaster property is <κ-productive.
Theorem (LH-L¨
ucke, ’18)
If the κ-Knaster property is ℵ0-productive, then κ is weakly
compact in L.
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Accessible cardinals
The results of the previous slide left open the question of whether
the κ-Knaster property can consistently be infinitely productive
for some accessible cardinal κ, e.g., κ = ℵ2 or κ = ℵω+1.
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Accessible cardinals
The results of the previous slide left open the question of whether
the κ-Knaster property can consistently be infinitely productive
for some accessible cardinal κ, e.g., κ = ℵ2 or κ = ℵω+1.
Theorem (LH-Rinot, [1])
Suppose that κ is a successor cardinal. Then there is a κ-Knaster
poset P such that Pℵ0 is not κ-cc.
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Accessible cardinals
The results of the previous slide left open the question of whether
the κ-Knaster property can consistently be infinitely productive
for some accessible cardinal κ, e.g., κ = ℵ2 or κ = ℵω+1.
Theorem (LH-Rinot, [1])
Suppose that κ is a successor cardinal. Then there is a κ-Knaster
poset P such that Pℵ0 is not κ-cc.
The proof of this theorem involved colorings c : [κ]2 → ω with
strong unboundedness properties and initiated a systematic
investigation of such colorings, their variations, and their
applications.
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References
[1] Chris Lambie-Hanson and Assaf Rinot, Knaster and friends I:
Closed colorings and precalibers, Algebra Universalis 79
(2018), no. 4, Art. 90, 39. MR 3878671
[2] , Knaster and friends II: The C-sequence number, J.
Math. Log. 21 (2021), no. 01, 2150002.
[3] , Knaster and friends III: Subadditive colorings,
(2021), In preparation.
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Chapter 1
Strongly unbounded colorings
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U(κ, µ, θ, χ)
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U(κ, µ, θ, χ)
Definition
U(κ, µ, θ, χ) asserts the existence of a coloring c : [κ]2 → θ such
that,
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U(κ, µ, θ, χ)
Definition
U(κ, µ, θ, χ) asserts the existence of a coloring c : [κ]2 → θ such
that, for every χ < χ and A ⊆ [κ]χ consisting of κ-many
pairwise disjoint sets,
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U(κ, µ, θ, χ)
Definition
U(κ, µ, θ, χ) asserts the existence of a coloring c : [κ]2 → θ such
that, for every χ < χ and A ⊆ [κ]χ consisting of κ-many
pairwise disjoint sets, and for every color i < θ,
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U(κ, µ, θ, χ)
Definition
U(κ, µ, θ, χ) asserts the existence of a coloring c : [κ]2 → θ such
that, for every χ < χ and A ⊆ [κ]χ consisting of κ-many
pairwise disjoint sets, and for every color i < θ, there is B ⊆ A
of size µ such that, for all a < b, both from B, we have
min(c[a × b]) > i.
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U(κ, µ, θ, χ)
Definition
U(κ, µ, θ, χ) asserts the existence of a coloring c : [κ]2 → θ such
that, for every χ < χ and A ⊆ [κ]χ consisting of κ-many
pairwise disjoint sets, and for every color i < θ, there is B ⊆ A
of size µ such that, for all a < b, both from B, we have
min(c[a × b]) > i.
U(κ, µ, θ, χ) can be seen as asserting a strong failure of Ramsey’s
theorem at κ.
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U(κ, µ, θ, χ)
Definition
U(κ, µ, θ, χ) asserts the existence of a coloring c : [κ]2 → θ such
that, for every χ < χ and A ⊆ [κ]χ consisting of κ-many
pairwise disjoint sets, and for every color i < θ, there is B ⊆ A
of size µ such that, for all a < b, both from B, we have
min(c[a × b]) > i.
U(κ, µ, θ, χ) can be seen as asserting a strong failure of Ramsey’s
theorem at κ.
Note that, for all µ ≤ µ and χ ≤ χ, U(κ, µ, θ, χ) implies
U(κ, µ , θ, χ ), but there is no such obvious monotonicity in the
third coordinate.
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Failure of infinite productivity
Lemma
Suppose that θ ≤ χ < κ are infinite, regular cardinals, κ is
(<χ)-inaccessible, and U(κ, κ, θ, χ) holds. Then there is a poset
P such that
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Failure of infinite productivity
Lemma
Suppose that θ ≤ χ < κ are infinite, regular cardinals, κ is
(<χ)-inaccessible, and U(κ, κ, θ, χ) holds. Then there is a poset
P such that
• P is χ-directed closed;
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Failure of infinite productivity
Lemma
Suppose that θ ≤ χ < κ are infinite, regular cardinals, κ is
(<χ)-inaccessible, and U(κ, κ, θ, χ) holds. Then there is a poset
P such that
• P is χ-directed closed;
• Pτ is κ-Knaster for all τ < θ;
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Failure of infinite productivity
Lemma
Suppose that θ ≤ χ < κ are infinite, regular cardinals, κ is
(<χ)-inaccessible, and U(κ, κ, θ, χ) holds. Then there is a poset
P such that
• P is χ-directed closed;
• Pτ is κ-Knaster for all τ < θ;
• Pθ is not κ-c.c.
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Failure of infinite productivity
Lemma
Suppose that θ ≤ χ < κ are infinite, regular cardinals, κ is
(<χ)-inaccessible, and U(κ, κ, θ, χ) holds. Then there is a poset
P such that
• P is χ-directed closed;
• Pτ is κ-Knaster for all τ < θ;
• Pθ is not κ-c.c.
Sketch of proof.
Let c : [κ]2 → θ witness U(κ, κ, θ, χ).
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Failure of infinite productivity
Lemma
Suppose that θ ≤ χ < κ are infinite, regular cardinals, κ is
(<χ)-inaccessible, and U(κ, κ, θ, χ) holds. Then there is a poset
P such that
• P is χ-directed closed;
• Pτ is κ-Knaster for all τ < θ;
• Pθ is not κ-c.c.
Sketch of proof.
Let c : [κ]2 → θ witness U(κ, κ, θ, χ). For all i < θ, let Pi be the
poset whose conditions are all sets x ∈ [κ]<χ such that
min(c“[x]2) > i, ordered by reverse inclusion.
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Failure of infinite productivity
Lemma
Suppose that θ ≤ χ < κ are infinite, regular cardinals, κ is
(<χ)-inaccessible, and U(κ, κ, θ, χ) holds. Then there is a poset
P such that
• P is χ-directed closed;
• Pτ is κ-Knaster for all τ < θ;
• Pθ is not κ-c.c.
Sketch of proof.
Let c : [κ]2 → θ witness U(κ, κ, θ, χ). For all i < θ, let Pi be the
poset whose conditions are all sets x ∈ [κ]<χ such that
min(c“[x]2) > i, ordered by reverse inclusion. Let P be the lottery
sum i<θ Pi .
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Failure of infinite productivity
Lemma
Suppose that θ ≤ χ < κ are infinite, regular cardinals, κ is
(<χ)-inaccessible, and U(κ, κ, θ, χ) holds. Then there is a poset
P such that
• P is χ-directed closed;
• Pτ is κ-Knaster for all τ < θ;
• Pθ is not κ-c.c.
Sketch of proof.
Let c : [κ]2 → θ witness U(κ, κ, θ, χ). For all i < θ, let Pi be the
poset whose conditions are all sets x ∈ [κ]<χ such that
min(c“[x]2) > i, ordered by reverse inclusion. Let P be the lottery
sum i<θ Pi . Now check that P works.
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Further proof sketch?
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Productivity at successor cardinals
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Productivity at successor cardinals
The principle U(· · · ) has been implicit in a variety of previous
work. Particularly notable for us is the following result of
Todorcevic.
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Productivity at successor cardinals
The principle U(· · · ) has been implicit in a variety of previous
work. Particularly notable for us is the following result of
Todorcevic.
Theorem (Todorcevic)
For every infinite cardinal λ, U(λ+, λ+, ℵ0, cf(λ)) holds.
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Productivity at successor cardinals
The principle U(· · · ) has been implicit in a variety of previous
work. Particularly notable for us is the following result of
Todorcevic.
Theorem (Todorcevic)
For every infinite cardinal λ, U(λ+, λ+, ℵ0, cf(λ)) holds.
Corollary
For every successor cardinal κ, the κ-Knaster property fails to be
ℵ0-productive.
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Provable instances of U(κ, µ, θ, χ)
Given a coloring c : [κ]2 → θ, an ordinal β < κ, and a color i < θ,
let Dc
≤i
(β) := {α < β | c(α, β) ≤ i}.
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Provable instances of U(κ, µ, θ, χ)
Given a coloring c : [κ]2 → θ, an ordinal β < κ, and a color i < θ,
let Dc
≤i
(β) := {α < β | c(α, β) ≤ i}. We say that c is closed if
Dc
≤i
(β) is a closed subset of β for all β < κ and i < θ.
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Provable instances of U(κ, µ, θ, χ)
Given a coloring c : [κ]2 → θ, an ordinal β < κ, and a color i < θ,
let Dc
≤i
(β) := {α < β | c(α, β) ≤ i}. We say that c is closed if
Dc
≤i
(β) is a closed subset of β for all β < κ and i < θ.
Theorem (LH-Rinot, [1])
Suppose that θ, χ < κ are regular cardinals. Any one of the
following entails U(κ, κ, θ, χ).
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Provable instances of U(κ, µ, θ, χ)
Given a coloring c : [κ]2 → θ, an ordinal β < κ, and a color i < θ,
let Dc
≤i
(β) := {α < β | c(α, β) ≤ i}. We say that c is closed if
Dc
≤i
(β) is a closed subset of β for all β < κ and i < θ.
Theorem (LH-Rinot, [1])
Suppose that θ, χ < κ are regular cardinals. Any one of the
following entails U(κ, κ, θ, χ).
1 There is a non-reflecting stationary subset of κ ∩ cof(≥ χ).
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Provable instances of U(κ, µ, θ, χ)
Given a coloring c : [κ]2 → θ, an ordinal β < κ, and a color i < θ,
let Dc
≤i
(β) := {α < β | c(α, β) ≤ i}. We say that c is closed if
Dc
≤i
(β) is a closed subset of β for all β < κ and i < θ.
Theorem (LH-Rinot, [1])
Suppose that θ, χ < κ are regular cardinals. Any one of the
following entails U(κ, κ, θ, χ).
1 There is a non-reflecting stationary subset of κ ∩ cof(≥ χ).
2 κ = λ+ and λ is regular.
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Provable instances of U(κ, µ, θ, χ)
Given a coloring c : [κ]2 → θ, an ordinal β < κ, and a color i < θ,
let Dc
≤i
(β) := {α < β | c(α, β) ≤ i}. We say that c is closed if
Dc
≤i
(β) is a closed subset of β for all β < κ and i < θ.
Theorem (LH-Rinot, [1])
Suppose that θ, χ < κ are regular cardinals. Any one of the
following entails U(κ, κ, θ, χ).
1 There is a non-reflecting stationary subset of κ ∩ cof(≥ χ).
2 κ = λ+ and λ is regular.
3 κ = λ+ and cf(λ) = θ.
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Provable instances of U(κ, µ, θ, χ)
Given a coloring c : [κ]2 → θ, an ordinal β < κ, and a color i < θ,
let Dc
≤i
(β) := {α < β | c(α, β) ≤ i}. We say that c is closed if
Dc
≤i
(β) is a closed subset of β for all β < κ and i < θ.
Theorem (LH-Rinot, [1])
Suppose that θ, χ < κ are regular cardinals. Any one of the
following entails U(κ, κ, θ, χ).
1 There is a non-reflecting stationary subset of κ ∩ cof(≥ χ).
2 κ = λ+ and λ is regular.
3 κ = λ+ and cf(λ) = θ.
4 (κ) holds.
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Provable instances of U(κ, µ, θ, χ)
Given a coloring c : [κ]2 → θ, an ordinal β < κ, and a color i < θ,
let Dc
≤i
(β) := {α < β | c(α, β) ≤ i}. We say that c is closed if
Dc
≤i
(β) is a closed subset of β for all β < κ and i < θ.
Theorem (LH-Rinot, [1])
Suppose that θ, χ < κ are regular cardinals. Any one of the
following entails U(κ, κ, θ, χ).
1 There is a non-reflecting stationary subset of κ ∩ cof(≥ χ).
2 κ = λ+ and λ is regular.
3 κ = λ+ and cf(λ) = θ.
4 (κ) holds.
Moreover, in all instances, U(κ, κ, θ, χ) is witnessed by a closed
coloring.
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Consistent failures of U(κ, µ, θ, χ)
Proposition (LH-Rinot, [1])
1 If κ is weakly compact, then U(κ, 2, θ, 2) fails for all θ < κ.
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Consistent failures of U(κ, µ, θ, χ)
Proposition (LH-Rinot, [1])
1 If κ is weakly compact, then U(κ, 2, θ, 2) fails for all θ < κ.
2 If λ is a singular limit of strongly compact cardinals, then
U(λ+, 2, θ, cf(λ)+) fails for all regular θ ∈ λ+ \ {cf(λ)}
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Consistent failures of U(κ, µ, θ, χ)
Proposition (LH-Rinot, [1])
1 If κ is weakly compact, then U(κ, 2, θ, 2) fails for all θ < κ.
2 If λ is a singular limit of strongly compact cardinals, then
U(λ+, 2, θ, cf(λ)+) fails for all regular θ ∈ λ+ \ {cf(λ)}
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
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Consistent failures of U(κ, µ, θ, χ)
Proposition (LH-Rinot, [1])
1 If κ is weakly compact, then U(κ, 2, θ, 2) fails for all θ < κ.
2 If λ is a singular limit of strongly compact cardinals, then
U(λ+, 2, θ, cf(λ)+) fails for all regular θ ∈ λ+ \ {cf(λ)}
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
• There is a forcing extension in which U(κ, κ, ω, ω) fails but κ
is not weakly compact.
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Consistent failures of U(κ, µ, θ, χ)
Proposition (LH-Rinot, [1])
1 If κ is weakly compact, then U(κ, 2, θ, 2) fails for all θ < κ.
2 If λ is a singular limit of strongly compact cardinals, then
U(λ+, 2, θ, cf(λ)+) fails for all regular θ ∈ λ+ \ {cf(λ)}
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
• There is a forcing extension in which U(κ, κ, ω, ω) fails but κ
is not weakly compact.
• For every infinite regular θ < κ, there is a forcing extension in
which U(κ, κ, θ, χ) holds for all χ < κ but U(κ, κ, θ , θ+) fails
for all regular θ = θ.
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Consistent failures of U(κ, µ, θ, χ)
Proposition (LH-Rinot, [1])
1 If κ is weakly compact, then U(κ, 2, θ, 2) fails for all θ < κ.
2 If λ is a singular limit of strongly compact cardinals, then
U(λ+, 2, θ, cf(λ)+) fails for all regular θ ∈ λ+ \ {cf(λ)}
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
• There is a forcing extension in which U(κ, κ, ω, ω) fails but κ
is not weakly compact.
• For every infinite regular θ < κ, there is a forcing extension in
which U(κ, κ, θ, χ) holds for all χ < κ but U(κ, κ, θ , θ+) fails
for all regular θ = θ.
2 If there is a supercompact cardinal, then there is a forcing
extension in which U(ℵω+1, 2, ℵk , ℵ1) fails for all 1 ≤ k < ω.
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Chapter 3
The C-sequence number
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C-sequences and weak
compactness
Definition
A C-sequence over κ is a sequence Cα | α < κ such that
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C-sequences and weak
compactness
Definition
A C-sequence over κ is a sequence Cα | α < κ such that
• Cα is a club in α for all α ∈ lim(κ);
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C-sequences and weak
compactness
Definition
A C-sequence over κ is a sequence Cα | α < κ such that
• Cα is a club in α for all α ∈ lim(κ);
• Cα+1 = {α} for all α < κ.
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C-sequences and weak
compactness
Definition
A C-sequence over κ is a sequence Cα | α < κ such that
• Cα is a club in α for all α ∈ lim(κ);
• Cα+1 = {α} for all α < κ.
Theorem (Todorcevic)
For every regular, uncountable cardinal κ, the following are
equivalent.
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C-sequences and weak
compactness
Definition
A C-sequence over κ is a sequence Cα | α < κ such that
• Cα is a club in α for all α ∈ lim(κ);
• Cα+1 = {α} for all α < κ.
Theorem (Todorcevic)
For every regular, uncountable cardinal κ, the following are
equivalent.
1 κ is weakly compact.
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C-sequences and weak
compactness
Definition
A C-sequence over κ is a sequence Cα | α < κ such that
• Cα is a club in α for all α ∈ lim(κ);
• Cα+1 = {α} for all α < κ.
Theorem (Todorcevic)
For every regular, uncountable cardinal κ, the following are
equivalent.
1 κ is weakly compact.
2 For every C-sequence Cα | α < κ , there is an unbounded
D ⊆ κ such that, for every α < κ, there is β < κ for which
D ∩ α = Cβ ∩ α.
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The C-sequence number
Motivated by Todorcevic’s characterization of weak compactness,
we introduce the notion of the C-sequence number of a cardinal
κ (denoted χ(κ)), which can be seen as measuring how far away
κ is from being weakly compact.
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The C-sequence number
Motivated by Todorcevic’s characterization of weak compactness,
we introduce the notion of the C-sequence number of a cardinal
κ (denoted χ(κ)), which can be seen as measuring how far away
κ is from being weakly compact.
Definition (The C-sequence number)
For every regular, uncountable cardinal κ, let χ(κ) = 0 if κ is
weakly compact.
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The C-sequence number
Motivated by Todorcevic’s characterization of weak compactness,
we introduce the notion of the C-sequence number of a cardinal
κ (denoted χ(κ)), which can be seen as measuring how far away
κ is from being weakly compact.
Definition (The C-sequence number)
For every regular, uncountable cardinal κ, let χ(κ) = 0 if κ is
weakly compact. Otherwise, let χ(κ) be the least cardinal χ such
that, for every C-sequence Cα | α < κ , there is an unbounded
D ⊆ κ such that, for every α < κ, there is b ∈ [κ]χ for which
D ∩ α ⊆ β∈b
Cβ.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ. In
particular, if λ is regular, then χ(λ+) = λ.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ. In
particular, if λ is regular, then χ(λ+) = λ.
3 If λ is a singular limit of strongly compact cardinals, then
χ(λ+) = cf(λ).
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ. In
particular, if λ is regular, then χ(λ+) = λ.
3 If λ is a singular limit of strongly compact cardinals, then
χ(λ+) = cf(λ).
4 If χ(κ) > 1, then χ(κ) ≥ ℵ0.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ. In
particular, if λ is regular, then χ(λ+) = λ.
3 If λ is a singular limit of strongly compact cardinals, then
χ(λ+) = cf(λ).
4 If χ(κ) > 1, then χ(κ) ≥ ℵ0.
5 If χ(κ) = 1, then κ is greatly Mahlo.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ. In
particular, if λ is regular, then χ(λ+) = λ.
3 If λ is a singular limit of strongly compact cardinals, then
χ(λ+) = cf(λ).
4 If χ(κ) > 1, then χ(κ) ≥ ℵ0.
5 If χ(κ) = 1, then κ is greatly Mahlo.
6 Every stationary subset of κ ∩ cof(> χ(κ)) reflects.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ. In
particular, if λ is regular, then χ(λ+) = λ.
3 If λ is a singular limit of strongly compact cardinals, then
χ(λ+) = cf(λ).
4 If χ(κ) > 1, then χ(κ) ≥ ℵ0.
5 If χ(κ) = 1, then κ is greatly Mahlo.
6 Every stationary subset of κ ∩ cof(> χ(κ)) reflects.
7 If µ < κ and (κ, < µ) holds, then χ(κ) ≥ ℵ0.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ. In
particular, if λ is regular, then χ(λ+) = λ.
3 If λ is a singular limit of strongly compact cardinals, then
χ(λ+) = cf(λ).
4 If χ(κ) > 1, then χ(κ) ≥ ℵ0.
5 If χ(κ) = 1, then κ is greatly Mahlo.
6 Every stationary subset of κ ∩ cof(> χ(κ)) reflects.
7 If µ < κ and (κ, < µ) holds, then χ(κ) ≥ ℵ0.
8 If (κ, < ω) holds, then χ(κ) is as large as possible.
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Some basic facts
Proposition
The C-sequence number satisfies the following properties.
1 χ(κ) ≤ κ.
2 For every infinite cardinal λ, cf(λ) ≤ χ(λ+) ≤ λ. In
particular, if λ is regular, then χ(λ+) = λ.
3 If λ is a singular limit of strongly compact cardinals, then
χ(λ+) = cf(λ).
4 If χ(κ) > 1, then χ(κ) ≥ ℵ0.
5 If χ(κ) = 1, then κ is greatly Mahlo.
6 Every stationary subset of κ ∩ cof(> χ(κ)) reflects.
7 If µ < κ and (κ, < µ) holds, then χ(κ) ≥ ℵ0.
8 If (κ, < ω) holds, then χ(κ) is as large as possible. I.e.,
χ(κ) = sup(κ ∩ Reg).
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Some consistency results
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
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Some consistency results
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
• There is a forcing extension in which χ(κ) = 1.
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Some consistency results
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
• There is a forcing extension in which χ(κ) = 1.
• For every infinite regular θ < κ, there is a forcing extension in
which χ(κ) = θ.
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Some consistency results
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
• There is a forcing extension in which χ(κ) = 1.
• For every infinite regular θ < κ, there is a forcing extension in
which χ(κ) = θ.
2 If there is a supercompact cardinal, then there is a forcing
extension in which χ(ℵω+1) = ℵ0.
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Some consistency results
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
• There is a forcing extension in which χ(κ) = 1.
• For every infinite regular θ < κ, there is a forcing extension in
which χ(κ) = θ.
2 If there is a supercompact cardinal, then there is a forcing
extension in which χ(ℵω+1) = ℵ0.
Note the similarity to the previous consistency result about
failures of U(κ, µ, θ, χ).
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Some consistency results
Theorem (LH-Rinot, [2])
1 Suppose that κ is weakly compact.
• There is a forcing extension in which U(κ, κ, ω, ω) fails but κ
is not weakly compact.
• For every infinite regular θ < κ, there is a forcing extension in
which U(κ, κ, θ, χ) holds for all χ < κ but U(κ, κ, θ , θ+) fails
for all regular θ = θ.
2 If there is a supercompact cardinal, then there is a forcing
extension in which U(ℵω+1, 2, ℵk , ℵ1) fails for all 1 ≤ k < ω.
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χ(κ) and closed colorings
Theorem (LH-Rinot, [2])
Suppose that ℵ0 ≤ χ ≤ θ = cf(θ) < κ,
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χ(κ) and closed colorings
Theorem (LH-Rinot, [2])
Suppose that ℵ0 ≤ χ ≤ θ = cf(θ) < κ, and suppose that there is
a closed witness to U(κ, 2, θ, χ).
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χ(κ) and closed colorings
Theorem (LH-Rinot, [2])
Suppose that ℵ0 ≤ χ ≤ θ = cf(θ) < κ, and suppose that there is
a closed witness to U(κ, 2, θ, χ). Then χ(κ) ≥ χ.
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χ(κ) and closed colorings
Theorem (LH-Rinot, [2])
Suppose that ℵ0 ≤ χ ≤ θ = cf(θ) < κ, and suppose that there is
a closed witness to U(κ, 2, θ, χ). Then χ(κ) ≥ χ.
Theorem (LH-Rinot, [2])
There is a closed witness to U(κ, κ, ω, χ(κ)).
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χ(κ) and closed colorings
Theorem (LH-Rinot, [2])
Suppose that ℵ0 ≤ χ ≤ θ = cf(θ) < κ, and suppose that there is
a closed witness to U(κ, 2, θ, χ). Then χ(κ) ≥ χ.
Theorem (LH-Rinot, [2])
There is a closed witness to U(κ, κ, ω, χ(κ)).
Theorem (LH-Rinot, [2])
There is a closed witness to U(κ, κ, χ(κ), χ(κ)).
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χ(κ) and closed colorings
Theorem (LH-Rinot, [2])
Suppose that ℵ0 ≤ χ ≤ θ = cf(θ) < κ, and suppose that there is
a closed witness to U(κ, 2, θ, χ). Then χ(κ) ≥ χ.
Theorem (LH-Rinot, [2])
There is a closed witness to U(κ, κ, ω, χ(κ)).
Theorem (LH-Rinot, [2])
There is a closed witness to U(κ, κ, χ(κ), χ(κ)).
Theorem (LH-Rinot, [2])
For infinite regular θ < κ, TFAE:
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χ(κ) and closed colorings
Theorem (LH-Rinot, [2])
Suppose that ℵ0 ≤ χ ≤ θ = cf(θ) < κ, and suppose that there is
a closed witness to U(κ, 2, θ, χ). Then χ(κ) ≥ χ.
Theorem (LH-Rinot, [2])
There is a closed witness to U(κ, κ, ω, χ(κ)).
Theorem (LH-Rinot, [2])
There is a closed witness to U(κ, κ, χ(κ), χ(κ)).
Theorem (LH-Rinot, [2])
For infinite regular θ < κ, TFAE:
1 there is a C-sequence C over κ such that χ(C) = θ;
2 there is a closed witness to U(κ, κ, θ, θ).
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Some conjectures
Conjectures
1 For all regular θ < κ, U(κ, κ, θ, χ(κ)) holds.
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Some conjectures
Conjectures
1 For all regular θ < κ, U(κ, κ, θ, χ(κ)) holds.
2 (Slightly less ambitious) For all regular θ ≤ χ(κ),
U(κ, κ, θ, χ(κ)) holds.
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Some conjectures
Conjectures
1 For all regular θ < κ, U(κ, κ, θ, χ(κ)) holds.
2 (Slightly less ambitious) For all regular θ ≤ χ(κ),
U(κ, κ, θ, χ(κ)) holds.
3 For all regular, uncountable κ, TFAE:
• χ(κ) ≤ 1;
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Some conjectures
Conjectures
1 For all regular θ < κ, U(κ, κ, θ, χ(κ)) holds.
2 (Slightly less ambitious) For all regular θ ≤ χ(κ),
U(κ, κ, θ, χ(κ)) holds.
3 For all regular, uncountable κ, TFAE:
• χ(κ) ≤ 1;
• the κ-Knaster property is ℵ0
-productive.
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Some conjectures
Conjectures
1 For all regular θ < κ, U(κ, κ, θ, χ(κ)) holds.
2 (Slightly less ambitious) For all regular θ ≤ χ(κ),
U(κ, κ, θ, χ(κ)) holds.
3 For all regular, uncountable κ, TFAE:
• χ(κ) ≤ 1;
• the κ-Knaster property is ℵ0
-productive.
4 For all regular, uncountable κ, TFAE:
• κ is weakly compact;
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Some conjectures
Conjectures
1 For all regular θ < κ, U(κ, κ, θ, χ(κ)) holds.
2 (Slightly less ambitious) For all regular θ ≤ χ(κ),
U(κ, κ, θ, χ(κ)) holds.
3 For all regular, uncountable κ, TFAE:
• χ(κ) ≤ 1;
• the κ-Knaster property is ℵ0
-productive.
4 For all regular, uncountable κ, TFAE:
• κ is weakly compact;
• U(κ, 2, θ, 2) fails for all infinite regular θ < κ;
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Some conjectures
Conjectures
1 For all regular θ < κ, U(κ, κ, θ, χ(κ)) holds.
2 (Slightly less ambitious) For all regular θ ≤ χ(κ),
U(κ, κ, θ, χ(κ)) holds.
3 For all regular, uncountable κ, TFAE:
• χ(κ) ≤ 1;
• the κ-Knaster property is ℵ0
-productive.
4 For all regular, uncountable κ, TFAE:
• κ is weakly compact;
• U(κ, 2, θ, 2) fails for all infinite regular θ < κ;
• U(κ, 2, ℵ0
, 2) fails.
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Chapter 3
Subadditive colorings
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Subadditivity
Definition
A coloring c : [κ]2 → θ is said to be subadditive if, for all
α < β < γ < κ,
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Subadditivity
Definition
A coloring c : [κ]2 → θ is said to be subadditive if, for all
α < β < γ < κ,
1 c(α, γ) ≤ max{c(α, β), c(β, γ)};
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Subadditivity
Definition
A coloring c : [κ]2 → θ is said to be subadditive if, for all
α < β < γ < κ,
1 c(α, γ) ≤ max{c(α, β), c(β, γ)};
2 c(α, β) ≤ max{c(α, γ), c(β, γ)}.
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Subadditivity
Definition
A coloring c : [κ]2 → θ is said to be subadditive if, for all
α < β < γ < κ,
1 c(α, γ) ≤ max{c(α, β), c(β, γ)};
2 c(α, β) ≤ max{c(α, γ), c(β, γ)}.
We say that c is subadditive of the first (resp. second) kind if it
satisfies 1 (resp. 2).
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Subadditivity
Definition
A coloring c : [κ]2 → θ is said to be subadditive if, for all
α < β < γ < κ,
1 c(α, γ) ≤ max{c(α, β), c(β, γ)};
2 c(α, β) ≤ max{c(α, γ), c(β, γ)}.
We say that c is subadditive of the first (resp. second) kind if it
satisfies 1 (resp. 2).
Subadditive witnesses to U(κ, µ, θ, χ) are particularly useful in
applications.
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Subadditivity
Definition
A coloring c : [κ]2 → θ is said to be subadditive if, for all
α < β < γ < κ,
1 c(α, γ) ≤ max{c(α, β), c(β, γ)};
2 c(α, β) ≤ max{c(α, γ), c(β, γ)}.
We say that c is subadditive of the first (resp. second) kind if it
satisfies 1 (resp. 2).
Subadditive witnesses to U(κ, µ, θ, χ) are particularly useful in
applications.
Theorem (LH-Rinot, [3])
Suppose that (κ) holds. Then, for every regular θ < κ, there is
a subadditive witness to U(κ, κ, θ, χ) for all χ < κ.
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Consistent negative results
Subadditive witnesses to U(κ, λ, θ, χ) do not necessarily exist,
though, even when U(κ, λ, θ, χ) holds.
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Consistent negative results
Subadditive witnesses to U(κ, λ, θ, χ) do not necessarily exist,
though, even when U(κ, λ, θ, χ) holds.
Theorem (LH-Rinot, [3])
If the P-ideal dichotomy holds, then, for every regular κ > ℵ1,
there is no subadditive witness to U(κ, 2, ℵ0, 2).
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Consistent negative results
Subadditive witnesses to U(κ, λ, θ, χ) do not necessarily exist,
though, even when U(κ, λ, θ, χ) holds.
Theorem (LH-Rinot, [3])
If the P-ideal dichotomy holds, then, for every regular κ > ℵ1,
there is no subadditive witness to U(κ, 2, ℵ0, 2).
Theorem (Shani, ’16, LH, ’17)
Relative to the consistency of large cardinals, there are
consistently regular cardinals κ for which (κ, 2) holds but, for
every regular θ < κ, there is no subadditive witness to
U(κ, 2, θ, 2).
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Consistent negative results
Subadditive witnesses to U(κ, λ, θ, χ) do not necessarily exist,
though, even when U(κ, λ, θ, χ) holds.
Theorem (LH-Rinot, [3])
If the P-ideal dichotomy holds, then, for every regular κ > ℵ1,
there is no subadditive witness to U(κ, 2, ℵ0, 2).
Theorem (Shani, ’16, LH, ’17)
Relative to the consistency of large cardinals, there are
consistently regular cardinals κ for which (κ, 2) holds but, for
every regular θ < κ, there is no subadditive witness to
U(κ, 2, θ, 2).
(Here κ can be arranged to be a successor of a regular cardinal, a
successor of a singular cardinal, or an inaccessible cardinal.)
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Tightness of Gδ-modifications
Definition
Let X be a topological space.
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Tightness of Gδ-modifications
Definition
Let X be a topological space.
1 The tightness of X, denoted t(X), is the least cardinal κ
such that, for every T ⊆ X and every x ∈ cl(T), there is
T ∈ [T]≤κ such that x ∈ cl(T ).
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Tightness of Gδ-modifications
Definition
Let X be a topological space.
1 The tightness of X, denoted t(X), is the least cardinal κ
such that, for every T ⊆ X and every x ∈ cl(T), there is
T ∈ [T]≤κ such that x ∈ cl(T ).
2 X is Fr´
echet if for every T ⊆ X and every x ∈ cl(T), there is
a (countable) sequence of elements of T converging to x.
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Tightness of Gδ-modifications
Definition
Let X be a topological space.
1 The tightness of X, denoted t(X), is the least cardinal κ
such that, for every T ⊆ X and every x ∈ cl(T), there is
T ∈ [T]≤κ such that x ∈ cl(T ).
2 X is Fr´
echet if for every T ⊆ X and every x ∈ cl(T), there is
a (countable) sequence of elements of T converging to x.
Note that, if X is Fr´
echet, then t(X) ≤ ℵ0.
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Tightness of Gδ-modifications
Definition
Let X be a topological space.
1 The tightness of X, denoted t(X), is the least cardinal κ
such that, for every T ⊆ X and every x ∈ cl(T), there is
T ∈ [T]≤κ such that x ∈ cl(T ).
2 X is Fr´
echet if for every T ⊆ X and every x ∈ cl(T), there is
a (countable) sequence of elements of T converging to x.
Note that, if X is Fr´
echet, then t(X) ≤ ℵ0.
3 The Gδ-modification of X, denoted Xδ, is the space with the
same underlying set whose topology is generated by the Gδ
sets of X.
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Tightness of Gδ-modifications
Definition
Let X be a topological space.
1 The tightness of X, denoted t(X), is the least cardinal κ
such that, for every T ⊆ X and every x ∈ cl(T), there is
T ∈ [T]≤κ such that x ∈ cl(T ).
2 X is Fr´
echet if for every T ⊆ X and every x ∈ cl(T), there is
a (countable) sequence of elements of T converging to x.
Note that, if X is Fr´
echet, then t(X) ≤ ℵ0.
3 The Gδ-modification of X, denoted Xδ, is the space with the
same underlying set whose topology is generated by the Gδ
sets of X.
Some recent work has been done studying the relationship
between t(X) and t(Xδ). Of particular interest is whether there is
an upper bound on t(Xδ) for countably tight (or stronger) spaces
X.
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Some results
Theorem (Dow-Juh´
asz-Soukup-Szentmikl´
ossy-Weiss, ’19)
If there is a non-reflecting stationary subset of κ ∩ cof(ω), then
there is a Fr´
echet space X such that t(Xδ) = κ.
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Some results
Theorem (Dow-Juh´
asz-Soukup-Szentmikl´
ossy-Weiss, ’19)
If there is a non-reflecting stationary subset of κ ∩ cof(ω), then
there is a Fr´
echet space X such that t(Xδ) = κ.
Theorem (DHSSW, ’19)
If λ is strongly compact and t(X) < λ, then t(Xδ) ≤ λ.
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Some results
Theorem (Dow-Juh´
asz-Soukup-Szentmikl´
ossy-Weiss, ’19)
If there is a non-reflecting stationary subset of κ ∩ cof(ω), then
there is a Fr´
echet space X such that t(Xδ) = κ.
Theorem (DHSSW, ’19)
If λ is strongly compact and t(X) < λ, then t(Xδ) ≤ λ.
Theorem (Chen-Mertens–Szeptycki, ’2X)
If (κ) holds, then there is a Fr´
echet α1-space X such that
t(Xδ) = κ.
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Some results
Theorem (Dow-Juh´
asz-Soukup-Szentmikl´
ossy-Weiss, ’19)
If there is a non-reflecting stationary subset of κ ∩ cof(ω), then
there is a Fr´
echet space X such that t(Xδ) = κ.
Theorem (DHSSW, ’19)
If λ is strongly compact and t(X) < λ, then t(Xδ) ≤ λ.
Theorem (Chen-Mertens–Szeptycki, ’2X)
If there is a subadditive witness to U(κ, 2, ℵ0, 2), then there is a
Fr´
echet α1-space X such that t(Xδ) = κ.
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Some results
Theorem (Dow-Juh´
asz-Soukup-Szentmikl´
ossy-Weiss, ’19)
If there is a non-reflecting stationary subset of κ ∩ cof(ω), then
there is a Fr´
echet space X such that t(Xδ) = κ.
Theorem (DHSSW, ’19)
If λ is strongly compact and t(X) < λ, then t(Xδ) ≤ λ.
Theorem (Chen-Mertens–Szeptycki, ’2X)
If there is a subadditive witness to U(κ, 2, ℵ0, 2), then there is a
Fr´
echet α1-space X such that t(Xδ) = κ.
Theorem (Chen-Mertens–Szeptycki, ’2X)
If PID holds and X is a Fr´
echet α1-space, then t(Xδ) ≤ ℵ1.
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An example from a failure of SCH
Theorem (LH-Rinot, [3])
Suppose that µ is a singular cardinal of countable cofinality and
SCH fails at µ. Then there is a Fr´
echet α1-space X such that
t(Xδ) = µ+.
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An example from a failure of SCH
Theorem (LH-Rinot, [3])
Suppose that µ is a singular cardinal of countable cofinality and
SCH fails at µ. Then there is a Fr´
echet α1-space X such that
t(Xδ) = µ+.
We’ll end by sketching a proof of this theorem.
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An example from a failure of SCH
Theorem (LH-Rinot, [3])
Suppose that µ is a singular cardinal of countable cofinality and
SCH fails at µ. Then there is a Fr´
echet α1-space X such that
t(Xδ) = µ+.
We’ll end by sketching a proof of this theorem.
Lemma (folklore)
For every infinite cardinal µ, there is a witness c to
U(µ+, 2, cf(µ), 2) such that
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An example from a failure of SCH
Theorem (LH-Rinot, [3])
Suppose that µ is a singular cardinal of countable cofinality and
SCH fails at µ. Then there is a Fr´
echet α1-space X such that
t(Xδ) = µ+.
We’ll end by sketching a proof of this theorem.
Lemma (folklore)
For every infinite cardinal µ, there is a witness c to
U(µ+, 2, cf(µ), 2) such that
• c is subadditive of the first kind;
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An example from a failure of SCH
Theorem (LH-Rinot, [3])
Suppose that µ is a singular cardinal of countable cofinality and
SCH fails at µ. Then there is a Fr´
echet α1-space X such that
t(Xδ) = µ+.
We’ll end by sketching a proof of this theorem.
Lemma (folklore)
For every infinite cardinal µ, there is a witness c to
U(µ+, 2, cf(µ), 2) such that
• c is subadditive of the first kind;
• c is locally small, i.e., |Dc
≤i
(β)| < µ for all i < cf(µ) and all
β < µ+.
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Proof sketch
Fix a coloring c : [µ+]2 → ω as in the Lemma.
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Proof sketch
Fix a coloring c : [µ+]2 → ω as in the Lemma. The underlying set
of X will be µ+ ∪ {∞}.
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Proof sketch
Fix a coloring c : [µ+]2 → ω as in the Lemma. The underlying set
of X will be µ+ ∪ {∞}. Every element of µ+ is isolated in X.
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Proof sketch
Fix a coloring c : [µ+]2 → ω as in the Lemma. The underlying set
of X will be µ+ ∪ {∞}. Every element of µ+ is isolated in X.
Basic open neighborhoods of ∞ in X are of the form
Ni,β := {∞} ∪ (µ+ \ Dc
≤i
(β)),
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Proof sketch
Fix a coloring c : [µ+]2 → ω as in the Lemma. The underlying set
of X will be µ+ ∪ {∞}. Every element of µ+ is isolated in X.
Basic open neighborhoods of ∞ in X are of the form
Ni,β := {∞} ∪ (µ+ \ Dc
≤i
(β)),
i.e., the closed subsets of µ+ are precisely the intersections of the
sets Dc
≤i
(β), so, if T ⊆ µ+, then
∞ ∈ cl(T) ⇔ ( ∃(i, β) ∈ ω × µ+)(T ⊆ Dc
≤i
(β))
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X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T).
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X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T). Suppose first that
|T| < µ. For each β > sup(T) + 1, define fβ : T → ω by
fβ(α) := c(α, β).
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X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T). Suppose first that
|T| < µ. For each β > sup(T) + 1, define fβ : T → ω by
fβ(α) := c(α, β). Since µ is strong limit, there are an unbounded
A ⊆ µ+ and a function f : T → ω such that fβ = f for all β ∈ A.
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X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T). Suppose first that
|T| < µ. For each β > sup(T) + 1, define fβ : T → ω by
fβ(α) := c(α, β). Since µ is strong limit, there are an unbounded
A ⊆ µ+ and a function f : T → ω such that fβ = f for all β ∈ A.
Since ∞ ∈ cl(T), for each i < ω, we can fix αi ∈ T such that
f (αi ) ≥ i.
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X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T). Suppose first that
|T| < µ. For each β > sup(T) + 1, define fβ : T → ω by
fβ(α) := c(α, β). Since µ is strong limit, there are an unbounded
A ⊆ µ+ and a function f : T → ω such that fβ = f for all β ∈ A.
Since ∞ ∈ cl(T), for each i < ω, we can fix αi ∈ T such that
f (αi ) ≥ i. Then every Dc
≤i
(β) contains only finitely many of the
αi ’s, so αi | i < ω converges to ∞.
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X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T). Suppose first that
|T| < µ. For each β > sup(T) + 1, define fβ : T → ω by
fβ(α) := c(α, β). Since µ is strong limit, there are an unbounded
A ⊆ µ+ and a function f : T → ω such that fβ = f for all β ∈ A.
Since ∞ ∈ cl(T), for each i < ω, we can fix αi ∈ T such that
f (αi ) ≥ i. Then every Dc
≤i
(β) contains only finitely many of the
αi ’s, so αi | i < ω converges to ∞.
Suppose now that |T| ≥ µ.
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X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T). Suppose first that
|T| < µ. For each β > sup(T) + 1, define fβ : T → ω by
fβ(α) := c(α, β). Since µ is strong limit, there are an unbounded
A ⊆ µ+ and a function f : T → ω such that fβ = f for all β ∈ A.
Since ∞ ∈ cl(T), for each i < ω, we can fix αi ∈ T such that
f (αi ) ≥ i. Then every Dc
≤i
(β) contains only finitely many of the
αi ’s, so αi | i < ω converges to ∞.
Suppose now that |T| ≥ µ. For each (i, β), we have
|Dc
≤i
(β)| < µ, so |[Dc
≤i
(β)]ℵ0 | < µ.
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X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T). Suppose first that
|T| < µ. For each β > sup(T) + 1, define fβ : T → ω by
fβ(α) := c(α, β). Since µ is strong limit, there are an unbounded
A ⊆ µ+ and a function f : T → ω such that fβ = f for all β ∈ A.
Since ∞ ∈ cl(T), for each i < ω, we can fix αi ∈ T such that
f (αi ) ≥ i. Then every Dc
≤i
(β) contains only finitely many of the
αi ’s, so αi | i < ω converges to ∞.
Suppose now that |T| ≥ µ. For each (i, β), we have
|Dc
≤i
(β)| < µ, so |[Dc
≤i
(β)]ℵ0 | < µ. Therefore, the number of
elements of [T]ℵ0 that are contained in some Dc
≤i
(β) is at most
µ+.
Slide 132
Slide 132 text
X is Fr´
echet
Suppose that T ⊆ µ+ and ∞ ∈ cl(T). Suppose first that
|T| < µ. For each β > sup(T) + 1, define fβ : T → ω by
fβ(α) := c(α, β). Since µ is strong limit, there are an unbounded
A ⊆ µ+ and a function f : T → ω such that fβ = f for all β ∈ A.
Since ∞ ∈ cl(T), for each i < ω, we can fix αi ∈ T such that
f (αi ) ≥ i. Then every Dc
≤i
(β) contains only finitely many of the
αi ’s, so αi | i < ω converges to ∞.
Suppose now that |T| ≥ µ. For each (i, β), we have
|Dc
≤i
(β)| < µ, so |[Dc
≤i
(β)]ℵ0 | < µ. Therefore, the number of
elements of [T]ℵ0 that are contained in some Dc
≤i
(β) is at most
µ+. Since SCH fails at µ, |[T]ℵ0 | > µ+, so we can find a
countable subset of T that is not contained in any Dc
≤i
(β), and
proceed as in the previous case.
Slide 133
Slide 133 text
t(Xδ
) = µ+
In X, every closed subset of µ+ is bounded in µ+.
Slide 134
Slide 134 text
t(Xδ
) = µ+
In X, every closed subset of µ+ is bounded in µ+. Since every
closed subset of µ+ in Xδ is contained in a countable union of
closed subsets of µ+ in X, this is also true in Xδ.
Slide 135
Slide 135 text
t(Xδ
) = µ+
In X, every closed subset of µ+ is bounded in µ+. Since every
closed subset of µ+ in Xδ is contained in a countable union of
closed subsets of µ+ in X, this is also true in Xδ. In particular, in
Xδ, ∞ ∈ cl(µ+).
Slide 136
Slide 136 text
t(Xδ
) = µ+
In X, every closed subset of µ+ is bounded in µ+. Since every
closed subset of µ+ in Xδ is contained in a countable union of
closed subsets of µ+ in X, this is also true in Xδ. In particular, in
Xδ, ∞ ∈ cl(µ+).
However, for every β < µ+, we have β = i<ω
Dc
≤i
(β).
Slide 137
Slide 137 text
t(Xδ
) = µ+
In X, every closed subset of µ+ is bounded in µ+. Since every
closed subset of µ+ in Xδ is contained in a countable union of
closed subsets of µ+ in X, this is also true in Xδ. In particular, in
Xδ, ∞ ∈ cl(µ+).
However, for every β < µ+, we have β = i<ω
Dc
≤i
(β). Since
Dc
≤i
(β) is closed in X, β is closed in Xδ.
Slide 138
Slide 138 text
t(Xδ
) = µ+
In X, every closed subset of µ+ is bounded in µ+. Since every
closed subset of µ+ in Xδ is contained in a countable union of
closed subsets of µ+ in X, this is also true in Xδ. In particular, in
Xδ, ∞ ∈ cl(µ+).
However, for every β < µ+, we have β = i<ω
Dc
≤i
(β). Since
Dc
≤i
(β) is closed in X, β is closed in Xδ. Therefore, ∞ is not in
the closure of any bounded subset of µ+ in Xδ.
Slide 139
Slide 139 text
t(Xδ
) = µ+
In X, every closed subset of µ+ is bounded in µ+. Since every
closed subset of µ+ in Xδ is contained in a countable union of
closed subsets of µ+ in X, this is also true in Xδ. In particular, in
Xδ, ∞ ∈ cl(µ+).
However, for every β < µ+, we have β = i<ω
Dc
≤i
(β). Since
Dc
≤i
(β) is closed in X, β is closed in Xδ. Therefore, ∞ is not in
the closure of any bounded subset of µ+ in Xδ. It follows that
t(Xδ) = µ+.
Slide 140
Slide 140 text
All artwork by Vera Moln´
ar.
Slide 141
Slide 141 text
Thank you!