1 -strongly compact cardinals and cardinal functions in topology - - PowerPoint PPT Presentation

ω1-strongly compact cardinals and cardinal functions in topology

Toshimichi Usuba

Waseda University

• Nov. 19, 2018

Reflections on Set Theoretic reflection, Sant Bernat

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 1 / 27

Reflection and Compactness

Reflection: LARGE to SMALL, GLOBAL to LOCAL Compactness: SMALL to LARGE, LOCAL to GLOBAL These are dual concepts!

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 2 / 27

ω1-strongly compact cardinal

Definition

An uncountable cardinal κ is strongly compact if for every κ-complete filter F over the set A there is a κ-complete ultrafilter over A extending F. Strongly compact cardinals give natural limitations or upper bounds in many contexts. Bagaria and Magidor introduced the notion of ω1-strongly compact cardinal to analyze and optimize it.

Definition (Bagaria-Magidor (2014))

An uncountable cardinal κ is ω1-strongly compact if for every κ-complete filter F over the set A there is a σ-complete ultrafilter over A extending F.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 3 / 27

ω1-strongly compact cardinal

Definition

An uncountable cardinal κ is strongly compact if for every κ-complete filter F over the set A there is a κ-complete ultrafilter over A extending F. Strongly compact cardinals give natural limitations or upper bounds in many contexts. Bagaria and Magidor introduced the notion of ω1-strongly compact cardinal to analyze and optimize it.

Definition (Bagaria-Magidor (2014))

An uncountable cardinal κ is ω1-strongly compact if for every κ-complete filter F over the set A there is a σ-complete ultrafilter over A extending F.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 3 / 27

Fact (Bagaria-Magidor)

1 If κ is ω1-strongly compact, then every cardinal ≥ κ is ω1-strongly

compact.

2 So the important one is the least ω1-strongly compact cardinal. 3 strongly compact ⇒ ω1-strongly compact. 4 If κ is ω1-strongly compact, then there is a measurable cardinal ≤ κ.

Fact (Bagaria-Magidor)

1 It is consistent that the least ω1-strongly compact is supercompact. 2 It is consistent that the least measurable is ω1-strongly compact (in

this case, the least measurable must be strongly compact).

3 It is consistent that the least ω1-strongly compact is singular.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 4 / 27

Fact (Bagaria-Magidor)

1 If κ is ω1-strongly compact, then every cardinal ≥ κ is ω1-strongly

compact.

2 So the important one is the least ω1-strongly compact cardinal. 3 strongly compact ⇒ ω1-strongly compact. 4 If κ is ω1-strongly compact, then there is a measurable cardinal ≤ κ.

Fact (Bagaria-Magidor)

1 It is consistent that the least ω1-strongly compact is supercompact. 2 It is consistent that the least measurable is ω1-strongly compact (in

this case, the least measurable must be strongly compact).

3 It is consistent that the least ω1-strongly compact is singular.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 4 / 27

Lindel¨

• f degree

Definition

X: topological space The Lindel¨

• f degree of X, L(X), is min{κ | every open cover of X has

a subcover of size ≤ κ}. X is Lindel¨

• f if L(X) = ω, that is, every open cover has a countable

subcover. Clearly |X| ≥ L(X).

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 5 / 27

Product of Lindel¨

• f spaces

Fact

1 (Tychonoff) The product of compact spaces is compact. 2 The product of Lindel¨

• f spaces need not to be Lindel¨
• f; If S is the

Sorgenfrey line, then S is Lindel¨

• f but L(S2) = 2ω.

Question (Classical question)

How large is the Lindel¨

• f degree of the product of Lindel¨
• f spaces?

Fact (Juh´ asz(?))

For Lindel¨

• f spaces Xi (i ∈ I),

L(∏

i∈I Xi) ≤ the least strongly compact cardinal.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 6 / 27

Product of Lindel¨

• f spaces

Fact

1 (Tychonoff) The product of compact spaces is compact. 2 The product of Lindel¨

• f spaces need not to be Lindel¨
• f; If S is the

Sorgenfrey line, then S is Lindel¨

• f but L(S2) = 2ω.

Question (Classical question)

How large is the Lindel¨

• f degree of the product of Lindel¨
• f spaces?

Fact (Juh´ asz(?))

For Lindel¨

• f spaces Xi (i ∈ I),

L(∏

i∈I Xi) ≤ the least strongly compact cardinal.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 6 / 27

Product of Lindel¨

• f spaces

Fact

1 (Tychonoff) The product of compact spaces is compact. 2 The product of Lindel¨

• f spaces need not to be Lindel¨
• f; If S is the

Sorgenfrey line, then S is Lindel¨

• f but L(S2) = 2ω.

Question (Classical question)

How large is the Lindel¨

• f degree of the product of Lindel¨
• f spaces?

Fact (Juh´ asz(?))

For Lindel¨

• f spaces Xi (i ∈ I),

L(∏

i∈I Xi) ≤ the least strongly compact cardinal.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 6 / 27

Product of Lindel¨

• f spaces

Fact

1 (Tychonoff) The product of compact spaces is compact. 2 The product of Lindel¨

• f spaces need not to be Lindel¨
• f; If S is the

Sorgenfrey line, then S is Lindel¨

• f but L(S2) = 2ω.

Question (Classical question)

How large is the Lindel¨

• f degree of the product of Lindel¨
• f spaces?

Fact (Juh´ asz(?))

For Lindel¨

• f spaces Xi (i ∈ I),

L(∏

i∈I Xi) ≤ the least strongly compact cardinal.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 6 / 27

Characterization: Product of Lindel¨

• f spaces

Fact (Bagaria-Magidor)

For uncountable cardinal κ, the following are equivalent:

1 κ is ω1-strongly compact. 2 For every family {Xi | i ∈ I} of Lindel¨

• f spaces and open cover U of

the product space ∏

i∈I Xi, U has a subcover of size < κ.

3 For every family {Xi | i ∈ I} of Lindel¨

• f spaces we have

L(∏

i∈I Xi) ≤ κ,

the least ω1-strongly compact= sup{L(∏

i∈I Xi) | Xi is Lindel¨

• f}.

(1) ⇒ (2): Use a standard argument. (2) ⇒ (1): If λ < the least ω1-strongly compact cardinal, then we can find κ ≥ λ with L(ωκ) ≥ λ (use infinite Abelian group theory).

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 7 / 27

Characterization: Product of Lindel¨

• f spaces

Fact (Bagaria-Magidor)

For uncountable cardinal κ, the following are equivalent:

1 κ is ω1-strongly compact. 2 For every family {Xi | i ∈ I} of Lindel¨

• f spaces and open cover U of

the product space ∏

i∈I Xi, U has a subcover of size < κ.

3 For every family {Xi | i ∈ I} of Lindel¨

• f spaces we have

L(∏

i∈I Xi) ≤ κ,

the least ω1-strongly compact= sup{L(∏

i∈I Xi) | Xi is Lindel¨

• f}.

(1) ⇒ (2): Use a standard argument. (2) ⇒ (1): If λ < the least ω1-strongly compact cardinal, then we can find κ ≥ λ with L(ωκ) ≥ λ (use infinite Abelian group theory).

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 7 / 27

Characterization: Product of Lindel¨

• f spaces

Fact (Bagaria-Magidor)

For uncountable cardinal κ, the following are equivalent:

1 κ is ω1-strongly compact. 2 For every family {Xi | i ∈ I} of Lindel¨

• f spaces and open cover U of

the product space ∏

i∈I Xi, U has a subcover of size < κ.

3 For every family {Xi | i ∈ I} of Lindel¨

• f spaces we have

L(∏

i∈I Xi) ≤ κ,

the least ω1-strongly compact= sup{L(∏

i∈I Xi) | Xi is Lindel¨

• f}.

(1) ⇒ (2): Use a standard argument. (2) ⇒ (1): If λ < the least ω1-strongly compact cardinal, then we can find κ ≥ λ with L(ωκ) ≥ λ (use infinite Abelian group theory).

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 7 / 27

Bagaria and Magidor obtained various characterizations of ω1-strongly compact cardinal in many fields: Infinite Abelian groups. Topological spaces. Reflection principles... In this talk we consider more characterizations of ω1-strongly compact cardinals.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 8 / 27

Characterization:Infinitary logic

Fact (Folklore)

κ is ω1-strongly compact ⇐ ⇒ κ is Lω1,ω-compact; that is, for every theory T of Lω1ω-sentences, T has a model if every subtheory of T with size < κ has a model.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 9 / 27

Characterization:Uniform filters

A filter over the infinite cardinal κ is uniform if |A| = κ for every A ∈ F.

Fact (Ketonen)

An uncountable cardinal κ is strongly compact if and only if for every regular λ ≥ κ, there is a κ-complete uniform ultrafilter over λ.

Proposition

An uncountable cardinal κ is ω1-strongly compact if and only if for every regular λ ≥ κ, there is a σ-complete uniform ultrafilter over λ. Proof is the same to Ketonen’s one.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 10 / 27

Characterization:Uniform filters

A filter over the infinite cardinal κ is uniform if |A| = κ for every A ∈ F.

Fact (Ketonen)

An uncountable cardinal κ is strongly compact if and only if for every regular λ ≥ κ, there is a κ-complete uniform ultrafilter over λ.

Proposition

An uncountable cardinal κ is ω1-strongly compact if and only if for every regular λ ≥ κ, there is a σ-complete uniform ultrafilter over λ. Proof is the same to Ketonen’s one.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 10 / 27

Arhangel’skii’s Problem

Definition

For a topological space X, the Gδ-midification Xδ is the space X with the topology generated by all Gδ-subsets of X. Xδ is finer than X. Xδ is a P-space, that is, every Gδ-set is open. If X is T1 and first countable, then Xδ is a discrete space, hence L(Xδ) = |X|. The unit interval [0, 1] is compact but L(([0, 1])δ) = 2ω.

Question (Arhangelskii, 1980’s)

Let X be a compact Hausdorff space. Is L(Xδ) ≤ 2ω?

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 11 / 27

Arhangel’skii’s Problem

Definition

For a topological space X, the Gδ-midification Xδ is the space X with the topology generated by all Gδ-subsets of X. Xδ is finer than X. Xδ is a P-space, that is, every Gδ-set is open. If X is T1 and first countable, then Xδ is a discrete space, hence L(Xδ) = |X|. The unit interval [0, 1] is compact but L(([0, 1])δ) = 2ω.

Question (Arhangelskii, 1980’s)

Let X be a compact Hausdorff space. Is L(Xδ) ≤ 2ω?

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 11 / 27

Arhangel’skii’s Problem

Definition

For a topological space X, the Gδ-midification Xδ is the space X with the topology generated by all Gδ-subsets of X. Xδ is finer than X. Xδ is a P-space, that is, every Gδ-set is open. If X is T1 and first countable, then Xδ is a discrete space, hence L(Xδ) = |X|. The unit interval [0, 1] is compact but L(([0, 1])δ) = 2ω.

Question (Arhangelskii, 1980’s)

Let X be a compact Hausdorff space. Is L(Xδ) ≤ 2ω?

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 11 / 27

Known partial answers

Fact (Gorelic (1996))

Let κ be a regular uncountable cardinal, and suppose there is no σ-complete uniform ultrafilter over κ. Then L(ω2κ) ≥ κ. (ω2κ)δ is a closed subspace of ((ω + 1)2κ)δ, hence (ω + 1)2κ is compact but L(((ω + 1)2κ)δ) ≥ L((ω2κ)δ) ≥ L(ω2κ) ≥ κ.

Corollary

If κ is strictly less than the least measurable cardinal, then there is a compact Hausdorff space X with L(Xδ) ≥ κ. Arhangeskii’s question for the Lindel¨

• f degree of the Gδ-modification has a

negative answer.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 12 / 27

Known partial answers

Fact (Gorelic (1996))

Let κ be a regular uncountable cardinal, and suppose there is no σ-complete uniform ultrafilter over κ. Then L(ω2κ) ≥ κ. (ω2κ)δ is a closed subspace of ((ω + 1)2κ)δ, hence (ω + 1)2κ is compact but L(((ω + 1)2κ)δ) ≥ L((ω2κ)δ) ≥ L(ω2κ) ≥ κ.

Corollary

If κ is strictly less than the least measurable cardinal, then there is a compact Hausdorff space X with L(Xδ) ≥ κ. Arhangeskii’s question for the Lindel¨

• f degree of the Gδ-modification has a

negative answer.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 12 / 27

Characterization: Lindel¨

• f degree of Gδ-modification

For Arhangelskii’s question, we have:

Theorem

For an uncountable cardinal κ, the following are equivalent:

1 κ is ω1-strongly compact. 2 L(Xδ) ≤ κ for every compact Hausdorff space X.

the least ω1-strongly compact = sup{L(Xδ) | X is compact Hausdorff}

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 13 / 27

Characterization: Lindel¨

• f degree of Gδ-modification

For Arhangelskii’s question, we have:

Theorem

For an uncountable cardinal κ, the following are equivalent:

1 κ is ω1-strongly compact. 2 L(Xδ) ≤ κ for every compact Hausdorff space X.

the least ω1-strongly compact = sup{L(Xδ) | X is compact Hausdorff}

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 13 / 27

L(Xδ) ≤ ω1-strongly compact

Lemma

Suppose κ is ω1-strongly compact. Let X be a Lindel¨

• f space, and U a

cover of Gδ-subsets of X. Then U has a subcover of size < κ. In particular L(Xδ) ≤ κ. Proof: Use a standard ultrapower argument.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 14 / 27

L(Xδ) ≥ ω1-strongly compact

We have several ways to construct a space X with L(Xδ) large.

1 Bagaria-Magidor already showed that for λ <the least ω1-strongly

compact, there is κ ≥ λ with L(ωκ) ≥ λ. Then (ω + 1)κ is compact but L(((ω + 1)κ)δ) ≥ L(ωκ) ≥ λ.

2 Gorelic showed that L(ω2λ) ≥ λ if there is no σ-complete uniform

ultrafilter over λ, and we have known that the least ω1-strongly compact= sup{λ | there is no σ-complete uniform ultrafilter over λ}.

3 More direct and informative construction; Variant of the Stone-ˇ

Cech space For an infinite set A, we identify A as the discrete space. Let βA be the Stone-ˇ Cech compactification of A: βA is the set of all ultrafilters of A, the topology is generated by {{U ∈ βκ | B ∈ U} | B ⊆ A}. βA is a compact Hausdorff space.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 15 / 27

L(Xδ) ≥ ω1-strongly compact

We have several ways to construct a space X with L(Xδ) large.

1 Bagaria-Magidor already showed that for λ <the least ω1-strongly

compact, there is κ ≥ λ with L(ωκ) ≥ λ. Then (ω + 1)κ is compact but L(((ω + 1)κ)δ) ≥ L(ωκ) ≥ λ.

2 Gorelic showed that L(ω2λ) ≥ λ if there is no σ-complete uniform

ultrafilter over λ, and we have known that the least ω1-strongly compact= sup{λ | there is no σ-complete uniform ultrafilter over λ}.

3 More direct and informative construction; Variant of the Stone-ˇ

Cech space For an infinite set A, we identify A as the discrete space. Let βA be the Stone-ˇ Cech compactification of A: βA is the set of all ultrafilters of A, the topology is generated by {{U ∈ βκ | B ∈ U} | B ⊆ A}. βA is a compact Hausdorff space.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 15 / 27

L(Xδ) ≥ ω1-strongly compact

We have several ways to construct a space X with L(Xδ) large.

1 Bagaria-Magidor already showed that for λ <the least ω1-strongly

compact, there is κ ≥ λ with L(ωκ) ≥ λ. Then (ω + 1)κ is compact but L(((ω + 1)κ)δ) ≥ L(ωκ) ≥ λ.

2 Gorelic showed that L(ω2λ) ≥ λ if there is no σ-complete uniform

ultrafilter over λ, and we have known that the least ω1-strongly compact= sup{λ | there is no σ-complete uniform ultrafilter over λ}.

3 More direct and informative construction; Variant of the Stone-ˇ

Cech space For an infinite set A, we identify A as the discrete space. Let βA be the Stone-ˇ Cech compactification of A: βA is the set of all ultrafilters of A, the topology is generated by {{U ∈ βκ | B ∈ U} | B ⊆ A}. βA is a compact Hausdorff space.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 15 / 27

L(Xδ) ≥ ω1-strongly compact

We have several ways to construct a space X with L(Xδ) large.

1 Bagaria-Magidor already showed that for λ <the least ω1-strongly

compact, there is κ ≥ λ with L(ωκ) ≥ λ. Then (ω + 1)κ is compact but L(((ω + 1)κ)δ) ≥ L(ωκ) ≥ λ.

2 Gorelic showed that L(ω2λ) ≥ λ if there is no σ-complete uniform

ultrafilter over λ, and we have known that the least ω1-strongly compact= sup{λ | there is no σ-complete uniform ultrafilter over λ}.

3 More direct and informative construction; Variant of the Stone-ˇ

Cech space For an infinite set A, we identify A as the discrete space. Let βA be the Stone-ˇ Cech compactification of A: βA is the set of all ultrafilters of A, the topology is generated by {{U ∈ βκ | B ∈ U} | B ⊆ A}. βA is a compact Hausdorff space.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 15 / 27

L(Xδ) ≥ ω1-strongly compact

We have several ways to construct a space X with L(Xδ) large.

1 Bagaria-Magidor already showed that for λ <the least ω1-strongly

compact, there is κ ≥ λ with L(ωκ) ≥ λ. Then (ω + 1)κ is compact but L(((ω + 1)κ)δ) ≥ L(ωκ) ≥ λ.

2 Gorelic showed that L(ω2λ) ≥ λ if there is no σ-complete uniform

ultrafilter over λ, and we have known that the least ω1-strongly compact= sup{λ | there is no σ-complete uniform ultrafilter over λ}.

3 More direct and informative construction; Variant of the Stone-ˇ

Cech space For an infinite set A, we identify A as the discrete space. Let βA be the Stone-ˇ Cech compactification of A: βA is the set of all ultrafilters of A, the topology is generated by {{U ∈ βκ | B ∈ U} | B ⊆ A}. βA is a compact Hausdorff space.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 15 / 27

For an infinite cardinal κ and a set A, let PκA = {x ⊆ A | |x| < κ}. A filter U over PκA is fine if for every a ∈ A, we have {x ∈ PκA | a ∈ x} ∈ U.

Fact (Bagaria-Magidor (2014))

An uncountable cardinal κ is ω1-strongly compact if and only if for every λ ≥ κ, there is a σ-complete fine ultrafilter over Pκλ. Suppose κ is not ω1-strongly compact. Then we can find λ ≥ κ such that there is no σ-complete fine ultrafilter over Pκλ. Let µ(Pκλ) be the subspace of β(Pκλ) consists of the all fine ultrafilters

• ver Pκλ. µ(Pκλ) is a closed subspace of β(Pκλ), hence is compact

Hausdorff.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 16 / 27

For an infinite cardinal κ and a set A, let PκA = {x ⊆ A | |x| < κ}. A filter U over PκA is fine if for every a ∈ A, we have {x ∈ PκA | a ∈ x} ∈ U.

Fact (Bagaria-Magidor (2014))

An uncountable cardinal κ is ω1-strongly compact if and only if for every λ ≥ κ, there is a σ-complete fine ultrafilter over Pκλ. Suppose κ is not ω1-strongly compact. Then we can find λ ≥ κ such that there is no σ-complete fine ultrafilter over Pκλ. Let µ(Pκλ) be the subspace of β(Pκλ) consists of the all fine ultrafilters

• ver Pκλ. µ(Pκλ) is a closed subspace of β(Pκλ), hence is compact

Hausdorff.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 16 / 27

For an infinite cardinal κ and a set A, let PκA = {x ⊆ A | |x| < κ}. A filter U over PκA is fine if for every a ∈ A, we have {x ∈ PκA | a ∈ x} ∈ U.

Fact (Bagaria-Magidor (2014))

An uncountable cardinal κ is ω1-strongly compact if and only if for every λ ≥ κ, there is a σ-complete fine ultrafilter over Pκλ. Suppose κ is not ω1-strongly compact. Then we can find λ ≥ κ such that there is no σ-complete fine ultrafilter over Pκλ. Let µ(Pκλ) be the subspace of β(Pκλ) consists of the all fine ultrafilters

• ver Pκλ. µ(Pκλ) is a closed subspace of β(Pκλ), hence is compact

Hausdorff.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 16 / 27

Covers of Gδ-subsets

Lemma

There is a cover U of µ(Pκλ) by Gδ-subsets such that U has no subcover

• f size < κ.

For a countable partition A of Pκλ, let SA = {U ∈ µ(Pκλ) | A / ∈ U for every A ∈ A}. A cover {SA | A is a countable partition} is as required.

Corollary

If κ is not ω1-strongly compact, then L((µ(Pκλ))δ) ≥ κ for some λ ≥ κ.

Corollary

the least ω1-strongly compact = sup{L(Xδ) | X is compact Hausdorff}.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 17 / 27

Covers of Gδ-subsets

Lemma

There is a cover U of µ(Pκλ) by Gδ-subsets such that U has no subcover

• f size < κ.

For a countable partition A of Pκλ, let SA = {U ∈ µ(Pκλ) | A / ∈ U for every A ∈ A}. A cover {SA | A is a countable partition} is as required.

Corollary

If κ is not ω1-strongly compact, then L((µ(Pκλ))δ) ≥ κ for some λ ≥ κ.

Corollary

the least ω1-strongly compact = sup{L(Xδ) | X is compact Hausdorff}.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 17 / 27

The tightness of Xδ

Definition

For a space X, the tightness number of X, t(X) is the minimal cardinal κ such that for every A ⊆ X and p ∈ A, there is B ⊆ A with |B| ≤ κ and p ∈ B. X is countably tight if t(X) = ω.

Question (Arhangel’skii)

If X is a countably tight space, how large is t(Xδ)?

Fact (Dow-Juh´ asz-Soukup-Szentmikl´

• ssy-Weiss (2018))

If X is regular Lindel¨

• f, then t(Xδ) ≤ 2t(X).

Fact (Dow-Juh´ asz-Soukup-Szentmikl´

• ssy-Weiss (2018))

Suppose V = L. Then for every cardinal κ, there is a Frechet space X (so t(X) = ω) with t(Xδ) ≥ κ.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 18 / 27

The tightness of Xδ

Definition

For a space X, the tightness number of X, t(X) is the minimal cardinal κ such that for every A ⊆ X and p ∈ A, there is B ⊆ A with |B| ≤ κ and p ∈ B. X is countably tight if t(X) = ω.

Question (Arhangel’skii)

If X is a countably tight space, how large is t(Xδ)?

Fact (Dow-Juh´ asz-Soukup-Szentmikl´

• ssy-Weiss (2018))

If X is regular Lindel¨

• f, then t(Xδ) ≤ 2t(X).

Fact (Dow-Juh´ asz-Soukup-Szentmikl´

• ssy-Weiss (2018))

Suppose V = L. Then for every cardinal κ, there is a Frechet space X (so t(X) = ω) with t(Xδ) ≥ κ.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 18 / 27

The tightness of Xδ

Definition

For a space X, the tightness number of X, t(X) is the minimal cardinal κ such that for every A ⊆ X and p ∈ A, there is B ⊆ A with |B| ≤ κ and p ∈ B. X is countably tight if t(X) = ω.

Question (Arhangel’skii)

If X is a countably tight space, how large is t(Xδ)?

Fact (Dow-Juh´ asz-Soukup-Szentmikl´

• ssy-Weiss (2018))

If X is regular Lindel¨

• f, then t(Xδ) ≤ 2t(X).

Fact (Dow-Juh´ asz-Soukup-Szentmikl´

• ssy-Weiss (2018))

Suppose V = L. Then for every cardinal κ, there is a Frechet space X (so t(X) = ω) with t(Xδ) ≥ κ.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 18 / 27

Characterization:Tightness of Gδ-modification

Theorem

1 If κ is ω1-strongly compact and X is a countable tight space, then

t(Xδ) ≤ κ.

2 Suppose there is no weakly Mahlo cardinal < 2ω (e.g., CH holds).

Then κ is ω1-strongly compact if and only if t(Xδ) ≤ κ for every countably tight Tychonoff space X. Assuming 2ω is not so large, the least ω1-strongly compact = sup{t(Xδ) | X is countably tight Tychonoff}

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 19 / 27

Characterization:Tightness of Gδ-modification

Theorem

1 If κ is ω1-strongly compact and X is a countable tight space, then

t(Xδ) ≤ κ.

2 Suppose there is no weakly Mahlo cardinal < 2ω (e.g., CH holds).

Then κ is ω1-strongly compact if and only if t(Xδ) ≤ κ for every countably tight Tychonoff space X. Assuming 2ω is not so large, the least ω1-strongly compact = sup{t(Xδ) | X is countably tight Tychonoff}

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 19 / 27

Characterization:Tightness of Gδ-modification

Theorem

1 If κ is ω1-strongly compact and X is a countable tight space, then

t(Xδ) ≤ κ.

2 Suppose there is no weakly Mahlo cardinal < 2ω (e.g., CH holds).

Then κ is ω1-strongly compact if and only if t(Xδ) ≤ κ for every countably tight Tychonoff space X. Assuming 2ω is not so large, the least ω1-strongly compact = sup{t(Xδ) | X is countably tight Tychonoff}

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 19 / 27

What is the condition “2ω is not so large”?

Fact (Arhangelskii-Pykateev)

For a Tychonoff space X, let Cp(X) be the continuous functions from X to R with the pointwise convergent topology. Then t(Cp(X)) = sup{L(X n) | n < ω}.

Lemma

Suppose there is no σ-complete fine ultrafilter over Pκλ, and suppose that for every fine ultrafilters Un over Pκλ (n < ω), we have that ∩

n<ω Un is

NOT σ-complete. Then there is a cover U by Gδ-subsets of X := µ(Pκλ) such that U witnesses t((Cp(X))δ) ≥ κ.

Lemma

Let S be an uncountable set, and Un (n < ω) be non-principal non-σ-complete ultrafilters over S. If F = ∩

n<ω Un is σ-complete, then F

is σ-complete ω1-saturated filter over S, and there is a weakly Mahlo < 2ω.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 20 / 27

What is the condition “2ω is not so large”?

Fact (Arhangelskii-Pykateev)

For a Tychonoff space X, let Cp(X) be the continuous functions from X to R with the pointwise convergent topology. Then t(Cp(X)) = sup{L(X n) | n < ω}.

Lemma

Suppose there is no σ-complete fine ultrafilter over Pκλ, and suppose that for every fine ultrafilters Un over Pκλ (n < ω), we have that ∩

n<ω Un is

NOT σ-complete. Then there is a cover U by Gδ-subsets of X := µ(Pκλ) such that U witnesses t((Cp(X))δ) ≥ κ.

Lemma

Let S be an uncountable set, and Un (n < ω) be non-principal non-σ-complete ultrafilters over S. If F = ∩

n<ω Un is σ-complete, then F

is σ-complete ω1-saturated filter over S, and there is a weakly Mahlo < 2ω.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 20 / 27

What is the condition “2ω is not so large”?

Fact (Arhangelskii-Pykateev)

For a Tychonoff space X, let Cp(X) be the continuous functions from X to R with the pointwise convergent topology. Then t(Cp(X)) = sup{L(X n) | n < ω}.

Lemma

Suppose there is no σ-complete fine ultrafilter over Pκλ, and suppose that for every fine ultrafilters Un over Pκλ (n < ω), we have that ∩

n<ω Un is

NOT σ-complete. Then there is a cover U by Gδ-subsets of X := µ(Pκλ) such that U witnesses t((Cp(X))δ) ≥ κ.

Lemma

Let S be an uncountable set, and Un (n < ω) be non-principal non-σ-complete ultrafilters over S. If F = ∩

n<ω Un is σ-complete, then F

is σ-complete ω1-saturated filter over S, and there is a weakly Mahlo < 2ω.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 20 / 27

Product of two Lindel¨

• f spaces

Fact (Bagaria-Magidor)

For an uncountable cardinal κ, the following are equivalent:

1 κ is ω1-strongly compact. 2 For every family {Xi | i ∈ I} of Lindel¨

• f spaces, we have

L(∏

i∈I Xi) ≤ κ,

Question (Still open)

Are there two Lindel¨

• f spaces X and Y with L(X × Y ) > 2ω?

Fact (Shelah (1980’s), Gorelic (1994))

It is consistent that there are regular Lindel¨

• f spaces X and Y such that

L(X × Y ) = 2ω1 > 2ω and 2ω1 is arbitrary large.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 21 / 27

Product of two Lindel¨

• f spaces

Fact (Bagaria-Magidor)

For an uncountable cardinal κ, the following are equivalent:

1 κ is ω1-strongly compact. 2 For every family {Xi | i ∈ I} of Lindel¨

• f spaces, we have

L(∏

i∈I Xi) ≤ κ,

Question (Still open)

Are there two Lindel¨

• f spaces X and Y with L(X × Y ) > 2ω?

Fact (Shelah (1980’s), Gorelic (1994))

It is consistent that there are regular Lindel¨

• f spaces X and Y such that

L(X × Y ) = 2ω1 > 2ω and 2ω1 is arbitrary large.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 21 / 27

Product of two Lindel¨

• f spaces

Fact (Bagaria-Magidor)

For an uncountable cardinal κ, the following are equivalent:

1 κ is ω1-strongly compact. 2 For every family {Xi | i ∈ I} of Lindel¨

• f spaces, we have

L(∏

i∈I Xi) ≤ κ,

Question (Still open)

Are there two Lindel¨

• f spaces X and Y with L(X × Y ) > 2ω?

Fact (Shelah (1980’s), Gorelic (1994))

It is consistent that there are regular Lindel¨

• f spaces X and Y such that

L(X × Y ) = 2ω1 > 2ω and 2ω1 is arbitrary large.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 21 / 27

Characterization: Product of two Lindel¨

• f spaces

Theorem

It is consistent that for an uncountable cardinal κ, κ is ω1-strongly compact if and only if L(X × Y ) ≤ κ for every regular Lindel¨

• f spaces X

and Y . Actually in the Cohen forcing extension, the least ω1-strongly compact = sup{L(X × Y ) | X and Y are regular Lindel¨

• f}
• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 22 / 27

Sketch of the proof

Suppose there is no σ-complete fine ultrafilers over Pκλ. In the Cohen forcing extension, we construct two finer topology T0 ,T1 on µ(Pκλ)V using a Cohen real so that

1 ⟨µ(Pκλ)V , T0⟩ and ⟨µ(Pκλ)V , T1⟩ are still Lindel¨

• f, but

2 The product ⟨µ(Pκλ)V , T0⟩ × ⟨µ(Pκλ)V , T1⟩ has large Lindel¨

• f

degree.

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 23 / 27

Summary

the least ω1-strongly compact = sup{L(∏

i∈I Xi) | Xi is Lindel¨

• f} (Bagaria-Magidor)

= sup{L(Xδ) | X is comapct T2} = sup{t(Xδ) | X is countably tight} (if 2ω is not so large) = sup{L(X × Y ) | X and Y are Lindel¨

• f}

(in the Cohen forcing extension) = the least Lω1ω-compact (folklore) = the least cardinal κ such that for every regular λ ≥ κ there is a σ-complete uniform ultrafilter over λ (essentially Ketonen) . . .

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 24 / 27

Questions

Question

Does ZFC prove the following?

1 the least ω1-strongly compact

= sup{t(Xδ) | X is countably tight (Tychonoff)}.

2 the least ω1-strongly compact

= sup{L(X × Y ) | X and Y are (regular) Lindel¨

• f}.

(1) is almost equivalent to: For κ ≤ λ, suppose there is no σ-complete fine ultrafilter over Pκλ. For fine ultrafilters Un (n < ω) over Pκλ, is the filter ∩

n<ω Un never σ-complete?

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 25 / 27

References

• J. Bagaria, M. Madigor, Group radicals and strongly compact cardinals.
• Trans. Am. Math. Soc. Vol. 366, No. 4 (2014), 1857–1877
• J. Bagaria, M. Madigor, On ω1-strongly compact cardinals. J. Symb. Logic
• Vol. 79, No. 1 (2014), 268–278.

Alan Dow, Istv´ an Juh´ asz, Lajos Soukup, Zolt´ an Szentmikl´

• ssy, William

Weiss, On the tightness of Gδ-modifications. Preprint.

• I. Gorelic, On powers of Lindel¨
• f spaces. Comment. Math. Univ. Carol. Vol.

35, No. 2 (1994), 383–401.

• I. Gorelic, The Gδ-topology and incompactness of ωκ. Commentat. Math.
• Univ. Carol. Vol. 37, No. 3 (1996), 613–616.
• S. Shelah, On some problems in general topology. Contemp. Math. 192

(1996), 91–101.

• T. Usuba, Gδ-topology and compact cardinals. To appear in Fundamenta

Matheaticae

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 26 / 27

Happy Birthday, Joan!

• T. Usuba (Waseda Univ.)

ω1-strongly compact Nov.19, 2018 27 / 27