Axiomatic set theory and its applications

Some applications of iterated
ultrapowers in countably compact
groupsUlises Ariet RAMOS GARCÍA
National Autonomous University of Mexico, Morelia
ariet@matmor.unam.mx

RIMS
November 2018

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20181 / 18

Joint with M. Hrušák, J. van Mill y S. Shelah.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20182 / 18

Contenido

1

Motivation

2

Solution to Comfort’s/van Douwen’s problem

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20183 / 18

Versions of compatness
Given an ultrafilter p on ω, a sequence hxn : n ∈ ωi and a point x contained in a
topological space X we say that x = p-limxn if {n ∈ ω : xn ∈ U} ∈ p for every
neighbourhood U ⊆ X of x.

Definition
Let X be a topological space and let p be an ultrafilter on ω.
X is compact if every open cover of X has a finite subcover.
X is p-compact if for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ Xsuch that x = p-limxn .
X is countably compact if every countable open cover of X has a finite subcover.
II

iff every infinite set has an accumulation point.
iff for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ X and an ultrafilter
p on ω such that x = p-limxn .

X is pseudo-compact if every continuous function f : X → R is bounded.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20184 / 18

Versions of compatness
Given an ultrafilter p on ω, a sequence hxn : n ∈ ωi and a point x contained in a
topological space X we say that x = p-limxn if {n ∈ ω : xn ∈ U} ∈ p for every
neighbourhood U ⊆ X of x.

Definition
Let X be a topological space and let p be an ultrafilter on ω.
X is compact if every open cover of X has a finite subcover.
X is p-compact if for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ Xsuch that x = p-limxn .
X is countably compact if every countable open cover of X has a finite subcover.
II

iff every infinite set has an accumulation point.
iff for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ X and an ultrafilter
p on ω such that x = p-limxn .

X is pseudo-compact if every continuous function f : X → R is bounded.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20184 / 18

Versions of compatness
Given an ultrafilter p on ω, a sequence hxn : n ∈ ωi and a point x contained in a
topological space X we say that x = p-limxn if {n ∈ ω : xn ∈ U} ∈ p for every
neighbourhood U ⊆ X of x.

Definition
Let X be a topological space and let p be an ultrafilter on ω.
X is compact if every open cover of X has a finite subcover.
X is p-compact if for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ Xsuch that x = p-limxn .
X is countably compact if every countable open cover of X has a finite subcover.
II

iff every infinite set has an accumulation point.
iff for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ X and an ultrafilter
p on ω such that x = p-limxn .

X is pseudo-compact if every continuous function f : X → R is bounded.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20184 / 18

Versions of compatness
Given an ultrafilter p on ω, a sequence hxn : n ∈ ωi and a point x contained in a
topological space X we say that x = p-limxn if {n ∈ ω : xn ∈ U} ∈ p for every
neighbourhood U ⊆ X of x.

Definition
Let X be a topological space and let p be an ultrafilter on ω.
X is compact if every open cover of X has a finite subcover.
X is p-compact if for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ Xsuch that x = p-limxn .
X is countably compact if every countable open cover of X has a finite subcover.
II

iff every infinite set has an accumulation point.
iff for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ X and an ultrafilter
p on ω such that x = p-limxn .

X is pseudo-compact if every continuous function f : X → R is bounded.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20184 / 18

Versions of compatness
Given an ultrafilter p on ω, a sequence hxn : n ∈ ωi and a point x contained in a
topological space X we say that x = p-limxn if {n ∈ ω : xn ∈ U} ∈ p for every
neighbourhood U ⊆ X of x.

Definition
Let X be a topological space and let p be an ultrafilter on ω.
X is compact if every open cover of X has a finite subcover.
X is p-compact if for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ Xsuch that x = p-limxn .
X is countably compact if every countable open cover of X has a finite subcover.
II

iff every infinite set has an accumulation point.
iff for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ X and an ultrafilter
p on ω such that x = p-limxn .

X is pseudo-compact if every continuous function f : X → R is bounded.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20184 / 18

Versions of compatness
Given an ultrafilter p on ω, a sequence hxn : n ∈ ωi and a point x contained in a
topological space X we say that x = p-limxn if {n ∈ ω : xn ∈ U} ∈ p for every
neighbourhood U ⊆ X of x.

Definition
Let X be a topological space and let p be an ultrafilter on ω.
X is compact if every open cover of X has a finite subcover.
X is p-compact if for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ Xsuch that x = p-limxn .
X is countably compact if every countable open cover of X has a finite subcover.
II

iff every infinite set has an accumulation point.
iff for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ X and an ultrafilter
p on ω such that x = p-limxn .

X is pseudo-compact if every continuous function f : X → R is bounded.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20184 / 18

Versions of compatness
Given an ultrafilter p on ω, a sequence hxn : n ∈ ωi and a point x contained in a
topological space X we say that x = p-limxn if {n ∈ ω : xn ∈ U} ∈ p for every
neighbourhood U ⊆ X of x.

Definition
Let X be a topological space and let p be an ultrafilter on ω.
X is compact if every open cover of X has a finite subcover.
X is p-compact if for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ Xsuch that x = p-limxn .
X is countably compact if every countable open cover of X has a finite subcover.
II

iff every infinite set has an accumulation point.
iff for every sequence hxn : n ∈ ωi ⊆ X there is a point x ∈ X and an ultrafilter
p on ω such that x = p-limxn .

X is pseudo-compact if every continuous function f : X → R is bounded.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20184 / 18

Versions of compatness and products(Tychonoff ’30/’35) Any product of compact spaces is compact.
(Ginsburg-Saks ’75) Any product of p-compact spaces is p-compact.
(Tereska ’52, Novák ’53) There are countably compact spaces whose square is
not even pseudo-compact.
(Comfort-Ross ’66) Any product of pseudo-compact topological groups is pseudocompact.

Problem (Comfort ’66)
Are there countably compact groups G and H such that G × H is not countablycompact?

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 2018

5 / 18

Versions of compatness and products(Tychonoff ’30/’35) Any product of compact spaces is compact.
(Ginsburg-Saks ’75) Any product of p-compact spaces is p-compact.
(Tereska ’52, Novák ’53) There are countably compact spaces whose square is
not even pseudo-compact.
(Comfort-Ross ’66) Any product of pseudo-compact topological groups is pseudocompact.

Problem (Comfort ’66)
Are there countably compact groups G and H such that G × H is not countablycompact?

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 2018

5 / 18

Versions of compatness and products(Tychonoff ’30/’35) Any product of compact spaces is compact.
(Ginsburg-Saks ’75) Any product of p-compact spaces is p-compact.
(Tereska ’52, Novák ’53) There are countably compact spaces whose square is
not even pseudo-compact.
(Comfort-Ross ’66) Any product of pseudo-compact topological groups is pseudocompact.

Problem (Comfort ’66)
Are there countably compact groups G and H such that G × H is not countablycompact?

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 2018

5 / 18

Comfort’s problem – consistent solutions
Problem (Comfort ’66)
Are there countably compact groups G and H such that G × H is not countablycompact?
(van Douwen ’80) (MA) There are two countably compact subgroups of 2cwhose product is not countably compact.
(van Mill-Hart ’91) (MActble ) There is a countably compact group whose square
is not countably compact.
(Tomita ’99) (MActble ) There is a group whose square is countably compact
but the cube is not.
(van Douwen ’80) Every countably compact Boolean group without (nontrivial) convergent sequences contains two countably compact subgroups whose
product is not countably compact.
(Hajnal-Juhász ’76) (CH) There is a countably compact Boolean group without
convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20186 / 18

Comfort’s problem – consistent solutions
Problem (Comfort ’66)
Are there countably compact groups G and H such that G × H is not countablycompact?
(van Douwen ’80) (MA) There are two countably compact subgroups of 2cwhose product is not countably compact.
(van Mill-Hart ’91) (MActble ) There is a countably compact group whose square
is not countably compact.
(Tomita ’99) (MActble ) There is a group whose square is countably compact
but the cube is not.
(van Douwen ’80) Every countably compact Boolean group without (nontrivial) convergent sequences contains two countably compact subgroups whose
product is not countably compact.
(Hajnal-Juhász ’76) (CH) There is a countably compact Boolean group without
convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20186 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18van Douwen’s problem – consistent solutions
Problem (van Douwen ’80)
Is there a countably compact Boolean group without non-trivial convergent
sequences?
(Kuz’minov ’58) Every compact topological group contains a non-trivial convergent sequence.
(Hajnal-Juhász ’76) Yes assuming CH.
(van Douwen ’80) Yes assuming MA.
(Tomita, ’99) Yes assuming MActble .
...
(Garcı́a Ferreira-Tomita-Watson ’05) Yes assuming the existence of a selective
ultrafilter. In fact, such group is p-compact.
(Szeptycki-Tomita ’09) Yes in the random real model.
All of these constructions describe subgroups of 2c .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20187 / 18

Contenido

1

Motivation

2

Solution to Comfort’s/van Douwen’s problem

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 20188 / 18

Main result

Theorem
There is (in ZFC) a countably compact Boolean group (subgroup of 2c ) without
non-trivial convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 2018

9 / 18Bohr group topologies
The Bohr topology on a group G is the weakest group topology making every
homomorphism Φ ∈ Hom(G , T) continuous. We let (G , τBohr ) denote G equippedwith the Bohr topology.
(Folklore) If G is an Abelian group, then (G , τBohr ) is homeomorphic (andisomorphic) to a subgroup of THom(G ,T) via the evaluation mapping.

Theorem
Let G be a (infinite) countable Abelian group. Then (G , τBohr ) has no (non-trivial)convergent sequences.

(Folklore) Let X be a countably compact regular space without (non-trivial)convergent sequences. Then every infinite subset of X has at least c accumulation points.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201810 / 18

Bohr group topologies
The Bohr topology on a group G is the weakest group topology making every
homomorphism Φ ∈ Hom(G , T) continuous. We let (G , τBohr ) denote G equippedwith the Bohr topology.
(Folklore) If G is an Abelian group, then (G , τBohr ) is homeomorphic (andisomorphic) to a subgroup of THom(G ,T) via the evaluation mapping.

Theorem
Let G be a (infinite) countable Abelian group. Then (G , τBohr ) has no (non-trivial)convergent sequences.

(Folklore) Let X be a countably compact regular space without (non-trivial)convergent sequences. Then every infinite subset of X has at least c accumulation points.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201810 / 18

Bohr group topologies
The Bohr topology on a group G is the weakest group topology making every
homomorphism Φ ∈ Hom(G , T) continuous. We let (G , τBohr ) denote G equippedwith the Bohr topology.
(Folklore) If G is an Abelian group, then (G , τBohr ) is homeomorphic (andisomorphic) to a subgroup of THom(G ,T) via the evaluation mapping.

Theorem
Let G be a (infinite) countable Abelian group. Then (G , τBohr ) has no (non-trivial)convergent sequences.

(Folklore) Let X be a countably compact regular space without (non-trivial)convergent sequences. Then every infinite subset of X has at least c accumulation points.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201810 / 18

Bohr group topologies
The Bohr topology on a group G is the weakest group topology making every
homomorphism Φ ∈ Hom(G , T) continuous. We let (G , τBohr ) denote G equippedwith the Bohr topology.
(Folklore) If G is an Abelian group, then (G , τBohr ) is homeomorphic (andisomorphic) to a subgroup of THom(G ,T) via the evaluation mapping.

Theorem
Let G be a (infinite) countable Abelian group. Then (G , τBohr ) has no (non-trivial)convergent sequences.

(Folklore) Let X be a countably compact regular space without (non-trivial)convergent sequences. Then every infinite subset of X has at least c accumulation points.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201810 / 18

Bohr group topologies
The Bohr topology on a group G is the weakest group topology making every
homomorphism Φ ∈ Hom(G , T) continuous. We let (G , τBohr ) denote G equippedwith the Bohr topology.
(Folklore) If G is an Abelian group, then (G , τBohr ) is homeomorphic (andisomorphic) to a subgroup of THom(G ,T) via the evaluation mapping.

Theorem
Let G be a (infinite) countable Abelian group. Then (G , τBohr ) has no (non-trivial)convergent sequences.

(Folklore) Let X be a countably compact regular space without (non-trivial)convergent sequences. Then every infinite subset of X has at least c accumulation points.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201810 / 18

Extensions of homomorphisms to ultrapowers
Given a group G and an ultrafilter p ∈ ω ∗ , denote by
ultp (G ) = (G )ω / ≡p , where f ≡p g iff {n : f (n) = g (n)} ∈ p.

By Lós theorem, ultp (G ) is a group with the same first order properties as G .
There is a natural embedding of G into ultp (G ) sending each g ∈ G to theequivalence class of the constant function with value g . We shall therefore
consider G as a subgroup of ultp (G ).
Every Φ ∈ Hom(G , T) naturally extends to a homomorphism Φ ∈ (ultp (G ), T) bylettingΦ([f ]) = p-limn∈ω Φ(f (n)).
We let τBohr denote the weakest topology making every Φ continuous, where Φ ∈Hom(G , T).

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201811 / 18

Extensions of homomorphisms to ultrapowers
Given a group G and an ultrafilter p ∈ ω ∗ , denote by
ultp (G ) = (G )ω / ≡p , where f ≡p g iff {n : f (n) = g (n)} ∈ p.

By Lós theorem, ultp (G ) is a group with the same first order properties as G .
There is a natural embedding of G into ultp (G ) sending each g ∈ G to theequivalence class of the constant function with value g . We shall therefore
consider G as a subgroup of ultp (G ).
Every Φ ∈ Hom(G , T) naturally extends to a homomorphism Φ ∈ (ultp (G ), T) bylettingΦ([f ]) = p-limn∈ω Φ(f (n)).
We let τBohr denote the weakest topology making every Φ continuous, where Φ ∈Hom(G , T).

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201811 / 18

Extensions of homomorphisms to ultrapowers
Given a group G and an ultrafilter p ∈ ω ∗ , denote by
ultp (G ) = (G )ω / ≡p , where f ≡p g iff {n : f (n) = g (n)} ∈ p.

By Lós theorem, ultp (G ) is a group with the same first order properties as G .
There is a natural embedding of G into ultp (G ) sending each g ∈ G to theequivalence class of the constant function with value g . We shall therefore
consider G as a subgroup of ultp (G ).
Every Φ ∈ Hom(G , T) naturally extends to a homomorphism Φ ∈ (ultp (G ), T) bylettingΦ([f ]) = p-limn∈ω Φ(f (n)).
We let τBohr denote the weakest topology making every Φ continuous, where Φ ∈Hom(G , T).

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201811 / 18

Extensions of homomorphisms to ultrapowers
Given a group G and an ultrafilter p ∈ ω ∗ , denote by
ultp (G ) = (G )ω / ≡p , where f ≡p g iff {n : f (n) = g (n)} ∈ p.

By Lós theorem, ultp (G ) is a group with the same first order properties as G .
There is a natural embedding of G into ultp (G ) sending each g ∈ G to theequivalence class of the constant function with value g . We shall therefore
consider G as a subgroup of ultp (G ).
Every Φ ∈ Hom(G , T) naturally extends to a homomorphism Φ ∈ (ultp (G ), T) bylettingΦ([f ]) = p-limn∈ω Φ(f (n)).
We let τBohr denote the weakest topology making every Φ continuous, where Φ ∈Hom(G , T).

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201811 / 18

Extensions of homomorphisms to ultrapowers
Given a group G and an ultrafilter p ∈ ω ∗ , denote by
ultp (G ) = (G )ω / ≡p , where f ≡p g iff {n : f (n) = g (n)} ∈ p.

By Lós theorem, ultp (G ) is a group with the same first order properties as G .
There is a natural embedding of G into ultp (G ) sending each g ∈ G to theequivalence class of the constant function with value g . We shall therefore
consider G as a subgroup of ultp (G ).
Every Φ ∈ Hom(G , T) naturally extends to a homomorphism Φ ∈ (ultp (G ), T) bylettingΦ([f ]) = p-limn∈ω Φ(f (n)).
We let τBohr denote the weakest topology making every Φ continuous, where Φ ∈Hom(G , T).

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201811 / 18

Extensions of homomorphisms to ultrapowers
Given a group G and an ultrafilter p ∈ ω ∗ , denote by
ultp (G ) = (G )ω / ≡p , where f ≡p g iff {n : f (n) = g (n)} ∈ p.

By Lós theorem, ultp (G ) is a group with the same first order properties as G .
There is a natural embedding of G into ultp (G ) sending each g ∈ G to theequivalence class of the constant function with value g . We shall therefore
consider G as a subgroup of ultp (G ).
Every Φ ∈ Hom(G , T) naturally extends to a homomorphism Φ ∈ (ultp (G ), T) bylettingΦ([f ]) = p-limn∈ω Φ(f (n)).
We let τBohr denote the weakest topology making every Φ continuous, where Φ ∈Hom(G , T).

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201811 / 18

Iterated ultrapowers
The process can, of course, be iterated: given a group G , an ultrafilter p ∈ ω ∗ and0 < α 6 ω1 letβultαp (G ) = ultp (ultp (G )) if α = β + 1,andultα
p (G ) =[ultβp (G ) if α is limit.

0<β<α1
The group that will be relevant for us is the group ultωp (G ), endowed with the topology τBohr induced by the homomorphisms in Φ ∈ Hom(G , T) extended recursively
1
all the way to ultωp (G ) by the same formula as before:Φ([f ]) = p-limn∈ω Φ(f (n)).
The (iterated) ultrapower with this topology is usually not Hausdorff, so we identify
1
the inseparable functions and denote by Ultωp (G ) this quotient.

Fact
1
Ultω
p (G ) is a Hausdorff p-compact topological group.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201812 / 18

Iterated ultrapowers
The process can, of course, be iterated: given a group G , an ultrafilter p ∈ ω ∗ and0 < α 6 ω1 letβultαp (G ) = ultp (ultp (G )) if α = β + 1,andultα
p (G ) =[ultβp (G ) if α is limit.

0<β<α1
The group that will be relevant for us is the group ultωp (G ), endowed with the topology τBohr induced by the homomorphisms in Φ ∈ Hom(G , T) extended recursively
1
all the way to ultωp (G ) by the same formula as before:Φ([f ]) = p-limn∈ω Φ(f (n)).
The (iterated) ultrapower with this topology is usually not Hausdorff, so we identify
1
the inseparable functions and denote by Ultωp (G ) this quotient.

Fact
1
Ultω
p (G ) is a Hausdorff p-compact topological group.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201812 / 18

Iterated ultrapowers
The process can, of course, be iterated: given a group G , an ultrafilter p ∈ ω ∗ and0 < α 6 ω1 letβultαp (G ) = ultp (ultp (G )) if α = β + 1,andultα
p (G ) =[ultβp (G ) if α is limit.

0<β<α1
The group that will be relevant for us is the group ultωp (G ), endowed with the topology τBohr induced by the homomorphisms in Φ ∈ Hom(G , T) extended recursively
1
all the way to ultωp (G ) by the same formula as before:Φ([f ]) = p-limn∈ω Φ(f (n)).
The (iterated) ultrapower with this topology is usually not Hausdorff, so we identify
1
the inseparable functions and denote by Ultωp (G ) this quotient.

Fact
1
Ultω
p (G ) is a Hausdorff p-compact topological group.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201812 / 18

Iterated ultrapowers
The process can, of course, be iterated: given a group G , an ultrafilter p ∈ ω ∗ and0 < α 6 ω1 letβultαp (G ) = ultp (ultp (G )) if α = β + 1,andultα
p (G ) =[ultβp (G ) if α is limit.

0<β<α1
The group that will be relevant for us is the group ultωp (G ), endowed with the topology τBohr induced by the homomorphisms in Φ ∈ Hom(G , T) extended recursively
1
all the way to ultωp (G ) by the same formula as before:Φ([f ]) = p-limn∈ω Φ(f (n)).
The (iterated) ultrapower with this topology is usually not Hausdorff, so we identify
1
the inseparable functions and denote by Ultωp (G ) this quotient.

Fact
1
Ultω
p (G ) is a Hausdorff p-compact topological group.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201812 / 18

The plan

Theorem
There is (in ZFC) a countably compact Boolean group (subgroup of 2c ) without
non-trivial convergent sequences.

1
Fix an ultrafilter p ∈ ω ∗ and consider Ultωp (G ). There is a problem:
1
Does Ultωp (G ) have (non-trivial) convergent sequences?

Theorem
For every countable Abelian group G there is an ultrafilter p ∈ ω ∗ s.t. Ultp (G )has a non-trivial convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201813 / 18

The plan

Theorem
There is (in ZFC) a countably compact Boolean group (subgroup of 2c ) without
non-trivial convergent sequences.

1
Fix an ultrafilter p ∈ ω ∗ and consider Ultωp (G ). There is a problem:
1
Does Ultωp (G ) have (non-trivial) convergent sequences?

Theorem
For every countable Abelian group G there is an ultrafilter p ∈ ω ∗ s.t. Ultp (G )has a non-trivial convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201813 / 18

The plan

Theorem
There is (in ZFC) a countably compact Boolean group (subgroup of 2c ) without
non-trivial convergent sequences.

1
Fix an ultrafilter p ∈ ω ∗ and consider Ultωp (G ). There is a problem:
1
Does Ultωp (G ) have (non-trivial) convergent sequences?

Theorem
For every countable Abelian group G there is an ultrafilter p ∈ ω ∗ s.t. Ultp (G )has a non-trivial convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201813 / 18

The plan

Theorem
There is (in ZFC) a countably compact Boolean group (subgroup of 2c ) without
non-trivial convergent sequences.

1
Fix an ultrafilter p ∈ ω ∗ and consider Ultωp (G ). There is a problem:
1
Does Ultωp (G ) have (non-trivial) convergent sequences?

Theorem
For every countable Abelian group G there is an ultrafilter p ∈ ω ∗ s.t. Ultp (G )has a non-trivial convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201813 / 18

The plan works for selective ultrafilters and for ([ω]<ω , 4)
Proposition
p is selective iff for every p-independent set {fn : n ∈ ω} of functions fn : ω → [ω]<ω ,there is a sequence hUn : n ∈ ωi ⊆ p such that fn  Un is one-to-one for every n ∈ ωand {fn (m) : n ∈ ω and m ∈ Un } is linearly independent.
Given a non-empty set I , we shall call a set {fi : i ∈ I } of functions fi : ω → [ω]<ω
p-independent if)(Xfi (n) = ∅ ∈/pn: a +i∈E

for every non-empty finite set E ⊂ I and every a ∈ [ω]<ω .

Corollary<ω1
If p is selective, then Ultω) is a Hausdorff p-compact topological groupp ([ω]
without (non-trvial) convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201814 / 18

The plan works for selective ultrafilters and for ([ω]<ω , 4)
Proposition
p is selective iff for every p-independent set {fn : n ∈ ω} of functions fn : ω → [ω]<ω ,there is a sequence hUn : n ∈ ωi ⊆ p such that fn  Un is one-to-one for every n ∈ ωand {fn (m) : n ∈ ω and m ∈ Un } is linearly independent.
Given a non-empty set I , we shall call a set {fi : i ∈ I } of functions fi : ω → [ω]<ω
p-independent if)(Xfi (n) = ∅ ∈/pn: a +i∈E

for every non-empty finite set E ⊂ I and every a ∈ [ω]<ω .

Corollary<ω1
If p is selective, then Ultω) is a Hausdorff p-compact topological groupp ([ω]
without (non-trvial) convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201814 / 18

The plan works for selective ultrafilters and for ([ω]<ω , 4)
Proposition
p is selective iff for every p-independent set {fn : n ∈ ω} of functions fn : ω → [ω]<ω ,there is a sequence hUn : n ∈ ωi ⊆ p such that fn  Un is one-to-one for every n ∈ ωand {fn (m) : n ∈ ω and m ∈ Un } is linearly independent.
Given a non-empty set I , we shall call a set {fi : i ∈ I } of functions fi : ω → [ω]<ω
p-independent if)(Xfi (n) = ∅ ∈/pn: a +i∈E

for every non-empty finite set E ⊂ I and every a ∈ [ω]<ω .

Corollary<ω1
If p is selective, then Ultω) is a Hausdorff p-compact topological groupp ([ω]
without (non-trvial) convergent sequences.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201814 / 18

Different, yet the same
Lemma
There is a sequence hpα : α < ci ⊂ ω ∗ such that for every D ∈ [c]ω and every sequence hfα : α ∈ Di ⊂ ([ω]<ω )ω of one-to-one enumerations of linearly independent
sets there are hUα : α ∈ Di such thatUα ∈ pα for every α ∈ D, and{fα (n) : α ∈ D and n ∈ Uα } is a linearly independent subset of [ω]<ω .
We shall construct a countably compact group topology on [c]<ω starting from([ω]<ω , τBohr ) as follows: fix an indexed family {fα : α ∈ [ω, c)} ⊂ ([c]<ω )ω of
one-to-one sequences s.t.
1
for every infinite set X ⊆ [c]<ω there is an α ∈ [ω, c) with rng(fα ) ⊆ X ,2each fα is an enumeration of a linearly independent set, and3rng(fα ) ⊂ [α]<ω for every α ∈ [ω, c).
For every Φ ∈ Hom([ω]<ω , T) = Hom([ω]<ω , 2) define its extension Φ ∈ Hom([c]<ω, 2) recursively by putting
Φ({α}) = pα -limn∈ω Φ(fα (n)).
with the group topology τBohr induced by {Φ : Φ ∈ Hom([ω]<ω , 2)} on [c]<ω .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201815 / 18

Different, yet the same
Lemma
There is a sequence hpα : α < ci ⊂ ω ∗ such that for every D ∈ [c]ω and every sequence hfα : α ∈ Di ⊂ ([ω]<ω )ω of one-to-one enumerations of linearly independent
sets there are hUα : α ∈ Di such thatUα ∈ pα for every α ∈ D, and{fα (n) : α ∈ D and n ∈ Uα } is a linearly independent subset of [ω]<ω .
We shall construct a countably compact group topology on [c]<ω starting from([ω]<ω , τBohr ) as follows: fix an indexed family {fα : α ∈ [ω, c)} ⊂ ([c]<ω )ω of
one-to-one sequences s.t.
1
for every infinite set X ⊆ [c]<ω there is an α ∈ [ω, c) with rng(fα ) ⊆ X ,2each fα is an enumeration of a linearly independent set, and3rng(fα ) ⊂ [α]<ω for every α ∈ [ω, c).
For every Φ ∈ Hom([ω]<ω , T) = Hom([ω]<ω , 2) define its extension Φ ∈ Hom([c]<ω, 2) recursively by putting
Φ({α}) = pα -limn∈ω Φ(fα (n)).
with the group topology τBohr induced by {Φ : Φ ∈ Hom([ω]<ω , 2)} on [c]<ω .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201815 / 18

Different, yet the same
Lemma
There is a sequence hpα : α < ci ⊂ ω ∗ such that for every D ∈ [c]ω and every sequence hfα : α ∈ Di ⊂ ([ω]<ω )ω of one-to-one enumerations of linearly independent
sets there are hUα : α ∈ Di such thatUα ∈ pα for every α ∈ D, and{fα (n) : α ∈ D and n ∈ Uα } is a linearly independent subset of [ω]<ω .
We shall construct a countably compact group topology on [c]<ω starting from([ω]<ω , τBohr ) as follows: fix an indexed family {fα : α ∈ [ω, c)} ⊂ ([c]<ω )ω of
one-to-one sequences s.t.
1
for every infinite set X ⊆ [c]<ω there is an α ∈ [ω, c) with rng(fα ) ⊆ X ,2each fα is an enumeration of a linearly independent set, and3rng(fα ) ⊂ [α]<ω for every α ∈ [ω, c).
For every Φ ∈ Hom([ω]<ω , T) = Hom([ω]<ω , 2) define its extension Φ ∈ Hom([c]<ω, 2) recursively by putting
Φ({α}) = pα -limn∈ω Φ(fα (n)).
with the group topology τBohr induced by {Φ : Φ ∈ Hom([ω]<ω , 2)} on [c]<ω .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201815 / 18

Different, yet the same
Lemma
There is a sequence hpα : α < ci ⊂ ω ∗ such that for every D ∈ [c]ω and every sequence hfα : α ∈ Di ⊂ ([ω]<ω )ω of one-to-one enumerations of linearly independent
sets there are hUα : α ∈ Di such thatUα ∈ pα for every α ∈ D, and{fα (n) : α ∈ D and n ∈ Uα } is a linearly independent subset of [ω]<ω .
We shall construct a countably compact group topology on [c]<ω starting from([ω]<ω , τBohr ) as follows: fix an indexed family {fα : α ∈ [ω, c)} ⊂ ([c]<ω )ω of
one-to-one sequences s.t.
1
for every infinite set X ⊆ [c]<ω there is an α ∈ [ω, c) with rng(fα ) ⊆ X ,2each fα is an enumeration of a linearly independent set, and3rng(fα ) ⊂ [α]<ω for every α ∈ [ω, c).
For every Φ ∈ Hom([ω]<ω , T) = Hom([ω]<ω , 2) define its extension Φ ∈ Hom([c]<ω, 2) recursively by putting
Φ({α}) = pα -limn∈ω Φ(fα (n)).
with the group topology τBohr induced by {Φ : Φ ∈ Hom([ω]<ω , 2)} on [c]<ω .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201815 / 18

Different, yet the same
Lemma
There is a sequence hpα : α < ci ⊂ ω ∗ such that for every D ∈ [c]ω and every sequence hfα : α ∈ Di ⊂ ([ω]<ω )ω of one-to-one enumerations of linearly independent
sets there are hUα : α ∈ Di such thatUα ∈ pα for every α ∈ D, and{fα (n) : α ∈ D and n ∈ Uα } is a linearly independent subset of [ω]<ω .
We shall construct a countably compact group topology on [c]<ω starting from([ω]<ω , τBohr ) as follows: fix an indexed family {fα : α ∈ [ω, c)} ⊂ ([c]<ω )ω of
one-to-one sequences s.t.
1
for every infinite set X ⊆ [c]<ω there is an α ∈ [ω, c) with rng(fα ) ⊆ X ,2each fα is an enumeration of a linearly independent set, and3rng(fα ) ⊂ [α]<ω for every α ∈ [ω, c).
For every Φ ∈ Hom([ω]<ω , T) = Hom([ω]<ω , 2) define its extension Φ ∈ Hom([c]<ω, 2) recursively by putting
Φ({α}) = pα -limn∈ω Φ(fα (n)).
with the group topology τBohr induced by {Φ : Φ ∈ Hom([ω]<ω , 2)} on [c]<ω .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201815 / 18

Different, yet the same
Lemma
There is a sequence hpα : α < ci ⊂ ω ∗ such that for every D ∈ [c]ω and every sequence hfα : α ∈ Di ⊂ ([ω]<ω )ω of one-to-one enumerations of linearly independent
sets there are hUα : α ∈ Di such thatUα ∈ pα for every α ∈ D, and{fα (n) : α ∈ D and n ∈ Uα } is a linearly independent subset of [ω]<ω .
We shall construct a countably compact group topology on [c]<ω starting from([ω]<ω , τBohr ) as follows: fix an indexed family {fα : α ∈ [ω, c)} ⊂ ([c]<ω )ω of
one-to-one sequences s.t.
1
for every infinite set X ⊆ [c]<ω there is an α ∈ [ω, c) with rng(fα ) ⊆ X ,2each fα is an enumeration of a linearly independent set, and3rng(fα ) ⊂ [α]<ω for every α ∈ [ω, c).
For every Φ ∈ Hom([ω]<ω , T) = Hom([ω]<ω , 2) define its extension Φ ∈ Hom([c]<ω, 2) recursively by putting
Φ({α}) = pα -limn∈ω Φ(fα (n)).
with the group topology τBohr induced by {Φ : Φ ∈ Hom([ω]<ω , 2)} on [c]<ω .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201815 / 18

Different, yet the same
Lemma
There is a sequence hpα : α < ci ⊂ ω ∗ such that for every D ∈ [c]ω and every sequence hfα : α ∈ Di ⊂ ([ω]<ω )ω of one-to-one enumerations of linearly independent
sets there are hUα : α ∈ Di such thatUα ∈ pα for every α ∈ D, and{fα (n) : α ∈ D and n ∈ Uα } is a linearly independent subset of [ω]<ω .
We shall construct a countably compact group topology on [c]<ω starting from([ω]<ω , τBohr ) as follows: fix an indexed family {fα : α ∈ [ω, c)} ⊂ ([c]<ω )ω of
one-to-one sequences s.t.
1
for every infinite set X ⊆ [c]<ω there is an α ∈ [ω, c) with rng(fα ) ⊆ X ,2each fα is an enumeration of a linearly independent set, and3rng(fα ) ⊂ [α]<ω for every α ∈ [ω, c).
For every Φ ∈ Hom([ω]<ω , T) = Hom([ω]<ω , 2) define its extension Φ ∈ Hom([c]<ω, 2) recursively by putting
Φ({α}) = pα -limn∈ω Φ(fα (n)).
with the group topology τBohr induced by {Φ : Φ ∈ Hom([ω]<ω , 2)} on [c]<ω .
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201815 / 18

Different, yet the same
Call a set D ∈ [c]<ω suitably closed if ω ⊆ D andSn∈ω fα (n) ⊆ D for every α ∈ D.

Proposition
The group topology τBohr contains no non-trivial convergent sequences iff∀D ∈ [c]ω suitably closed ∀X ∈ [D]ω ∃Ψ ∈ Hom([D]<ω , 2) s.t.
1

∀α ∈ D Ψ({α}) = pα -limn∈ω Ψ(fα (n)), and2|X ∩ Ker(Ψ)| = |X \ Ker(Ψ)| = ω.

Now, if this happens (and it does by our choice of the ultrafilters) then, in particular,\
K=
Ker({Φ)Φ∈Hom([ω]<ω ,2)

is finite, and [c]<ω /K with the quotient topology is the Hausdorff countably compact
group without non-trivial convergent sequences we want.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201816 / 18

Different, yet the same
Call a set D ∈ [c]<ω suitably closed if ω ⊆ D andSn∈ω fα (n) ⊆ D for every α ∈ D.

Proposition
The group topology τBohr contains no non-trivial convergent sequences iff∀D ∈ [c]ω suitably closed ∀X ∈ [D]ω ∃Ψ ∈ Hom([D]<ω , 2) s.t.
1

∀α ∈ D Ψ({α}) = pα -limn∈ω Ψ(fα (n)), and2|X ∩ Ker(Ψ)| = |X \ Ker(Ψ)| = ω.

Now, if this happens (and it does by our choice of the ultrafilters) then, in particular,\
K=
Ker({Φ)Φ∈Hom([ω]<ω ,2)

is finite, and [c]<ω /K with the quotient topology is the Hausdorff countably compact
group without non-trivial convergent sequences we want.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201816 / 18

Different, yet the same
Call a set D ∈ [c]<ω suitably closed if ω ⊆ D andSn∈ω fα (n) ⊆ D for every α ∈ D.

Proposition
The group topology τBohr contains no non-trivial convergent sequences iff∀D ∈ [c]ω suitably closed ∀X ∈ [D]ω ∃Ψ ∈ Hom([D]<ω , 2) s.t.
1

∀α ∈ D Ψ({α}) = pα -limn∈ω Ψ(fα (n)), and2|X ∩ Ker(Ψ)| = |X \ Ker(Ψ)| = ω.

Now, if this happens (and it does by our choice of the ultrafilters) then, in particular,\
K=
Ker({Φ)Φ∈Hom([ω]<ω ,2)

is finite, and [c]<ω /K with the quotient topology is the Hausdorff countably compact
group without non-trivial convergent sequences we want.

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201816 / 18

Final remarks and questions
Theorem
For every n ∈ ω there is a group G such that G n is countably compact while G n+1
is not.

Questions1

Is there a countably compact group G without convergent sequences which is
not a torsion group, i.e., contains a copy of Z?

2(Wallace ’55) Is there a Hausdorff countably compact semigroup with bothsided cancellation which is not a topological group?

Yes to 1 implies Yes to 2.

Questions
Is there consistently a countably compact group G without convergent sequences of weight less than c?
Is there (in ZFC) a p-compact group G without convergent sequences?
Is there (in ZFC) a p ∈ ω ∗ and a group G s.t. Ultp (G ) has no (non-trivial)convergent sequences?
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201817 / 18

Final remarks and questions
Theorem
For every n ∈ ω there is a group G such that G n is countably compact while G n+1
is not.

Questions1

Is there a countably compact group G without convergent sequences which is
not a torsion group, i.e., contains a copy of Z?

2(Wallace ’55) Is there a Hausdorff countably compact semigroup with bothsided cancellation which is not a topological group?

Yes to 1 implies Yes to 2.

Questions
Is there consistently a countably compact group G without convergent sequences of weight less than c?
Is there (in ZFC) a p-compact group G without convergent sequences?
Is there (in ZFC) a p ∈ ω ∗ and a group G s.t. Ultp (G ) has no (non-trivial)convergent sequences?
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201817 / 18

Final remarks and questions
Theorem
For every n ∈ ω there is a group G such that G n is countably compact while G n+1
is not.

Questions1

Is there a countably compact group G without convergent sequences which is
not a torsion group, i.e., contains a copy of Z?

2(Wallace ’55) Is there a Hausdorff countably compact semigroup with bothsided cancellation which is not a topological group?

Yes to 1 implies Yes to 2.

Questions
Is there consistently a countably compact group G without convergent sequences of weight less than c?
Is there (in ZFC) a p-compact group G without convergent sequences?
Is there (in ZFC) a p ∈ ω ∗ and a group G s.t. Ultp (G ) has no (non-trivial)convergent sequences?
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201817 / 18

Final remarks and questions
Theorem
For every n ∈ ω there is a group G such that G n is countably compact while G n+1
is not.

Questions1

Is there a countably compact group G without convergent sequences which is
not a torsion group, i.e., contains a copy of Z?

2(Wallace ’55) Is there a Hausdorff countably compact semigroup with bothsided cancellation which is not a topological group?

Yes to 1 implies Yes to 2.

Questions
Is there consistently a countably compact group G without convergent sequences of weight less than c?
Is there (in ZFC) a p-compact group G without convergent sequences?
Is there (in ZFC) a p ∈ ω ∗ and a group G s.t. Ultp (G ) has no (non-trivial)convergent sequences?
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201817 / 18

Final remarks and questions
Theorem
For every n ∈ ω there is a group G such that G n is countably compact while G n+1
is not.

Questions1

Is there a countably compact group G without convergent sequences which is
not a torsion group, i.e., contains a copy of Z?

2(Wallace ’55) Is there a Hausdorff countably compact semigroup with bothsided cancellation which is not a topological group?

Yes to 1 implies Yes to 2.

Questions
Is there consistently a countably compact group G without convergent sequences of weight less than c?
Is there (in ZFC) a p-compact group G without convergent sequences?
Is there (in ZFC) a p ∈ ω ∗ and a group G s.t. Ultp (G ) has no (non-trivial)convergent sequences?
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201817 / 18

Final remarks and questions
Theorem
For every n ∈ ω there is a group G such that G n is countably compact while G n+1
is not.

Questions1

Is there a countably compact group G without convergent sequences which is
not a torsion group, i.e., contains a copy of Z?

2(Wallace ’55) Is there a Hausdorff countably compact semigroup with bothsided cancellation which is not a topological group?

Yes to 1 implies Yes to 2.

Questions
Is there consistently a countably compact group G without convergent sequences of weight less than c?
Is there (in ZFC) a p-compact group G without convergent sequences?
Is there (in ZFC) a p ∈ ω ∗ and a group G s.t. Ultp (G ) has no (non-trivial)convergent sequences?
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201817 / 18

Final remarks and questions
Theorem
For every n ∈ ω there is a group G such that G n is countably compact while G n+1
is not.

Questions1

Is there a countably compact group G without convergent sequences which is
not a torsion group, i.e., contains a copy of Z?

2(Wallace ’55) Is there a Hausdorff countably compact semigroup with bothsided cancellation which is not a topological group?

Yes to 1 implies Yes to 2.

Questions
Is there consistently a countably compact group G without convergent sequences of weight less than c?
Is there (in ZFC) a p-compact group G without convergent sequences?
Is there (in ZFC) a p ∈ ω ∗ and a group G s.t. Ultp (G ) has no (non-trivial)convergent sequences?
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201817 / 18

Final remarks and questions
Theorem
For every n ∈ ω there is a group G such that G n is countably compact while G n+1
is not.

Questions1

Is there a countably compact group G without convergent sequences which is
not a torsion group, i.e., contains a copy of Z?

2(Wallace ’55) Is there a Hausdorff countably compact semigroup with bothsided cancellation which is not a topological group?

Yes to 1 implies Yes to 2.

Questions
Is there consistently a countably compact group G without convergent sequences of weight less than c?
Is there (in ZFC) a p-compact group G without convergent sequences?
Is there (in ZFC) a p ∈ ω ∗ and a group G s.t. Ultp (G ) has no (non-trivial)convergent sequences?
U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201817 / 18

Thank you for your attention!

U. A. Ramos Garcı́a (UNAM)Countably compact groups

November 201818 / 18