Countably compact groups without non-trivial convergent sequences
Abstract.
We construct, in ZFC, a countably compact subgroup of without non-trivial convergent sequences, answering an old problem of van Douwen. As a consequence we also prove the existence of two countably compact groups and such that the product is not countably compact, thus answering a classical problem of Comfort.
Key words and phrases:
Products of countably compact groups, -compact groups, ultrapowers, countably compact groups without convergent sequences2010 Mathematics Subject Classification:
Primary 22A05, 03C20; Secondary 03E05, 54H111. Introduction
The celebrated Comfort-Ross theorem [11, 7] states that any product of pseudo-compact topological groups is pseudo-compact, in stark contrast with the examples due to Novák [28] and Terasaka [34] who constructed pairs of countably compact spaces whose product is not even pseudo-compact. This motivated Comfort [9] (repeated in [8]) to ask:
Question 1.1 (Comfort [8]).
Are there countably compact groups such that is not countably compact?
The first consistent positive answer was given by van Douwen [45] under MA, followed by Hart-van Mill [21] under MActble. In his paper van Douwen showed that every Boolean countably compact group without non-trivial convergent sequences contains two countably compact subgroups whose product is not countably compact, and asked:
Question 1.2 (van Douwen [45]).
Is there a countably compact group without non-trivial convergent sequences?
In fact, the first example of such a group was constructed by Hajnal and Juhász [20] a few years before van Douwen’s [45] assuming CH. Recall, that every compact topological group contains a non-trivial convergent sequence, as an easy consequence of the classical and highly non-trivial theorem of Ivanovskiĭ-Vilenkin-Kuz’minov (see [25]) that every compact topological group is dyadic, i.e., a continuous image of for some cardinal number .
Both questions have been studied extensively in recent decades, providing a large variety of sufficient conditions for the existence of examples to these questions, much work being done by Tomita and collaborators [17, 18, 23, 30, 33, 40, 41, 42, 37, 43, 44, 38], but also others [10, 13, 14, 27, 35]. The questions are considered central in the theory of topological groups [1, 2, 7, 8, 15, 32, 36].
Here we settle both problems by constructing in ZFC a countably compact subgroup of without non-trivial convergent sequences.
The paper is organized as follows: In Section 2 we fix notation and review basic facts concerning ultrapowers, Fubini products of ultrafilters and Bohr topology. In Section 3 we study van Douwen’s problem in the realm of -compact groups. We show how iterated ultrapowers can be used to give interesting partial solutions to the problem. In particular, we show that an iterated ultrapower of the countable Boolean group endowed with the Bohr topology via a selective ultrafilter produces a -compact subgroup of without non-trivial convergent sequences. This on the one hand raises interesting questions about ultrafilters, and on the other hand serves as a warm up for Section 4, where the main result of the paper is proved by constructing a countably compact subgroup of without non-trivial convergent sequences using not a single ultrafilter, but rather a carefully constructed -sized family of ultrafilters.
2. Notation and terminology
Recall that an infinite topological space is countably compact if every infinite subset of has an accumulation point. Given a nonprincipal ultrafilter on (for short, ), a point and a sequence we say (following [5]) that - if for every open containing the set . It follows that a space is countably compact if and only if every sequence has a -limit in for some ultrafilter . Given an ultrafilter , a space is -compact if for every sequence there is an such that -.
For introducing the following definition, we fix a bijection , and for a limit ordinal , we pick an increasing sequence of smaller ordinals with supremum . Given an ultrafilter , the iterated Fubini powers or Frolík sums [16] of are defined recursively as follows:
The choice of the ultrafilter depends on (the arbitrary) choice of and the choice of the sequence , however, the type of does not (see e.g., [16, 19]).
For our purposes we give an alternative definition of the iterated Fubini powers of : given we fix a well-founded tree such that
-
(i)
, where denotes the rank function on ;
-
(ii)
For every , if then for all .
For , let and .
If , then will be used to denote the collection of all trees such that for every the set belongs to . Notice that each is also a well-founded tree with . Moreover, the family forms a base of an ultrafilter on which has the same type of . If and , denotes the tree in for which for all .
Next we recall the ultrapower construction from model theory and algebra. Given a group and an ultrafilter , denote by
The Theorem of Łós [26] states that for any formula with parameters , if and only if . In particular, is a group with the same first order properties as .
There is a natural embedding of into sending each to the equivalence class of the constant function with value . We shall therefore consider as a subgroup of . Also, without loss of generality, we can assume that for every .
Recall that the Bohr topology on a group is the weakest group topology making every homomorphism continuous, where the circle group carries the usual compact topology. We let denote equipped with the Bohr topology.
Finally, our set-theoretic notation is mostly standard and follows [24]. In particular, recall that an ultrafilter is a -point if every function on is finite-to-one or constant when restricted to some set in the ultrafilter and, an ultrafilter is a -point if every finite-to-one function on becomes one-to-one when restricted to a suitable set in the ultrafilter. The ultrafilters which are P-point and Q-point are called selective ultrafilters. For more background on set-theoretic aspects of ultrafilters see [6].
3. Iterated ultrapowers as -compact groups
In this section we shall give a canonical construction of a -compact group for every ultrafilter . This will be done by studying the iterated ultrapower construction.
Fix a group and put . Given an ordinal with , let
where denotes the direct limit of the direct system with the following properties:
-
(1)
is the identity function on , and
-
(2)
is the canonical embedding of into , defined recursively by the constant function with value , and the direct limit of for a limit ordinal .
In what follows, we will abbreviate for . Moreover, we will treat as and, in such case, we put for every . This is, of course, formally wrong, but is facilitated by our identification of with a subgroup of . In this way we can avoid talking about direct limit constructions.
We now consider . Having fixed an ultrafilter , this topology naturally lifts to a topology on as follows: Every naturally extends to a homomorphism by letting
(3.1) |
By Łós’s theorem, is indeed a homomorphism from to and hence the weakest topology making every continuous, where , is a group topology on . This topology will be denoted by .
The following is a trivial, yet fundamental fact:
Lemma 3.1.
For every , - in .
Proof.
This follows directly from the definition of and the identification of with a subgroup of . ∎
The group that will be relevant for us is the group , endowed with the topology induced by the homomorphisms in extended recursively all the way to by the same formula (3.1).
The (iterated) ultrapower with this topology is usually not Hausdorff (see [12, 3]), so we identify the inseparable functions and denote by this quotient. More explicitly,
where . The natural projection will be denoted by
The main reason for considering the iterated Fubini powers here is the following simple and crucial fact:
Proposition 3.2.
Let be an ultrafilter.
-
(1)
for , and
-
(2)
is a Hausdorff -compact topological group.
Proof.
To prove (1), fix an . For given , recursively define a tree and a function so that
-
•
and , where ;
-
•
if is defined say , then and for every .
We define given by
Claim 3.3.
is an isomorphism.
Proof of the claim. To see that is a surjection, let be such that for some . Consider the function defined recursively by
-
•
and,
-
•
if , then .
Notice that the function satisfies that for every . In particular, and, a routine calculation shows that .
To see that is injective, suppose that . Then there exists a tree such that
If set , then we can verify recursively that . Therefore, is a one-to-one function.
Finally, using again a recursive argument, one can check that preserves the group structure.
To prove (2) note that by definition is a Hausdorff topological group. To see that is -compact, since is a continuous image of , it suffices to check that is -compact. Let be a sequence and let . So , that is, there exists such that . Thus, for every there exists such that and hence for . Then - in as by the construction - for every . This gives us the -compactness of . ∎
The plan for our construction is as follows: fix an ultrafilter , find a suitable topological group without convergent sequences and consider . The remaining issue is: Does have non-trivial convergent sequences?
While our approach is applicable to an arbitrary group , in the remainder of this paper we will be dealing exclusively with Boolean groups, i.e., groups where each element is its own inverse.111The general case will be dealt with in a separate paper. These groups are, in every infinite cardinality , isomorphic to the group with the symmetric difference as the group operation and as the neutral element. Every Boolean group is a vector space over the trivial -element field which we identify with . Hence, we can talk, e.g., about linearly independent subsets of a Boolean group. Also, since every homomorphism from a Boolean group into the torus takes at most two values (in the unique subgroup of of size ) we may and will identify with to highlight the fact that there are only two possible values. Hence also is a Boolean group and a vector space over the same field.
The following theorem is the main result of this section.
Theorem 3.4.
Let be a selective ultrafilter. Then is a Hausdorff -compact topological Boolean group without non-trivial convergent sequences.
In order to prove this theorem, we apply the first step of our plan.
Proposition 3.5.
The group endowed with the topology is a non-discrete Hausdorff topological group without non-trivial convergent sequences.
Proof.
It is well-known and easy to see that is a non-discrete Hausdorff group topology (e.g., see [2] Section 9.9). To see that has no non-trivial convergent sequences, assume that is a non-trivial sequence. Then is an infinite set. Find an infinite linearly independent set and split it into two infinite pieces and , and take such that for every . Therefore, is a witness that the sequence does not converge. ∎
We say that a sequence is -separated if for every there is a such that . In other words, a sequence is -separated if and only if its elements represent distinct elements of
where and is the corresponding projection.
We next show that, in general, the plan does not work for all .
Lemma 3.6.
The following are equivalent:
-
(1)
There exists a such that has non-trivial convergent sequences.
-
(2)
There exist a sequence and a mapping
such that for every the familyis centered.
Proof.
Let us prove (1) implies (2). Let be a non-trivial sequence, say () where . Without loss of generality we can assume that is a one-to-one function converging to , here denotes the constant sequence where each term is . So is a -separated sequence -converging to , where for . By taking a subsequence if necessary, we may assume that for every there is a such that . Now, by -convergence of , there is a mapping such that for each and each it follows that . Now we will check that for every the family is centered.222For a subset of the group , . For this, since is without non-trivial convergent sequences and , we may assume that for every , that is, is infinite for all . Now, fix and let be a finite set such that for every . Then for every and hence there exists such that for every . Since , there exists such that for every . Put . Then , so we are done.
To prove (2) implies (1), first we observe that there is a sequence such that for each and every there exists such that for all . Now, define and for all .
Fix and as in (2).
Claim 3.7.
The collection forms a centered family which generates a free filter .
Proof of the claim. To show that such family is centered, let and for every fix a finite set . Then, considering all choice functions
we can ensure that
is an infinite set.
To see that the filter is free, let . If there is an such that , then . In another case, since does not -converge to , there exists such that for infinitely many . Then pick one of such with and, .
Let extend . By Claim 3.7, it follows that
-
(i)
, for every .
-
(ii)
The sequence converges to , for every , i.e., is a -convergent sequence to .
Finally, taking a subsequence if necessary, we can assume that is -separated and, hence is a non-trivial convergent sequence in . ∎
Remark 3.8.
Note that the filter is actually an -filter.
Theorem 3.9.
There exists a such that has non-trivial convergent sequences.
Proof.
We will show that the second clause of the Lemma 3.6 holds. To see this, choose any countable linearly independent set . Let be a vector subspace of such that . We define the mapping as follows:
Now, let and fix a finite set . In order to show that
is infinite, we shall need a fact concerning linear functionals on a vector space.
Fact 3.10 ([31], p. 124).
Let be a vector space and linear functionals on . Then the following statements are equivalent:
-
(1)
.
-
(2)
.
Using this fact, and noting that , one sees that
Pick an arbitrary and put
Then is an infinite set, and hence is an infinite set too. But
so we are done. ∎
Corollary 3.11 ().
There is a P-point such that has non-trivial convergent sequences.
Proof.
It is well-known (e.g., see [22]) that assuming every -filter can be extended to a P-point. ∎
As is a topological subgroup of there are ultrafilters (even P-points assuming CH) such that has a non-trivial convergent sequence.
Selective ultrafilters and Q-points, have immediate combinatorial reformulations relevant in our context. Given a non-empty set and a Boolean group, we shall call a set of functions -independent if
for every non-empty finite set and every . Note that, in particular, a function is not constant on an element of if and only if is -independent. Now, we will say that a function is linearly independent if is one-to-one and is a linearly independent set and, a function is -independent if is one-to-one and is a -independent set, where for .
Proposition 3.12.
Let be an ultrafilter. Then:
-
(1)
is a Q-point if and only if for every finite-to-one function there is a set such that is linearly independent.
-
(2)
The following are equivalent
-
(a)
is selective;
-
(b)
for every function which is not constant on an element of there is a set such that is linearly independent;
-
(c)
for every -independent set of functions , there is a set and a function so that is one-to-one for , if , and
is linearly independent.333Here denotes the disjoint union.
-
(a)
Proof.
Let us prove (1). Suppose first that is a Q-point. Let be a finite-to-one function. Recursively define a strictly increasing sequence of elements of and a strictly increasing sequence of finite subgroups of so that
-
(i)
for all , and
-
(ii)
, for all .
Then partitioning into the union of even intervals, and the union of odd intervals, one of them is in , say
Applying Q-pointness we can assume that there exists a such that
and . By item (ii) and since is a strictly increasing sequence, it follows that is one-to-one and is linearly independent.
Suppose now that for every finite-to-one function there is a such that is one-to-one and is linearly independent. Let be a partition of into finite sets. Define a finite-to-one function by putting for each . Then there is an such that is one-to-one and is linearly independent. Note that necessarily for every and therefore is a Q-point.
(2) To see (a) implies (b), let be a function which is not constant on an element of . Using P-pointness, we may assume without loss of generality that is a finite-to-one function. So, by item (1), there is an such that is one-to-one and is linearly independent.
To see (b) implies (a), let be a function which is not constant on an element of . By item (b), there is an such that is one-to-one and is linearly independent, and hence is a P-point. To verify that is a Q-point, notice that every finite-to-one function is not constant on an element of . Thus, by clause (1) we get the desired conclusion.
To prove (a) implies (c), assume that is a -independent set of functions .
Fact 3.13.
Given a finite -independent set , and a finite linearly independent set , the set of all such that is linearly independent, belongs to .
Using Fact 3.13, we can recursively construct a -branching tree such that for every , it follows that
where .
By Galvin-Shelah’s theorem ([4, Theorem 4.5.3]), let be a branch such that . Thus, if we put and for , we get the properties as in (c).
Finally, notice that (b) is a particular instance of (c) when . Therefore, (c) implies (b). ∎
Remark 3.14.
In the previous theorem, it is possible to change the group to any arbitrary Boolean group and, the conclusions of the theorem remain true.
For technical reasons, it will be necessary to reformulate the notion of -independence.
Lemma 3.15.
Let be a Boolean group and . Then:
-
(1)
A set of functions is -independent if and only if the function
defined by for is linearly independent, where denotes the natural projection.
-
(2)
A set of functions is -independent if and only if the set of functions is a -independent set, where each is defined by for and
denotes the natural projection.
Proof.
To see (1), note that
iff
for every non-empty finite set and every .
To see (2). Let be a non-empty finite set and and, notice that
iff
iff
where for some we have that for . ∎
Note also that if for , then is not constant on an element of (equivalently, is -independent).
Lemma 3.16.
Let , and a selective ultrafilter. If is not constant on an element of , then there is a tree with such that is linearly independent.444Here, we are using the notation from the proof of Proposition 3.2 (1).
Proof.
First, if , then the conclusion of the lemma follows from Proposition 3.12 (2) (b). Thus, we may assume that .
We plan to construct a tree with , so that the following hold for any :
-
•
if , then forms a -independence sequence;
-
•
if , then forms a linearly independent sequence.
In order to do this, first, we recursively construct a tree with , so that the following hold for any with :
-
•
if , then forms a linearly independent sequence;
-
•
if with , then forms a -independent sequence;
-
•
if is a limit ordinal, then is a strictly increasing sequence of non-zero ordinals.
At step . If and is not constant on an element of , then and applying Proposition 3.12 (2) (b) there exists with such that is linearly independent. Therefore, in this case we put .
If with and is not constant on an element of , then consider the sequence
defined by for . Since is not constant on an element of , by Lemma 3.15 (2), the sequence is not constant on an element of . Therefore, applying Proposition 3.12 (2) (b) and Remark 3.14, we can find an element with such that is linearly independent. Thus, by Lemma 3.15 (1), putting we can conclude that forms a -independent sequence.
If is a limit ordinal, then for every we set . Then
where each . The selectiveness of implies that there is an such that for every . Thus, in this case put . This concludes recursive construction of .
Notice that for every . Now given a tree with , we can canonically list its members as so that
-
•
entails ;
-
•
, , is a limit ordinal, and entails ;
-
•
, , is a successor ordinal, and entails .
Choose a sufficiently large regular cardinal and a countable elementary submodel of containing all the relevant objects such as and . Fix so that is a pseudo-intersection of . Put and for .
We unfix , and construct by recursion on the required condition with , as well as an auxiliary function and sets for such that the following are satisfied:
-
(a)
for all (by definition).
-
(b)
For all ,
-
•
if , then
forms a linearly independent sequence;
-
•
if with , then
forms a -independence sequence;
-
•
if is a limit ordinal, then
forms an one-to-one sequence, and
-
•
Before describing the construction let us recall a simple fact from linear algebra:
Fact 3.17.
Let and be linearly independent sets in a Boolean group with a finite set. Then there is such that and is linearly independent.
Proof of the fact. Let and . Then , so there exists a set such that . Therefore, and is linearly independent.
Basic step . So . We put and hence . The conditions (a) and (b) are immediate.
Recursion step . Assume (for ) as well as have been defined so as to satisfy (a) and (b). In particular, we know already , for it is of the form for some where . Put and assume . Note that, since (b) is satisfied for , we must have and
is a -independent sequence. Put
and .
If , then applying Proposition 3.12 (2) (c) there exists and a function such that
is a linearly independent sequence. Using the elementarity of and our assumption about we conclude that there exists a function such that
is a linearly independent sequence. Note that and for . Since is a finite set, using Fact 3.17, we can find a natural number so that
forms a linearly independent sequence, as required.
For the case with , we will proceed in a similar way as the previous case. Given , let
be defined by for . By Lemma 3.15 (2), is a -independent set. Thus, applying Proposition 3.12 (2) (c) and Remark 3.14, we can find an element and a function such that
is a linearly independent sequence. By elementarity of and the property of we have that there exists a function such that
is a linearly independent sequence. Since is a finite set, and for , using Fact 3.17, we can find a natural number so that
forms a linearly independent sequence and, by Lemma 3.15 (1), this means that
forms a -independent sequence, as required.
If is a limit ordinal, then applying Proposition 3.12 (2) (c) there exists and a function such that
is a linearly independent sequence. Thus, proceeding as previous cases, it is possible to find a function and a natural number so that
forms a linearly independent sequence. In particular,
forms an one-to-one sequence and, since is a limit ordinal, one sees that without loss of generality, we may assume that
as required. ∎
Now we are ready to prove the main theorem of this section.
Proof of the Theorem 3.4. According to Proposition 3.2, is a Hausdorff -compact topological group. It remains therefore only to show that contains no non-trivial convergent sequences to . To see this, let be a non-trivial sequence, say () where . Without loss of generality we can assume that is a one-to-one function. Thus, since
there exists so that and is not constant on an element of . By Lemma 3.16, there is a tree with such that is linearly independent. Note that . Take so that . So and hence . Thus, is a witness that the sequence does not -converge to and, since is one-to-one, in fact does not converge to .
4. Countably compact group without convergent sequences
In this section we develop the ideas introduced in the previous section into a ZFC construction of a countably compact subgroup of without non-trivial convergent sequences. Recall that any boolean group of size (in particular ) is isomorphic to . In fact, the extension of homomorphisms produces a (topological and algebraic) embedding of into defined by
Similarly to the ultrapower construction, we shall extend the Bohr topology on to a group topology on to obtain the result. The difference is that rather than using a single ultrafilter, we shall use a carefully constructed -sized family of ultrafilters.
Theorem 4.1.
There is a Hausdorff countably compact topological Boolean group without non-trivial convergent sequences.
Proof.
We shall construct a countably compact topology on starting from as follows:
Fix an indexed family of one-to-one sequences such that
-
(1)
for every infinite there is an with ,
-
(2)
each is a sequence of linearly independent elements, and
-
(3)
for every .
Given a sequence define for every its extension recursively by putting
Note that together with the independent set generate the group so the above definition uniquely extends to a homomorphism .
This allows us to define the topology induced by on as the weakest topology making all continuous (for ), or equivalently, the group topology having as a subbasis of the filter of neighbourhoods of the neutral element . It follows directly from the above observation that independently of the choice of the ultrafilters the topology is a countably compact group topology on . Indeed, for every , in fact .
Call a set suitably closed if and for every . The following claim shows that the construction is locally countable.
Claim 4.2.
The topology contains no non-trivial convergent sequences if and only if
-
(1)
;
-
(2)
.
Proof of the claim. Given an infinite there is an such that . Let be suitably closed with , and let be the given homomorphism. It follows directly from the definition, and property (1) of , that, if then in turn , which implies that (and hence also ) is not a convergent sequence as takes both values and infinitely often on the set .
The reverse implication is even more trivial (and not really necessary for the proof).
Note that if this happens then, in particular,
is finite, and with the quotient topology is the Hausdorff countably compact group without non-trivial convergent sequences we want.
Hence to finish the proof it suffices to produce a suitable family of ultrafilters.
Claim 4.3.
There is a family of free ultrafilters on such that for every and such that each is an one-to-one enumeration of linearly independent elements of there is a sequence such that
-
(1)
is a family of pairwise disjoint subsets of ,
-
(2)
for every , and
-
(3)
is a linearly independent subset of .
Proof of the claim. Fix a partition of into finite sets such that
and let
Note that is a centered family, and denote by the filter it generates. Note also, that if is an infinite subset of then .
Let be any almost disjoint family of size of infinite subsets of , and let, for every , be any ultrafilter on extending .
To see that this works, let and a family of one-to-one sequences of linearly independent elements of be given. Let be a partition of such that for every , and recursively define a set such that, ,
for every , and
In order to obtain the set we recursively use Fact 3.17 to construct a sequence of finite sets such that:
-
•
;
-
•
If , then
-
(i)
,
-
(ii)
, and
-
(iii)
is linearly independent, where is such that .
-
(i)
Put . By (ii), it follows that . Since and , is clear that . By (iii), it follows that is linearly independent. Therefore, the sequence is as required.
Now, use this family of ultrafilters as the parameter in the construction of the topology described above. By Claim 4.2 it suffices to show that given a suitably closed and there is a homomorphism such that
-
(1)
-
(2)
.
By Claim 4.3, there is a sequence such that
-
(1)
is a family of pairwise disjoint subsets of ,
-
(2)
for every , and
-
(3)
is a linearly independent subset of .
Enumerate as so that . Recursively define a function so that
-
(1)
takes both values and infinitely often on ,
-
(2)
-, and
-
(3)
if is in the subgroup generated by then , and making sure that
-
(4)
Where is a homomorphism defined on the subgroup generated by
extending . Then let be any homomorphism extending . Doing this is straightforward given that the set
is linearly independent.
Finally, note that if we, for , let
then is a continuous homomorphism from to whose kernel is the same group , which defines a homeomorphism (and isomorphism) of onto a subgroup of . ∎
5. Concluding remarks and questions
Even though the results of the paper solve longstanding open problems, they also open up very interesting new research possibilities. In Theorem 3.4 we showed that if is a selective ultrafilter then is a -compact group without non-trivial convergent sequences. This raises the following two interesting questions, the first of which is the equivalent of van Douwen’s problem for -compact groups.
Question 5.1.
Is there in ZFC a Hausdorff -compact topological group without a non-trivial convergent sequence?
A closely related problem asks how much can the property of being selective be weakened in Theorem 3.4. Recall that by Corollary 3.11 it is consistent that there is a P-point for which does contain a non-trivial convergent sequence. On the other hand, for every , so there are consistently non-P-points for which contains no non-trivial convergent sequences.
Question 5.2.
Is the existence of an ultrafilter such that contains no non-trivial convergent sequences equivalent to the existence of a selective ultrafilter?
Question 5.3.
Is it consistent with ZFC that contains a non-trivial convergent sequence for every ultrafilter ?
Assuming contains no non-trivial convergent sequences, it is easy to construct for every a subgroup of , such that is countably compact while is not. It should be possible to modify the construction in Theorem 4.1 to construct such groups in ZFC. These issues will be dealt with in a separate paper.
Another interesting question is:
Question 5.4.
Is it consistent with that there is a Hausdorff countably compact topological group without non-trivial convergent sequences of weight ?
Finally, let us recall a 1955 problem of Wallace:
Question 5.5 (Wallace [46]).
Is every both-sided cancellative countably compact topological semigroup necessarily a group?
It is well known that a counterexample can be recursively constructed inside of any non-torsion countably compact topological group without non-trivial convergent sequences [29, 39]. The fact that we do not know how to modify (in ZFC) the construction in Theorem 4.1 to get a non-torsion example of a countably compact group seems surprising. Also the proof of Theorem 3.4 does not seem to easily generalize to non-torsion groups. Hence:
Question 5.6.
Is there, in ZFC , a non-torsion countably compact topological group without non-trivial convergent sequences?
Question 5.7.
Assume is a selective ultrafilter. Does contain no non-trivial convergent sequence?
Here the is defined as before as the weakest topology on which makes all extensions of homomorphisms from to continuous, and the group with being the intersection of all kernels of the extended homomorphisms.
Acknowledments. The authors would like to thank Alan Dow and Osvaldo Guzmán for stimulating conversations. The authors also wish to thank the anonymous referee for a thorough reading of the text and for helpful suggestions.
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