A van Douwen-like ZFC theorem for small powers of countably compact groups without non-trivial convergent sequences
aa r X i v : . [ m a t h . GN ] S e p A VAN DOUWEN-LIKE ZFC THEOREM FOR SMALLPOWERS OF COUNTABLY COMPACT GROUPSWITHOUT NON-TRIVIAL CONVERGENTSEQUENCES
ARTUR HIDEYUKI TOMITA
To the Memory of Professor W. W. Comfort
Abstract.
We show that if κ ≤ ω and there exists a group topol-ogy without non-trivial convergent sequences on an Abelian group H such that H n is countably compact for each n < κ then thereexists a topological group G such that G n is countably compactfor each n < κ and G κ is not countably compact. If in addition H is torsion, then the result above holds for κ = ω . Combining withother results in the literature, we show that: a ) Assuming c incomparable selective ultrafilters, for each n ∈ ω ,there exists a group topology on the free Abelian group G such that G n is countably compact and G n +1 is not countably compact. (Itwas already know for ω ). b ) If κ ∈ ω ∪ { ω } ∪ { ω } , there exists in ZFC a topological group G such that G γ is countably compact for each cardinal γ < κ and G κ is not countably compact. Introduction
Countably compact groups without non-trivial conver-gent sequences and van Douwen’s theorem.
In 1980, van Douwen[4] showed in ZFC that if there exists a countably compact group with-out non-trivial convergent sequences that is a subgroup of 2 c , thenthere exist two countably compact subgroups whose product is notcountably compact. In the same paper, van Douwen produced a count-ably compact subgroup of 2 c , without non-trivial convergent sequences The author has received support from FAPESP Aux´ılio regular de pesquisa2012/01490-9 and CNPq Produtividade em Pesquisa 305612/2010-7 during the re-search that led to this work. The author received support from CNPq Produtivi-dade em Pesquisa 307130/2013-4 and CNPq Projeto Universal N. 483734/2013-6Universal during the preparation and revision of the manuscript. Final revisionunder support from FAPESP 2016/26216-8.2010
Mathematics Subject Classification.
Primary 54H11, 22A05; Secondary54A35, 54G20.
Key words and phrases. countable compactness, van Douwen, topological group,Tychonoff product, ZFC, non trivial convergent sequences. from Martin’s Axiom (a CH example was earlier produced by H´ajnaland Juh´asz [7]). Combining his two results, it follows that countablecompactness in the class of topological groups is not productive whenMartin’s Axiom holds.In 1991, Hart and van Mill [8] produced from Martin’s Axiom forcountable posets the first countably compact group whose square isnot countably compact. The approach differs from van Douwen’s astheir example contains many convergent sequences.In 2004, Tomita and Watson [22] constructed p -compact groups (inparticular countably compact groups) from a selective ultrafilter p .With this technique, Garcia-Ferreira, Tomita and Watson [5] obtaineda countably compact group without non-trivial convergent sequencesfrom a single selective ultrafilter.Szeptycki and Tomita [12] showed that in the Random model thereexists a countably compact group without non-trivial convergent se-quences, giving the first example that does not depend on selectiveultrafilters. In [13], it was showed that the countable power of thisexample is countably compact.Recently, Hruˇsak, van Mill, Ramos and Shelah showed in ZFC thatthere exists a Boolean group without non-trivial convergent sequences,answering a question of van Douwen from 1980. Using van Douwen’stheorem, they also answer in ZFC a 1966 question of Comfort aboutthe non-productivity of countable compactness in the class of topolog-ical groups. Hruˇsak has informed that the construction can be easilymodified to obtain G ω countably compact.In 2005, Tomita [19] improved van Douwen’s theorem showing thatif there exists a countably compact Abelian group without non-trivialconvergent sequences then there exists a countably compact Abeliangroup whose square is not countably compact.Tkachenko [14] showed that under CH, there exists a countably com-pact free Abelian group. Koszmider, Tomita and Watson [9] obtainedone from Martin’s Axiom for countable posets and Madariaga Garciaand Tomita [10] obtained an example from the existence of c selectiveultrafilters.1.2. Powers of countably compact groups and Comfort’s Ques-tion.
Inspired by the example of Hart and van Mill and a result ofGinsburg and Saks [6], Comfort asked the following question in theOpen Problems in Topology [3]:
Question 1. (Question 477, Open Problems in Topology, 1990) Isthere, for every (not necessarily infinite) cardinal number κ ≤ c , a MALL POWERS OF COUNTABLY COMPACT GROUPS 3 topological group G such that G γ is countably compact for all cardinals γ < κ , but G κ is not countably compact? The result in [6] implies that the question above only makes sensefor cardinals not greater than 2 c .Partial results for some finite cardinals other than 2 were obtainedin [15] and [17] using Martin’s Axiom for countable posets. Under thesame axiom, finite cardinals were solved in [18].In [20] the question was answered consistently. It was showed undersome cardinal restriction and assuming the existence of 2 c incomparableselective ultrafilters, there exists a topological group as in Comfort’squestion for every cardinal ≤ c . The examples obtained are subgroupsof a product of copies of 2.Sanchis and Tomita [11] showed that if there exists a selective ul-trafilter then there exists a topological group of order 2 without non-trivial convergent sequences as in Comfort’s question, for each cardinal α ≤ ω .In [9] and [10], it was showed that there exist a countably compactgroup topology on the free Abelian group of cardinality c whose squareis not countably compact from Martin’s Axiom and the existence of c selective ultrafilters, respectively.Tomita [21] showed that, assuming the existence of c incomparableselective ultrafilters, there exists a topological free Abelian group with-out non-trivial convergent sequences whose finite powers are countablycompact, improving the square result by Boero and Tomita [1].1.3. A van Douwen like theorem for Comfort’s Question.
Inthis work we obtain a van Douwen like theorem for Abelian groups with-out non-trivial convergent sequences whose small powers are countablycompact.We show that given a cardinal κ ≤ ω and there exists a topologicalAbelian group of finite order such that H γ is countably compact foreach cardinal γ < κ , then there exists a topological group G as inComfort’s question 477 for κ .Applying the recent result of Hruˇsak, van Mill, Ramos and Shelah, itfollows that Comfort’s Question 477 is settled in ZFC for every cardinal ≤ ω .We also show that if κ ≤ ω and there exists a non-torsion topologicalAbelian group such that G γ is countably compact for each γ < κ , thenthere exists a topological free Abelian group as in Comfort’s questionfor κ . Applying this result in the example in [21], it follows from the A. H. TOMITA existence of c selective ultrafilters that there exists, for each finite car-dinal M , a topological free Abelian group G such that G M is countablycompact, but G M +1 is not.2. Countably compact non torsion Abelian groups
Family of sequences that suffice to keep countable com-pactness in small powers.
We start defining the families that willhelp us obtain the countable compactness in small powers.
Notation 1.
Given m ∈ ω and a free Abelian group G , define F ( G, m ) ,as the set of m -uples ( f , . . . , f m − ) , where A is an infinite subset of ω , with f i : A −→ G for each i < m and for each nonzero func-tion s : m −→ Z there exists a finite set F s such that the sequence ( P i Given F ∈ Z ( c × ω ) and a family { z α,i : ( α, i ) ∈ supp F } ofelements of some group H , z F denotes the sum P ( α,i ) ∈ supp F F ( α, i ) .z α,i .If A is a subset of ω , g : A −→ Z ( c × ω ) and { z α,i : ( α, i ) ∈ S { supp g ( n ) : n ∈ A }} , we denote by z g the function with domain A and range H such that z g ( n ) = z g ( n ) , for each n ∈ A . Given M ∈ ω , the set F ( < M + 1) will be used often to index F ( K, < M + 1), when K is a free Abelian group: Lemma 3. Let X = { x α,i : α < c and i < ω } be a set of generatorsfor an Abelian group G .Then F ( G, < M + 1) ⊆ { ( x g i ) i Since Φ is a function, it follows that( P i Let M be a positive integer, G be a topological free Abeliangroup and ( f i ) i We will prove some auxiliary results toshow that sequences associated to F ( < k ) have many accumulationpoints that are related to independent sets. This property is used topreserve accumulation points after refining the topology. Lemma 5. Let H be a topological Abelian group without non-trivialconvergent sequences such that H M is countably compact for some pos-itive integer M , A and A are infinite subsets of ω and { y i,n : i 2. By hypothesis, the sequence (( y i,n ) i 2. We can fix c , c ∈ H such that c = c = 0 = c and c j is an accumulation point of MALL POWERS OF COUNTABLY COMPACT GROUPS 7 { P i ∈ supp s j s j ( i ) .y i,n : n ∈ C j } , for each j < 2. Let W j be a neigh-borhood of c j , for each j < 2, such that W ∩ W = ∅ and 0 / ∈ W ∪ W .Let W ∗ j be an open set such that c j ∈ W ∗ j ⊆ W ∗ j ⊆ W j , for each j < D j = { n ∈ C j : P i ∈ supp s j s j ( i ) .y i,n ∈ W ∗ j } is infinite and(( y i,n ) i 2, it follows that conditions 3) and 4) aresatisfied. The sets V j for j < V ∗ j be an open set such that ( b j,i ) i 2. The set B j = { n ∈ D j : ( y i,n ) i 2, satisfiescondition 2). (cid:3) Proposition 6. Let H be a topological Abelian group without non-trivial convergent sequences, M be a positive integer such that H M iscountably compact and (( y i,n ) i 2, set A p ∧ = A p ∧ = A p and V p ∧ = V p ∧ = V p .Enumerate as { (( s j , p j ) , ( s j , p j )) : j < t } all the pairs (( s , p ) , ( s , p ))such that p i ∈ m p = p and s u : M −→ [ − m, m ] ∩ Z whosesupport is nonempty, for each u < A jp and V jp for each j < l ≤ t satisfying: I ) V j +1 p ⊆ V jp and A j +1 p ⊆ A jp if p ∈ { p j , p j } and j + 1 < l ; II ) V j +1 p = V jp and A j +1 p = A jp if p ∈ m \ { p j , p j } and j + 1 < l ; III ) { ( y i,n ) i 1. Conditions IV ) and V ) for j = l − iii ) and iv ).Now, define V p = V tp and A p = A tp . We will check that conditions a )- e ) are satisfied. For each p ∈ m 2, there exists j < t such that p = p j .By I ) and II ) it follows that V p ⊆ V j +1 p ⊆ V jp ⊆ V p | m − and a ) holds.It follows from condition III ) that { ( y i,n ) i 2. Thus condition b ) is satisfied.Fix a pair ( s , p ) , ( s , p ) with p = p and | p | = | p | = m , s r : M −→ [ − m, m ] ∩ Z whose support is nonempty, for each r < 2. Let j < t be such that ( s r , p r ) = ( s rj , p rj ), for each r < 2. By condition IV ) P i ∈ supp s j s j ( i ) .V p j +1 ,i ∩ P i ∈ supp s j s j ( i ) .V p j +1 ,i = ∅ . By condition I )and II ) it follows that V p r = V p rt ⊆ V rp j +1 , for each r < 2. Thus c ) issatisfied.It follows from condition V ) that0 / ∈ P i ∈ supp s j s j ( i ) .V j +1 p j ,i ∪ P i ∈ supp s j s j ( i ) .V j +1 p j ,i . By I ) and II ) it followsthat condition d ) is satisfied.Condition I ), II ) and A p = A p | m − imply that A p ⊆ A p | m − , for each p ∈ S m<ω m j < 2, therefore condition e ) is satisfied. (cid:3) Proposition 7. Let M be a positive integer and { ( g α,i ) i A countably compact Abelian non torsion group is such that itsfree rank is at least c . In particular, we can find an independent set { x m,i : m, i < ω } of infinite order.Suppose by induction that { x β,i : β < α and i < ω } is linearlyindependent with α < c . Let K be the group generated by X α = { x β,i : β < α and i < ω } .Then, the sequence { ( x g α,i ( n )) i 2. By cardinalityarguments, it follows that there exists two distinct f and f in ω P i Suppose that there exists a non torsion topological Abeliangroup H without non-trivial convergent sequences such that the group H m is countably compact, for each m < ω . Then, there exists a topo-logical free Abelian group G of cardinality c such that G m is countablycompact for each m < ω and G ω is not countably compact.Proof. Let { g α,i : ω ≤ α < c and i < M α } be an enumeration of F ( < ω ) such that S i The next result is used to refine the orig-inal topology. A lemma for groups of order 2 appears in Tomita [20]. MALL POWERS OF COUNTABLY COMPACT GROUPS 11 There are some differences as we are dealing with a group of infiniteorder. For the sake of completeness we give a detailed proof. Proposition 9. Let { ( g α,i ) i 2. Fix m j ∈ ω such that m j ∈ F j and B , B are disjoint subsetsof ω such that B j ∈ U j for j < 2. Set the function E : ω × ω −→ T such that E ( n, i ) = a j if ( n, i ) ∈ { m j } × B j for j < E ( n, i ) = 0otherwise. Let ν < c be such that E = D µ .It follows that U -lim ( P n ∈ F r ( n ) .z n,i ( ν ) : i ∈ ω ) = r ( m ) .a = r ( m ) .a = U -lim ( P n ∈ F r ( n ) .z n,i ( ν ) : i ∈ ω ). Hence condition v )is also satisfied. (cid:3) We will now take a subgroup of the refinement to obtain the desiredtopology for finite powers. Theorem 10. Suppose that there exists an infinite non torsion topo-logical group H without non-trivial convergent sequences and such that H M is countably compact for some positive integer M . Then there ex-ists a free Abelian group G such that G M is countably compact and G M +1 is not countably compact.Proof. Let { g α,i : ω ≤ α < c and i < M α } be an enumeration of F ( < M + 1) such that S i 2, then U -lim( P n ∈ F r ( n ) .z n,i : i ∈ ω ) = U -lim ( P n ∈ F r ( n ) .z n,i : i ∈ ω ).Let U be the set of all ultrafilters U such that the U -limit of ( z n,i : i ∈ ω ) is an element of the group h Z i , for every n < M + 1. Note thatif ( a n ) n We will now obtain inductively a subset I α ⊆ c × ω for each α ∈ [ ω, c [and the desired example will be h{ z β,i : ( β, i ) ∈ S α< c I α }i .The induction will satisfy the following conditions:1) | I α | ≤ | α | + ω for each α < c and { I α : α < c } is a ⊆ -increasingchain in P ( c × ω );2) U α is the set of all ultrafilters U ∈ U for which there exists r : F −→ Z \ { } with ∅ 6 = F ⊆ M + 1 such that the U -limit of { P m ∈ F r ( m ) .z m,i : i ∈ ω } is an element of h{ z β,i : ( β, i ) ∈ I α }i ;3) | U α | ≤ | α | + ω for each α < c and { U α : α < c } is ⊆ -increasingchain;4) J α is a subset of c × ω and for each U ∈ U α , there exists m < M +1such that U -lim { z m,i : i ∈ ω } / ∈ h{ z β,i : ( β, i ) ∈ c \ J α }i ;5) | J α | ≤ | α | for each ω ≤ α < c and { J α : α < c } is a ⊆ -increasingchain in P ( c × ω );6) θ α ∈ [ ω, c [ is the least ordinal θ ∈ c \ { θ β : β < α } such that S n ∈ A θ ,i Case 2 [ α ≥ ω ].Let θ α be the least ordinal for which S i Assume the existence of c incomparable selective ultra-filters. For each M < ω , there exists a topological group topology onthe free Abelian group G of cardinality c such that G M is countablycompact and G M +1 is not countably compact.Proof. Assuming the existence of c incomparable selective ultrafilters,Tomita [21] showed that there exists a topological free Abelian groupwithout non-trivial convergent sequences whose every finite power iscountably compact. Applying Theorem 10, we obtain the example. (cid:3) Countably compact groups of finite order The proof for the finite case is very similar to the non torsion casebut we have the infinite case that did not occur for the non torsioncase. A key difference is that the ultrapower of a free Abelian groupis no longer a free Abelian group, whereas an ultrapower of a vectorspace over the field Z p is a vector space over the field Z p . This last factis used to show that an arbitrary countable family of sequences can beassociated to a family of sufficiently independent sequences for whichwe can refine the topology as in the torsion case.We will sketch the proofs for the torsion case and point out somedifferences with the non torsion case. Notation 12. Given a prime number P , an Abelian group G of or-der P and λ ≤ ω , let F P ( G, λ ) be the set of all sequences ( f i ) i<λ from some infinite subset A ⊆ ω into G such that, for a nonemptysupport function s ∈ ( Z P ) ( λ ) , there exists F s finite such that the set { P i ∈ supp s s ( i ) .f i ( n ) : n ∈ A \ F s } is one-to-one.Define F P ( G, < κ ) = S λ<κ F P ( G, λ ) , for each κ ≤ ω .If G = Z ( c × ω ) P then we will write F P ( < κ ) and F P ( κ ) , respectively. Notation 13. Given a prime number P , F ∈ Z ( c × ω ) P , a group H oforder P and a family { z α,i : ( α, i ) ∈ supp F } ⊆ H , z F denotes the sum P ( α,i ) ∈ supp F F ( α, i ) .z α,i .Given A a subset of ω , g : A −→ Z ( c × ω ) P and { z α,i : ( α, i ) ∈ S { supp g ( n ) : n ∈ A }} ⊆ H , we denote by z g the function with domain A and range H such that z g ( n ) = z g ( n ) , for each n ∈ A . Lemma 14. Let κ ≤ ω be a cardinal and H be a topological groupwhose torsion subgroup is of uncountable cardinality and such that H γ is countably compact, for each cardinal γ < κ . Then there exists aprime number P and a group topology without non-trivial convergentsequences on ( Z P ) ( c ) such that its γ -th power is countably compact, foreach γ < κ . MALL POWERS OF COUNTABLY COMPACT GROUPS 17 Proof. By hypothesis, the torsion subgroup of H is uncountable. Hence,there exists m such that m.g = 0 for uncountably many g ∈ H . Let K = { g ∈ H : m.g = 0 } . The group K is a bounded torsion group,therefore, a direct sum of cyclic groups of some order p l , p prime, and p l divisor of m . Then, there exists a prime P that divides m and un-countably many summands that are copies of Z P l for some l such that P l divides m . It follows that K = { g ∈ H : P.g = 0 } is uncontable.The group K is a closed subgroup of H , thus the γ -th power of K is countably compact, for each cardinal γ < κ . Using closing off argu-ments and the fact that γ < κ ≤ ω , we can find a subgroup K of K of cardinality c such that K γ is countably compact for each γ < κ .The group H does not have non-trivial convergent sequences, there-fore, its subgroup K also inherits this property. Since K is a group oforder prime P , it is isomorphic to ( Z P ) ( c ) . Use the isomorphism to givethe desired group topology on ( Z P ) ( c ) . (cid:3) Lemma 15. Let X = { x α,i : α < c and i < ω } be a set of generatorsfor an Abelian group G of prime order P .Then F P ( G, < κ ) ⊆ { ( x g i ) i<λ : ( g i ) i<λ ∈ F P ( < κ ) } .Proof. The proof is as in Lemma 3. (cid:3) Lemma 16. Let κ ≤ ω and G be a topological Abelian group of primeorder P such that { ( f i ( n )) i<λ : n ∈ A } has an accumulation point in G λ for each ( f i ) i<λ ∈ F P ( G, < κ ) with dom f i = A for each i < λ .Then G γ is countably compact, for each γ < κ .Proof. Fix α < κ and arbitrary functions g i : ω −→ G , for each i < α .We want to show that (( g i ( n )) i<α : n ∈ ω ) has an accumulation point in G α . Fix a free ultrafilter U on ω . Then, the Abelian group K generatedby { [ g i ] U : i < α } is a vector space of order P of finite (possibly 0) orcountable infinite dimension. If the dimension is 0 then the U -limit of( g i ( n ) : n ∈ ω ) is 0, for each i < α and we are done. Assume that thedimension is positive and let K be the subgroup of K of the classes ofconstant sequences. Since K is a vector space over the field Z P , thereexists a subgroup K of K such that K is the direct sum of K and K . Let { t ji : i < λ j } be such that { [ t ji ] U : i < λ j } is a basis for K j , foreach j < 2. We can choose t i to be constant functions, for each i < λ .Let C m be an element of U for each m < α , be such that thereexists s jm : λ j −→ Z with finite support for each j < g m ( n ) = P l ∈ supp s m s m ( l ) .t l ( n ) + P l ∈ supp s m s m ( l ) .t l ( n ), for each n ∈ C m . Without loss of generality, we can assume that { C m : m ∈ ω } isa ⊆ -decreasing sequence. Enumerate as { r m : m ∈ ω } the set of all functions r : α −→ Z P with nonempty finite support. Note that the set { P j ∈ supp r r ( j ) .t j ( n ) : n ∈ C } is infinite, for each C ∈ U .Choose inductively n q > max { n , . . . , n q − } such that n q ∈ T a,b ≤ q ∧ p Let H be a topological group of prime order P withoutnon-trivial convergent sequences such that H λ is countably compact forsome λ ≤ ω , A and A infinite subsets of ω and { y i,n : i < λ and n ∈ A ∪ A } be a subset of H . Suppose that the function s j : λ −→ Z P hasfinite support and the finite set F j are such that { P i ∈ supp s j s j ( i ) .y i,n : n ∈ A j \ F j } is one-to-one, for each j < .If { ( y i,n ) i<λ : n ∈ A j } has an accumulation point in an open set U j of H λ then there exist B j an infinite subset A j and V j = Q i<λ V ji abasic open subset of U j for each j < such that (1) V j ⊆ U j for each j < ; (2) { ( y i,n ) i<λ : n ∈ B j } ⊆ V j , for each j < ; (3) P i ∈ supp s s ( i ) .V ,i ∩ P i ∈ supp s s ( i ) .V ,i = ∅ and (4) 0 / ∈ P i ∈ supp s s ( i ) .V ,i ∪ P i ∈ supp s s ( i ) .V ,i .Proof. The proof is as in Lemma 5 if we replace M by λ . (cid:3) Proposition 18. Let H be a topological group of order P without non-trivial convergent sequences such that H γ is countably compact, forsome cardinal < γ ≤ ω . Let (( y i,n ) i<λ : n ∈ A ) ∈ F P ( H, λ ) , for λ ≤ γ . There exists a family { a i,f : i < λ and f ∈ ω } ⊆ H such that A ) { a i,f : i < λ } is an independent set whose elements have order P , for each f ∈ ω ; B ) ( a i,f ) i<λ is an accumulation point of { ( y i,n ) i<λ : n ∈ A } , for each f ∈ ω and C ) P i ∈ supp s s ( i ) .a i,f = P i ∈ supp s s ( i ) .a i,f , for s j : λ −→ Z P withnonempty finite support for j < and f , f ∈ ω with f = f .Proof. We modify the proof of Proposition 6 and give here a sketchand indicate where changes had to be made.We will construct a tree of basic open subsets of H λ , { V p : p ∈ S k<ω k } and a tree of subsets of A , { A p : p ∈ S k<ω k } satisfying MALL POWERS OF COUNTABLY COMPACT GROUPS 19 conditions a ) − d ). Conditions c ) and d ) are different from the proof ofProposition 6. a ) V p ∧ j ⊆ V p , for each p ∈ S k<ω k j < b ) { ( y i,n ) i<λ : n ∈ A p } ⊆ V p , for each p ∈ S k<ω k c ) and d ) below, we recall that V p,i is the i -th coordinate of thebasic open set V p , for each i < λ : c ) P i ∈ supp s s ( i ) .V p ,i ∩ P i ∈ supp s s ( i ) .V p ,i = ∅ , whenever | p | = | p | = k , s j : λ −→ Z P has non-empty support contained in k and p = p ; d ) 0 / ∈ P i ∈ supp s s ( i ) .V p,i , whenever p ∈ S k<ω k s : λ −→ Z P has nonempty support contained in k ∩ λ and e ) A p ∧ j ⊆ A p , for each p ∈ S k<ω k j < a )- e ). Bycondition e ), we can find A f ⊆ A such that A f \ A f | n is finite, for each f ∈ ω n ∈ ω .Since H γ is countably compact and λ ≤ γ , there exists ( a i,f ) i<λ in H λ that is an accumulation point of { ( y i,n ) i<λ : n ∈ A f } . It thenfollows that condition B ) is satisfied.Fix a function s : λ −→ Z whose support is a nonempty finiteset. Choose k ∈ N such that dom s ⊆ k . Then P i ∈ supp s s ( i ) .a i,f ∈ P i ∈ supp s s ( i ) .V f | k ,i by condition b ). The same argument in the proof ofProposition 6 shows that condition A ) is satisfied.Fix s j : λ −→ Z with nonempty finite support for j < f , f ∈ ω f = f . Let k ∈ N be such that dom s ∪ dom s ⊆ k and f | k = f | k . The same argument now shows that condition C ) issatisfied.Hence all conditions A )- C ) are satisfied.We return to the construction of the sets satisfying condition a ) − e ).Set V ∅ = G and A ∅ = A . Clearly all conditions are satisfied.Suppose that V p and A p are constructed for each p ∈ S k 2, set A p ∧ = A p ∧ = A p and V p ∧ = V p ∧ = V p .Enumerate as { (( s j , p j ) , ( s j , p j )) : j < t } all the pairs (( s , p ) , ( s , p ))such that p i ∈ m p = p and s l : λ ∩ m −→ Z P , for each l < A jp and V jp for each j < l ≤ t satisfying: I ) V j +1 p ⊆ V jp and A j +1 p ⊆ A jp if p ∈ { p j , p j } and j + 1 < l ; II ) V j +1 p = V jp and A j +1 p = A jp if p ∈ m \ { p j , p j } and j + 1 < l ; III ) { ( y i,n ) i<λ : n ∈ A jp } ⊆ V jp , for each p ∈ m j < l ; IV ) P i ∈ supp s j s j ( i ) .V j +1 p j ,i ∩ P i ∈ supp s j s j ( i ) .V j +1 p j ,i = ∅ , for each j +1 < l and V ) 0 / ∈ P i ∈ supp s j s j ( i ) .V j +1 p j ,i ∪ P i ∈ supp s j s j ( i ) .V j +1 p j ,i .By III ), we can apply Lemma 17 on A r = A l − p rl − and U r = V l − p rl − toobtain A lp rl − ⊆ A l − p rl − and a basic open set V lp rl − , for each r < 2, thatsatisfy: i ) V lp rl − ⊆ V l − p rl − , for each r < ii ) { ( y i,n ) i<λ : n ∈ A lp rl − } ⊆ V lp rl − , for each r < iii ) P i ∈ supp s l − s l − ( i ) .V lp l − ,i ∩ P i ∈ supp s l − s l − ( i ) .V lp l − ,i = ∅ and iv ) 0 / ∈ P i supp s l − s l − ( i ) .V lp l − ,i ∪ P i ∈ supp s l − s l − ( i ) .V lp l − ,i .Conditions I ) − V ) for l follow as in the proof of Proposition 6 usingconditions i ) − iv ).Now, define V p = V tp and A p = A tp . We will check that conditions a )- e ) are satisfied. For each p ∈ m 2, there exists j < t such that p = p j .Conditions a ) and b ) follow from conditions I ) − V ) as in the proofof Proposition 6.Fix a pair ( s , p ) , ( s , p ) with p = p and | p | = | p | = m , s r : λ −→ Z P whose support is a finite nonempty set, for each r < 2. Let j < t be such that ( s r , p r ) = ( s rj , p rj ), for each r < 2. Condition c ) holdsapplying the same argument as in Proposition 6.Condition d ) and e ) also follow from the same arguments in Propo-sition 6. (cid:3) We state now the topology refinement for groups of prime order P . Proposition 19. Let L be a nonempty subset of [ ω, c [ such that c \ L hascardinality c . Let κ be a cardinal ≤ ω and { ( g α,i ) i<λ α : α ∈ [ ω, c [ \ L } be an enumeration of F P ( < κ ) such that S i<λ α supp g α,i ⊆ α × ω foreach α < c . Let X = { x α,i : i < ω and α < c } be a basis for a groupof prime order P such that ( x α,i ) i<λ α is an accumulation point of thesequence { ( x g α,i ( n ) ) i<λ α : n ∈ ω } , for each infinite ordinal α ∈ L . Thereexists { z α,i : α < c and i ∈ ω } ⊆ h X i × Z c P such that i ) z α,i extends x α,i , for each α < c and i < ω ; ii ) ( z α,i ) i<λ α is an accumulation point of the sequence { ( z g α,i ( n ) ) i<λ α : n ∈ ω } , for each ordinal α ∈ [ ω, c [ \ L ; iii ) For each function D : L × ω −→ ( Z P ) ω , there exists a coordinate µ such that z n,i ( µ ) = D ( n, i ) , for each ( n, i ) ∈ L × ω ; MALL POWERS OF COUNTABLY COMPACT GROUPS 21 iv ) If U is an ultrafilter, F j is nonempty finite subsets of L for j < and r j : F j −→ Z P \ { } for j < k are distinct, then the U -lim ( P n ∈ F j r j ( n ) .z n,i : i ∈ ω ) = U -lim ( P n ∈ F l r l ( n ) .z n,i : i ∈ ω ) and v ) If U and U are distinct ultrafilters, F j is a nonempty finite subsetof L for j < and r j : F j −→ Z P \ { } for each j < , then U -lim ( P n ∈ F r ( n ) .z n,i : i ∈ ω ) = U -lim ( P n ∈ F r ( n ) .z n,i : i ∈ ω ) .Proof. It suffices to make minor changes in the proof of Proposition 9,replacing Z for Z P , T by ( Z P ) ω and ω × ω by L × ω . One can checkthat, following the proof of Proposition 9, if D ( β, i ) = 0 for β ∈ L ∩ η and i ∈ ω and D = D µ then z β,i ( µ ) = 0, for every β < η and i ∈ ω . (cid:3) We use ( Z P ) ω to get sufficiently large countable linearly independentsubsets to mimick the proof using T for free Abelian groups. Theorem 20. Suppose that there exists an infinite topological group H without non-trivial convergent sequences of prime order P and suchthat H γ is countably compact, for every γ < κ with κ ≤ ω . Thenthere exists a topological group G of order P such that G γ is countablycompact for each γ < κ and G κ is not countably compact.Proof. Let κ ≤ ω and L be a subset of cardinality κ . The only case wecould not start with L = κ is if κ = ω and the Continuum Hypothesisholds. Either κ < ω or κ = ω and CH does not hold .Take L = κ .Let Z = { z α,i : α < c and i ∈ ω } be the family in Proposition 19.Let U be the set of all ultrafilters U such that there exists ξ ∈ κ suchthat U -limit of { z ξ,i : i ∈ ω } is an element of the group h Z i .As in the proof of Theorem 10, U has cardinality at most c .We will now obtain inductively a subset I α ⊆ c × ω for each α ∈ [ ω, c [and the desired example will be h{ z β,i : ( β, i ) ∈ S α< c I α }i .The sets I α will satisfy the following conditions:1) | I α | ≤ | α | + κ for each α < c and { I α : α < c } is a ⊆ -increasingchain in P ( c × ω );2) U α is the set of all ultrafilters U ∈ U for which there existsa nonempty finite support function r : κ −→ Z P such that U -lim( P ξ ∈ supp r r ( ξ ) .z ξ,i : i ∈ ω ) is an element of h{ z β,i : ( β, i ) ∈ I α }i ;3) | U α | ≤ | α | + κ for each α < c and { U α : α < c } is ⊆ -increasingchain;4) J α is a subset of c × ω and for each U ∈ U α , there exists ξ < κ such that U -limit of ( z ξ,i : i ∈ ω ) / ∈ h{ z β,i : ( β, i ) ∈ c \ J α }i ;5) | J α | ≤ | α | for each κ ≤ α < c and { J α : α < c } is a ⊆ -increasingchain in P ( c × ω ); θ α ∈ [ ω, c [ is the least ordinal θ ∈ c \ { θ β : β < α } such that S n ∈ ω and i<λ θ supp g θ,i ( n ) ⊆ S µ<α I µ , for each α ∈ [ κ, c [;7) ρ α ∈ [ ω, c [ is such that g θ α = g ρ α for each α ∈ [ κ, c [;8) { ρ α } × ω ⊆ c \ ( S µ<α ( J µ ∪ I µ )), for each α ∈ [ κ, c [;9) { ρ α } × λ ρ α ⊆ I α , for each α ∈ [ κ, c [ and10) J α ∩ I α = ∅ , for each α < c .Once I α , J α , θ α , ρ α , U α are constructed for κ ≤ α < c satisfyingthe conditions above, then G = h{ z β,i : i ∈ ω and β ∈ S α< c I α }i is asrequired.Suppose that the inductive construction is complete. The argumentto show that the powers smaller than κ are countably compact andthat the κ -th power is not are as in the proof of Theorem 10.We will now proceed with the inductive construction: Case 1 [ α < κ ]. Set I α = κ × ω , for each α < κ .The same argument used in the proof of Theorem 10 works if wemake changes in notation and we obtain U α = J α = ∅ , for each α < κ . Case 2 [ α ≥ κ ].Basically, the same proof works. The only observation:Given z U ,n the U -limit of { z ξ,i : i ∈ ω } , for each ξ < κ . As before, thecardinality of the set is bigger than the cardinality of the set added atthe stage, thus, there is n U < κ such that z U ,n U / ∈ h{ z β,i : ( β, i ) ∈ I α }i ,for each U ∈ U α \ S µ<α U µ .3) The Continuum Hypothesis holds and κ = ω .We use ω ⊆ L ⊆ ω such that L and ω \ L are unbounded in ω .Let Z = { z α,i : α < c and i ∈ ω } be the family in Proposition 19.Let U be the set of all ultrafilters U such that for each ξ ∈ κ suchthat U -limit of { z ξ,i : i ∈ ω } is an element of the group h Z i .As in the proof of Theorem 10, U has cardinality at most c .We will now obtain inductively a subset I α ⊆ c × ω for each α ∈ [ ω, c [and the desired example will be h{ z β,i : ( β, i ) ∈ S α< c I α ∪ ( L × ω ) }i .The sets I α will satisfy the following conditions:1) | I α | < ω = c , for each α < ω , I α ⊃ ( L ∩ α ) × ω and { I α : α < ω } is a ⊆ -increasing chain in P ( c × ω );2) U α is the set of all ultrafilters U ∈ U for which there exists r : L ∩{ β : { β } × ω ∩ I α = ∅} −→ Z P with nonempty finite support such that U -lim ( P ξ ∈ F r ( ξ ) .z ξ,i : i ∈ ω ) is an element of h{ z β,i : ( β, i ) ∈ I α }i ;3) | U α | < ω for each α < ω and { U α : α < c } is ⊆ -increasing chain;4) J α is a subset of c × ω and for each U ∈ U α , there exists ξ ∈ L such that U -limit of ( z ξ,i : i ∈ ω ) / ∈ h{ z β,i : ( β, i ) ∈ c \ J α }i ;5) | J α | ≤ | α | for each ω ≤ α < c and { J α : α < c } is a ⊆ -increasingchain in P ( c × ω ); MALL POWERS OF COUNTABLY COMPACT GROUPS 23 θ α ∈ [ ω, c [ is the least ordinal θ ∈ c \ { θ β : β < α } such that S n ∈ ω and i<λ θ supp g θ,i ( n ) ⊆ S µ<α I µ , for each α ∈ [ ω, c [;7) ρ α ∈ [ ω, c [ is such that g θ α = g ρ α , whenever ω ≤ α < c ;8) { ρ α } × ω ⊆ c \ ( S µ<α ( J µ ∪ I µ )), for each ω ≤ α < c ;9) { ρ α } × λ ρ α ⊆ I α , for each ω ≤ α < c and10) J α ∩ ( L ∪ I α ) = ∅ , for each α < c .Before proving the conditions 1) − G de-fined as h{ z β,i : ( β, i ) ∈ S α< c I α ∪ ( L × ω ) }i is as required. Using similararguments as before, one can show that G ω is countably compact and G ω is not countably compact. Case 1 . I α = ( ω × ω ) ∪ ( L ∩ ω ) × ω = ( L ∩ ω ) × ω . Then U α = J α = ∅ ,for each α < ω . The argument is similar as before.Assume α ≥ ω and that the induction has been carried out for µ < α . Case 2 . The definition of θ α and ρ α are as before and conditions5) − 8) are satisfied.Fix ν α ∈ L such that ν α > ρ α and ν α > β for each ( β, i ) ∈ S µ<α I µ ∪ J µ This will be used to define J α later.Define I α = S µ<α I µ ∪ ( { ρ α } × λ ρ α ) ∪ ( { ν α } × ω ). Then condition 1)and 9) are satisfied.Define U α as in condition 2). Similarly as before, condition 3) issatisfied.The definition of J α is different and uses ν α . Fix U ∈ U α . Let z U bethe U -limit of ( z ν α ,i ) : i ∈ ω ). Let r ∈ Z ( c × ω ) be the such that z = z r . Claim 1 . The support of r is not a subset of ν α × ω . If that isthe case, fix a ∈ ( Z P ) ω of order P and define D : L × ω −→ ( Z P ) ω such that D ( β, i ) = a if ( β, i ) ∈ ( { ν α } × ω ) \ supp r and D ( β, i ) = 0otherwise. Let µ be such that z β,i ( µ ) = D ( β, i ), for each ( β, i ) ∈ L × ω .By the fact that z β,i ( µ ) = 0 for each ( β, i ) ∈ ( L × ω ) ∪ ( S β ∈ ν α \ L { β } × ( ω \ λ β ), it follows by induction that z β,i ( µ ) is an accumulation pointfor a sequence of 0’s. Thus, z β,i ( µ ) = 0, for each β ∈ ν α \ L and i < λ β .Hence, z β,i ( µ ) = 0 for each ( β, i ) ∈ ( ν α × ω ) ∪ supp r and z β,i ( µ ) = a for each ( β, i ) ∈ ( { ν α } × ω ) \ supp r .Therefore, the supp r \ ν α × ω = ∅ . Let β ∗ be the largest ordinal forwhich F ∗ = supp r ∩ ( { β ∗ } × ω ) = ∅ . Claim 2 . β ∗ L Suppose by contradiction that β ∗ ∈ L . Fix an independent set ( a β ∗ ,i :( β ∗ , i ) ∈ F ∗ ) contained in ( Z P ) ω . Define E : L × ω −→ ( Z P ) ω suchthat E ( β, i ) = a β,i if ( β, i ) ∈ F ∗ and E ( β, i ) = 0 otherwise. Let µ ∗ bethe coordinate for which z β,i ( µ ∗ ) = E ( β, i ), for each ( β, i ) ∈ L × ω . Then, the U -limit of ( z ν α ,i : i ∈ ω ) = 0 and z r ( µ ∗ ) = z r | F ∗ ( µ ∗ ) = P ( β ∗ ,i ) ∈ F ∗ r ( β ∗ , i ) a β ∗ ,i = 0, which is a contradiction.It follow from Claim 1 and 2 that there exist ( β U , i U ) ∈ supp r \ ( L × ω )such that β U > ν α .Set J α = S µ<α J µ ∪ { ( β U , i U ) : U ∈ U α } . The points ( β U , i U ) showthat condition 4) is satisfied and the choice of ν α and β U > ν α implythat 10) is satisfied.Thus all conditions 1) − 10) are satified. (cid:3) Example 21. In the Random model, for each cardinal κ ≤ ω , thereexists a topological group topology on the Boolean group G of cardinality c such that G γ is countably compact for each γ < κ and G κ is notcountably compact.Proof. Szeptycki and Tomita [13] showed that the Boolean group H ofcardinality c admits a group topology without non-trivial convergentsequences such that H ω is countably compact. Applying Theorem 20we obtain the desired example. (cid:3) As earlier mentioned, the example of Hruˇsak, van Mill, Ramos andShelah give: Example 22. In ZFC, for each cardinal κ ≤ ω , there exists a topolog-ical group topology on the Boolean group G of cardinality c such that G γ is countably compact for each γ < κ and G κ is not countably compact. Questions and Remarks This work has been inspired by questions on surveys of ProfessorComfort. We list some variations of his questions that remain open.A natural question is the preservation of the algebraic structure ofthe example. The following questions are in this direction: Question 2. Let G be a torsion Abelian group of cardinality c and κ some cardinal not greater than ω . Suppose that G is a topologicalgroup without non-trivial convergent sequences such that G α is count-ably compact, for each α < κ with a group topology τ . Is there a grouptopology σ on G which in addition makes its κ -th power not countablycompact? Under Martin’s Axiom, an Abelian group of cardinality c admits acountably compact group topology if and only if it admits a countablycompact group topology without non-trivial convergent sequences. Question 3. Assume Martin’s Axiom and let G be an Abelian groupof cardinality c . Does it hold that G has a countably compact group MALL POWERS OF COUNTABLY COMPACT GROUPS 25 topology if and only if for each κ ≤ ω there is a group topology on G such that κ is the least cardinal for which G κ fail to be countablycompact? We do not know a model without selective ultrafilters for which thereexists a countably compact free Abelian group. Question 4. Is there a model without selective ultrafilters in whichthere exists a group topology without non-trivial convergent sequencesin some torsion-free Abelian group whose finite powers are countablycompact? In [2] the authors showed that under Martin’s Axiom there exists atopology without non-trivial convergent sequences in the real line thatmakes its square countably compact, but not its cube: Question 5. Is there a topological group without non-trivial conver-gent sequences in a torsion-free Abelian group such that every power iscountably compact? In particular, the algebraic group R or T has suchgroup topology? After the result of Hruˇsak, van Mill, Ramos and Shelah, some newquestions arise in ZFC. The author would like to thank ProfessorHruˇsak for explaining the main idea of their construction before thecompletion of their preprint. Question 6. a ) Which Abelian groups G admit in ZFC a countablycompact group topology without non-trivial convergent sequences? Inparticular, is there a non torsion Abelian group without non-trivial con-vergent sequences that is countably compact? b ) Classify, in ZFC, the Abelian groups of size c admitting a countablycompact group topology. c ) Is there in ZFC a free ultrafilter p and a p -compact group withoutnon-trivial convergent sequences? Is there p for which such topolgoydoes not exist? d ) Is there a Wallace semigroup in ZFC? (A countably compactboth-sided cancellative topological semigroup that is not algebraicallya group). We note that a positive solution for the second question in a ) yieldsa positive solution for d ).The author would like to thank the referee for comments that im-proved the presentation of this work.I first met Professor Comfort as the external examiner for my doc-toral thesis and in all few occasions I saw him in conferences or asked him something by e-mail, I felt his kindness and good spirits and willalways remember him for this. References [1] A. C. Boero and A. H. 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Tomita and S.Watson, Ultraproducts, p -limits and antichains on the com-fort group order , Topology Appl. 143 (2004), 147–157. Instituto de Matem´atica e Estat´ıstica, Universidade de S˜ao Paulo,Rua do Mat˜ao, 1010, CEP 05508-090, S˜ao Paulo, Brazil E-mail address :: v ≥ q then n u ∈ C \ { n ∈ ω : P j