aa r X i v : . [ m a t h . G R ] S e p PSEUDOCOMPACT PARATOPOLOGICAL GROUPS
ALEX RAVSKY
Abstract.
We obtain many results and solve some problems about pseudocompactparatopological groups. In particular, we obtain necessary and sufficient conditions whensuch a group is topological. (2-)pseudocompact paratopological groups that are not topo-logical are constructed. It is proved that each 2-pseudocompact paratopological groupis pseudocompact and that each Hausdorff σ -compact pseudocompact paratopologicalgroup is a compact topological group. Our particular attention is devoted to periodicand topologically periodic groups. In addition, we prove that the product of a nonemptyfamily of pseudocompact paratopological groups is pseudocompact. We show that aparatopological group G is pseudocompact provided it has a normal subgroup N suchthat both N and the quotient paratopological group G/N are pseudocompact. Introduction
Let G be a group endowed with a topology τ . A pair ( G, τ ) is called a semitopologicalgroup provided the multiplication · : G × G → G is separately continuous. Moreover, if themultiplication is continuous then ( G, τ ) is called a paratopological group and the topology τ is called a semigroup topology on G . Moreover, if the inversion ( · ) − : G → G iscontinuous with respect to the topology τ , then ( G, τ ) is a topological group . A standardexample of a paratopological group failing to be a topological group is the Sorgenfreyline, that is the real line endowed with the Sorgenfrey topology (generated by the baseconsisting of half-intervals [ a, b ), a < b ).In this paper we search conditions when a paratopological group is topological. Aninterested reader can find known results on this subject in [AlaSan, Introduction], in thesurvey [Rav3, Section 5.1], in Section 3 of the survey [Tka2] and in the sequel.2. Definitions
In this paper the word ”space” means ”topological space”.We recall the following definitions. A space X is called • sequentially compact if each sequence of X contains a convergent subsequence, • ω -bounded if each countable subset of X has the compact closure, • totally countably compact if each sequence of X contains a subsequence with thecompact closure, • countably compact at a subset A of X if each infinite subset B of A has an accu-mulation point x in the space X (the latter means that each neighborhood of x contains infinitely many points of the set B ), • countably compact if X is countably compact at itself, Mathematics Subject Classification.
Key words and phrases. paratopological group, continuity of the inverse, totally countably com-pact paratopological group, countably compact paratopological group, saturated paratopological group,topologically periodic paratopological group, periodic paratopological group, product of paratopologicalgroups, pseudocompact topological group, countably compact topological group, compact space, count-ably pracompact space, 2-pseudocompact space. • countably pracompact if X is countably compact at a dense subset of X , • pseudocompact if each locally finite family of nonempty open subsets of the space X is finite, • finally compact if each open cover of X has a countable subcover.The following inclusions hold. • Each compact space is ω -bounded. • Each ω -bounded space is totally countably compact. • Each totally countably compact space is countably compact. • Each sequentially compact space is countably compact. • Each countably compact space is countably pracompact. • Each countably pracompact space is pseudocompact.In these terms, a space X is compact if and only if X is countably compact and finallycompact. A Tychonoff space X is pseudocompact if and only if each continuous real-valued function on X is bounded.A space X is quasiregular if each nonempty open subset A of X contains the closure ofsome nonempty open subset B of X .A semitopological group G is left (resp. right) precompact if for each neighborhood U of the unit of G there exists a finite subset F of G such that F U = G (resp. U F = G ).A paratopological group is left precompact if and only if it is right precompact. So weshall call left precompact paratopological groups precompact . A sequence { U n : n ∈ ω } of subsets of a space X is non-increasing if U n ⊃ U n +1 for each n ∈ ω . A semitopologicalgroup G is if T U − n = ∅ for each non-increasing sequence { U n : n ∈ ω } ofnonempty open subsets of G . Clearly, each countably compact semitopological group is 2-pseudocompact. A semitopological group G is left ω -precompact if for each neighborhood U of the unit of G there exists a countable subset F of G such that F U = G .A semitopological group G is saturated if for each nonempty open subset U of G thereexists a nonempty open subset V of G such that V − ⊂ U .Denote by B G the open base at the unit of a semitopological group G .Suppose A is a subset of a group G . Denote by h A i ⊂ G the subgroup generated bythe set A .3. Pseudocompact paratopological groups that are topological
Proposition 1.
Each totally countably compact paratopological group is a topologicalgroup.Proof.
Let G be such a group. Put B = T B G . From Pontrjagin conditions [Rav, 1.1] itfollows that B is a semigroup. Put B ′ = { e } = { x ∈ G : ( ∀ U ∈ B G )( xU ∋ e ) } = { x ∈ G : xB ∋ e } = B − . Hence B ′ is a closed subsemigroup of the group G . Since B ′ is a compactsemigroup, there is a minimal closed right ideal H ⊂ B ′ . Let x be an arbitrary element of H . Then x H = H ∋ x . Hence x − ∈ H and e ∈ H . Consequently H = B ′ and xB ′ = B ′ for each element x ∈ B ′ . Therefore B ′ is a group and B ′ = B . Since B = T B G , we seethat g − Bg ⊂ B for each g ∈ G . Hence B is a normal subgroup of the group G . Since B has the antidiscrete topology, B is a topological group.Since the set B is closed, we see that the quotient group G/B is a T -space. Now let { x n : n ∈ ω } be an arbitrary sequence of the group G/B and let π : G → G/B be thequotient map. Choose a sequence { x ′ n : n ∈ ω } of the group G such that π ( x ′ n ) = x n foreach n ∈ ω . Since G is totally countable compact, we see that there is a subsequence A SEUDOCOMPACT PARATOPOLOGICAL GROUPS 3 of { x ′ n : n ∈ ω } such that the closure A is compact. Since the set A is closed, A = AB .Hence the closed compact set π ( A ) contains a subsequence of the sequence { x n : n ∈ ω } .Therefore G/B is a T totally countably compact paratopological group. Consequently,from [AlaSan, Cor 2.3] it follows that G/B is a topological group. Hence both B and G/B are topological groups. Then by [Rav4, 1.3] or [Rav3, Pr. 5.3], G is a topologicalgroup too. (cid:3) Proposition 1 generalizes Corollary 2.3 in [AlaSan] and Lemma 5.4 in [Rav3].
Lemma 1. [Rav2, Pr. 1.7]
Each saturated paratopological group is quasiregular.
Given a topological space (
X, τ ) Stone [Sto] and Katetov [Kat] consider the topology τ r on X generated by the base consisting of all canonically open sets of the space ( X, τ ).This topology is called the regularization of the topology τ . If ( X, τ ) is a paratopologicalgroup then (
X, τ r ) is a T paratopological group [Rav2, Ex. 1.9], [Rav3, p. 31], and[Rav3, p. 28]. We remind that a paratopological group G is topologically periodic if foreach x ∈ G and a neighborhood U ⊂ G of the unit there is a number n ≥ x n ∈ U , see [BokGur]. Lemma 2. [RavRez]
Suppose ( G, τ ) is a quasiregular paratopological group such that ( G, τ r ) is a topological group; then ( G, τ ) is a topological group. The following lemma is quite easy and probably is known.
Lemma 3.
Let ( X, τ ) be a topological space. Then ( X, τ ) is pseudocompact if and onlyif the regularization ( X, τ r ) is pseudocompact. (cid:3) Lemma 4.
Each topologically periodic Baire paratopological group is saturated.Proof.
In [AlaSan], Alas and Sanchis proved that each T topologically periodic Baireparatopological group is saturated. This result answers an author’s Question 2 from[Rav5]. Moreover, from their proof follows that each topologically periodic Baire paratopo-logical group is saturated. (cid:3) Lemma 5.
Each 2-pseudocompact paratopological group is a Baire space.Proof.
In [AlaSan, Th. 2.2], Alas and Sanchis proved that each T (cid:3) Lemma 6.
Each pseudocompact topological group is precompact. (cid:3)
Lemma 7. [Rav3, Pr.3.1]
Each precompact paratopological group is saturated. (cid:3)
Lemma 8. (also, see [AlaSan, Th. 2.5] ) Each Baire left ω -precompact paratopologicalgroup is saturated. (cid:3) Lemma 9.
Each paratopological group that is a dense G δ -subset of a T pseudocompactspace is a topological group.Proof. In [ArkRez, Th. 1.7], Arkhangel’skii and Reznichenko proved that for each paratopo-logical group G such that G is a dense G δ -subset of a regular pseudocompact space, itfollows that G is a topological group. Their proof with respective changes also proves thislemma. (cid:3) ALEX RAVSKY
Lemma 10.
Each T Let G be such a group. Suppose G is not a topological group; then there existsa neighborhood U ∈ B G such that V U − ⊂ ( U − ) for each V ∈ B G . By inductionwe can build a sequence { U i : i ∈ ω } of open neighborhoods of the unit e of G suchthat U ⊂ U and U i +1 ⊂ U i for each i ∈ ω . Put F = T { U i : i ∈ ω } . Then F is asubsemigroup of G . Since the group G is 2-pseudocompact, there is a point x ∈ G suchthat x ∈ T ( U i \ U − ) − ⊂ T U − i ⊂ T ( U − i ) ⊂ F − . Moreover, x ∈ ( U \ U − ) − ⊂ ( U \ U − ) − ⊂ U − \ U ⊂ G \ U . Since x U there is a neighborhood W ∈ B G such that W ⊂ U and W x ∩ U = ∅ . Let m, n ∈ ω and n > m ≥
1. If ∅ 6 = W x n ∩ W x m then ∅ 6 = W x ∩ W x m +1 − n ⊂ W x ∩ U F ⊂ W x ∩ U ⊂ W x ∩ U = ∅ . Therefore { W x i : i ≥ } is an infinite family of disjoint sets. Choose a neighborhood V ∈ B G such that V ⊂ W .Since G is 2-pseudocompact, there is a point y ∈ T n ≥ ( S i ≥ n V x i ) − . There exist numbers m, n ∈ ω such that n > m ≥ yV ∩ ( V x m ) − = ∅ and yV ∩ ( V x n ) − = ∅ . Then y − ∈ V x m ∩ V x n ⊂ W x m ∩ W x n . This contradiction proves that G is a topologicalgroup. (cid:3) Each countably compact paratopological group and each pseudocompact topologicalgroup are both 2-pseudocompact.
Proposition 2.
Each 2-pseudocompact paratopological group is pseudocompact.Proof.
Let (
G, τ ) be such a group. The regularization (
G, τ r ) of the group ( G, τ ) is a T G, τ r ) is a topological group.By Lemma 3, ( G, τ ) is pseudocompact. (cid:3)
An opposite inclusion does not hold. Manuel Sanchis and Mikhail Tkachenko con-structed a Hausdorff pseudocompact Baire not 2-pseudocompact paratopological group[SanTka, Th.2], answering author’s questions from the previous version of the manuscript.
Problem 1.
Is each countably pracompact Baire paratopological group 2-pseudocompact?
The answer to Problem 1 is positive provided the group is quasiregular or saturated ortopologically periodic or left ω -precompact. Proposition 3.
Suppose G is a paratopological group; then the following conditions areequivalent:1. The group G is pseudocompact quasiregular;2. The group G is pseudocompact saturated;3. The group G is pseudocompact topologically periodic Baire;4. The group G is pseudocompact left ω -precompact Baire;5. The group G is pseudocompact precompact;6. The group G is a pseudocompact topological group.Proof. It is easy to prove that Condition 6 implies all other conditions.(1 ⇒
6) The regularization (
G, τ r ) of the group ( G, τ ) is a T pseudocompact paratopo-logical group. By Lemma 9, ( G, τ r ) is a topological group. Then by Lemma 2, ( G, τ ) is atopological group too.(2 ⇒
1) By Lemma 1, the group G is quasiregular.(3 ⇒
1) By Lemma 4, the group G is saturated. Then by Lemma 1, the group G isquasiregular.(4 ⇒
2) By Lemma 8, the group G is saturated. SEUDOCOMPACT PARATOPOLOGICAL GROUPS 5 (5 ⇒
2) By Lemma 7, the group G is saturated. (cid:3) Implication (1 ⇒
6) generalizes Proposition 2 from [RavRez]. Implication (4 ⇒ ⇒
6) answers Question C from[AlaSan] and generalizes Theorem 3 from [BokGur].A group has two basic operations: the multiplication and the inversion. The firstoperation is continuous on a paratopological group, but the second may be not continuous.But there are continuous operations on a paratopological group besides the multiplication.For instance, a power. Now suppose that there exists an open subset U of a paratopologicalgroup G such that the inversion on the set U coincides with a continuous operation. Then,since the inversion on the set U is continuous, it follows that the group G is topological. Letus consider a trivial application of this idea. Let G be a paratopological group of boundedexponent. There is a number n such that the inversion on the group G coincides with the n -th power. Thus the group G is topological. Another applications are formulated in twonext propositions. Proposition 4.
Each Baire periodic paratopological group is a topological group.Proof.
Let G be such a group. Put B ′ = T { U − : U ∈ B G } . Then B ′ is closed subset of G . Since B ′ is a periodic semigroup, we see that B ′ is a group. For each n > G n = { x ∈ G : x n ∈ B ′ } . Since the group G is Baire, there is a number n > G n has the nonempty interior.Therefore there are a point x ∈ G n and a neighborhood U of the unit of G such that( xu ) n ∈ B ′ for each u ∈ U . Then V − ⊂ B ′ ( xV ) n − x for each subset V of U . Let U bean arbitrary neighborhood of the unit of G . Then x n ∈ B ′ U . By the continuity of themultiplication there is a neighborhood W ⊂ U of the unit such that ( xW ) n − x ⊂ B ′ U .Since B ′ ⊂ U , we see that W − ⊂ B ′ U ⊂ U . (cid:3) A Hausdorff topological group G is called Raikov-complete , provided it is complete withrespect to the upper uniformity which is defined as the least upper bound
L ∨ R of theleft and the right uniformities on G . Recall that the sets { ( x, y ) : x − y ∈ U } , where U runs over a base at unit of G , constitute a base of entourages for the left uniformity L on G . In the case of the right uniformity R , the condition x − y ∈ U is replaced by yx − ∈ U .The Raikov completion ˆ G of a Hausdorff topological group G is a completion of G withrespect to the upper uniformity L ∨ R . For every Hausdorff topological group G the spaceˆ G has a natural structure of a topological group. The group ˆ G can be defined as a unique(up to an isomorphism) Raikov complete group containing a Hausdorff topological group G as a dense subspace. Proposition 5.
Each Hausdorff pseudocompact periodic paratopological group is a topo-logical group.Proof.
Let (
G, τ ) be such a group and let (
G, τ r ) be the regularization of the group ( G, τ ).Then (
G, τ r ) is a regular pseudocompact periodic topological group. Let ˆ G be the Raikovcompletion of the group ( G, τ r ).We claim that the group ˆ G is periodic. Assume the converse. Suppose there exists anon-periodic element x ∈ ˆ G . Then for each positive integer n there exists a neighborhood V n ∋ x such that V nn e . Since the group G is G δ -dense in the group ˆ G , we see that thereexists a point y ∈ G ∩ T V n . Then there is a positive integer n such that y n = e . Thiscontradiction proves the periodicity of the group ˆ G . ALEX RAVSKY
For each n > G n = { x ∈ ˆ G : x n = e } . Since the group ˆ G is Baire, there isa number n > G n has the nonempty interior. Since the group G isdense in ˆ G , we see that there are a point x ∈ G and a neighborhood U of the unit of G such that ( xu ) n = e for each u ∈ U .Let y be an arbitrary point of G and V be an arbitrary neighborhood of the point y − .Then x n y − = y − ∈ V . By the continuity of the multiplication there is a neighborhood W ∈ τ of the unit such that W ⊂ U and ( xW ) n − xy − ∈ V . Let z ∈ yW be an arbitrarypoint. Then ( xy − z ) n = e . Therefore z − = ( xy − z ) n − xy − ∈ ( xW ) n − xy − ⊂ V . Thusthe inversion on the group G is continuous. (cid:3) T. Banakh build the next example. This result answers some author’s questions andshows that in general, Proposition 5 cannot be generalized for T groups (even for count-ably pracompact and periodic groups). Moreover, in [AlaSan], Alas and Sanchis provedthat each topologically periodic T paratopological group is T . Example 1 shows thatthere is a T periodic paratopological group G such that G is not Hausdorff. Example 1.
There exists a T periodic paratopological group G such that each power of G is countably pracompact but G is not a topological group.Proof. For each positive integer n let C n be the set { , . . . , n − } endowed with thediscrete topology and the binary operation ”+ ′ ” such that x + ′ y ≡ x + y (mod n ) for each x, y ∈ C n . Let G = ∞ L n =1 C n be the direct sum. Let F be the family of all non-decreasingunbounded functions from ω \{ } to ω . For each f ∈ F put O f = { } ∪ { ( x n ) ∈ G : ( ∃ m ≥ ∀ n > m )( x n = 0)&(0 < x m < f ( m ))) } . It is easy to check that the family { O f : f ∈ F } is a base at the zero of a T semigrouptopology on G .For each positive integer m ≥ a m = ( a mn ) ∈ G , where a mn = 1 if n = m ; a mn = 0in the opposite case. Let f, g ∈ F be arbitrary functions and let x ∈ G be an arbitraryelement. There exists a number m such that f ( m ) ≥ g ( m ) ≥
2, and x n = 0 for each n ≥ m . Then a m ∈ O g and x + a m ∈ O f . Therefore x ∈ O f . Hence O f = G for each f ∈ F . Thus G is not Hausdorff. Since each two nonempty open subsets of G intersects,we see that each two nonempty open subsets of each power of G in the box topologyintersects too. Therefore each power of G in the box topology is pseudocompact. Put A = { } ∪ { a m : m ≥ } . Then A is compact and dense in G . Let κ be an arbitrarycardinal. Then A κ is compact and dense in G κ . Therefore the space G κ is countablycompact at a dense subset of G κ . Thus the space G κ is countably pracompact. (cid:3) Problem 2.
Suppose a paratopological group G satisfies one of the following conditions: • G is Hausdorff countably pracompact topologically periodic left ω -precompact, • G is Hausdorff countably pracompact topologically periodic, • G is Hausdorff pseudocompact topologically periodic left ω -precompact, • G is Hausdorff pseudocompact topologically periodic,Is G a topological group? If a paratopological group G satisfies one of Problem 2 Conditions 1-4 and the group G is quasiregular or saturated or Baire then G is a topological group.Denote by TT the following axiomatic assumption: there is an infinite torsion-freeabelian countably compact topological group without non-trivial convergent sequences. SEUDOCOMPACT PARATOPOLOGICAL GROUPS 7
The first example of such a group constructed by M. Tkachenko under the ContinuumHypothesis [Tka]. Later, the Continuum Hypothesis weakened to the Martin Axiom for σ -centered posets by Tomita in [Tom2], for countable posets in [KosTomWat], and finallyto the existence continuum many incomparable selective ultrafilters in [MadTom]. Yet,the problem of the existence of a countably compact group without convergent sequencesin ZFC seems to be open, see [DikSha].The proof of [BanDimGut, Lemma 6.4] implies the following Lemma 11. (TT) Let G be a free abelian group generated by the set c . There exists aHausdorff group topology on G such that for each countable infinite subset M of the group G there exists an element α ∈ M ∩ c such that M ⊂ h α i . (cid:3) Example 2. (TT) There exists a functionally Hausdorff countably compact free abelianparatopological group ( G, σ ) such that ( G, σ ) is not a topological group.Proof. Let G be a free abelian group generated by the set c . Lemma 11 implies that thereexists a Hausdorff group topology τ on the group G such that for each countable infinitesubset M of the group G there exists an element α ∈ M ∩ c such that M ⊂ h α i . Eachelement x ∈ G has a unique representation x = P α ∈ A x n α α such that the integer number n α is non-zero for each α ∈ A x . Put S = { x ∈ G : n sup A x < S = S ∪{ } . Clearly, S is a subsemigroup of the group G . It is easy to check that the family { U ∩ S : U ∈ τ, U ∋ } is a base at the zero of a semigroup topology on G . Denote this semigroup topology by σ . Let M be an arbitrary countable infinite subset of the group G . There exists an element α ∈ M τ ∩ c such that M ⊂ h α i . Since M ⊆ α + S , we see that α ∈ M σ . Hence ( G, σ ) isa countably compact Hausdorff paratopological group.Suppose (
G, σ ) is a topological group; then there exists a neighborhood U ∈ τ of thezero such that U ∩ S ⊂ S ∩ ( − S ) = { } . Fix any element α ∈ c . Then − α ∈ S . Sincethe topological group ( G, τ ) is countably compact, there exists a number n < nα ∈ U . Then 0 = nα ∈ U ∩ S . This contradiction proves that ( G, σ ) is not a topologicalgroup. (cid:3)
Example 2 negatively answers Problem 1 from [Gur] under TT. Since each countablycompact left ω -precompact paratopological group is a topological group [Rav3, p. 82],we see that Example 2 implies the negative answers to Question A from [AlaSan] and toProblem 2 from [Gur] under TT. Problem 3.
Is there a ZFC-example of a paratopological group G such that G is not atopological group but G satisfies one of the following conditions: • G is Hausdorff countably compact, • G is Hausdorff countably pracompact Baire, • G is T countably compact. If G is such a group then G is not topologically periodic, not saturated, not quasiregular,and not left ω -precompact. Moreover, G × G is not countably compact. By Theorem 2.2from [AlaSan], a group ( G, σ ) from Example 2 is Baire. Therefore under TT there is agroup G such that G satisfies all conditions listed in Problem 3.The following proposition answers Questions B from [AlaSan]. Also this and the nextpropositions positively answer a special case for 2-pseudocompact groups of Problem 8.1from [ACK]. ALEX RAVSKY
Proposition 6.
Each 2-pseudocompact paratopological group of countable pseudocharacteris a topological group.Proof.
Let G be such a group. By induction we can build a sequence { V i : i ∈ ω } of openneighborhoods of the unit e of G such that V i +1 ⊂ V i for each i ∈ ω and T { V i : i ∈ ω } = { e } . Suppose G is not a topological group; then there exists a neighborhood U ∈ B G suchthat V i U − ⊂ ( U − ) for each i ∈ ω . Since the group G is 2-pseudocompact, there isa point x ∈ G such that xW − ∩ ( V i \ U − ) = ∅ for each W ∈ B G and each i >
0. Then xW − ∩ ( V i \ U − ) = ∅ and W x − ∩ ( V − i \ U ) = ∅ . Therefore x − ∈ V − i ⊂ V − i − . Hence x = e . But e ∈ U − and U − ∩ ( V \ U − ) = ∅ . This contradiction proves that G is atopological group. (cid:3) We shall use the following lemmas to obtain a counterpart of Proposition 6 for Hausdorff2-pseudocompact paratopological groups.
Lemma 12. [Rez, Th. 0.5]
A compact Hausdorff semigroup with separately continouousmultiplication and two-sides cancellations is a topological group.
Lemma 13.
Let K be a compact non-empty subset of a Hausdorff semitopological group G . Then the set G K = { x ∈ G : xK ⊂ K } is a compact topological group.Proof. Clearly, G K is a subsemigroup of the group G . If x ∈ G \ G K then there exists apoint y ∈ K such that xy K . Since the group G is Hausdorff, the set G \ K is open.Since the multiplication on the group G is separately continuous, there exists an openneighborhood O x of the point x such that O x y ⊂ G \ K . Thus Ox ⊂ G \ G K . Therefore G K is a closed subset of the space G . Let y ∈ K be an arbitrary element. If x ∈ G K then x ∈ xKy − ⊂ Ky − . Therefore G K ⊂ Ky − . Since the set Ky − is compact, the set G K is compact too. So, G K is a Hausdorff compact cancellative semitopological semigroup.By Lemma 12, G K is a topological group. (cid:3) Remark 1.
Lemma 13 does not necessarily hold for T paratopological groups, as thefollowing example shows. Let G = ( Z , +) be the group of integers endowed with a topologywith the base {{ x } ∪ { z ∈ Z : z ≥ y } : x, y ∈ Z } . It is easy to check that G is a T paratopological group. Put K = { x ∈ Z : x ≥ } . Then K is a compact subset of thegroup G and G K = K is not a group. Lemma 14. [Rav, Pr. 1.8]
Let G be a paratopological group, K ⊂ G be a compactsubspace, F ⊂ G be a closed set, and K ∩ F = ∅ . Then there exists a neighborhood V ofthe unit such that V K ∩ F = ∅ . Proposition 7.
Each 2-pseudocompact Hausdorff paratopological group containing a com-pact non-empty G δ -set is a topological group.Proof. Let G be such a group and K be a compact non-empty G δ -set of the space G .Suppose G is not a topological group; then there exists a neighborhood U ∈ B G such that V U − ⊂ ( U − ) for each neighborhood V ∈ B G .By induction using Lemma 14 we can build a sequence { V i : i ∈ ω } of open neighbor-hoods of the unit e of G such that V = U , V i +1 ⊂ V i for each i ∈ ω and T { V i K : i ∈ ω } = K . Put H = T { V i : i ∈ ω } . The construction of H implies that H is a semigroup.Moreover, since V − i +1 ⊂ V − i +1 V − i +1 ⊂ V − i for each i ∈ ω , we see that H − is a closed subsetof G . Moreover, HK ⊂ T { V i K : i ∈ ω } = K , so H ⊂ G K . Since G K is a group then H − ⊂ G K too, so H − is a compact cancellative topological semigroup. By Lemma 12, H − is a group. SEUDOCOMPACT PARATOPOLOGICAL GROUPS 9
We have V i U − for each i ∈ ω . Since the group G is 2-pseudocompact, there is apoint x ∈ G such that xW − ∩ ( V i \ U − ) = ∅ for each W ∈ B G and each i >
0. Then xW − ∩ ( V i \ U − ) = ∅ and W x − ∩ ( V − i \ U ) = ∅ . Therefore x − ∈ V − i ⊂ V − i − . Hence x ∈ H − . But H − = H ⊂ U ∩ U − and U − ∩ ( V \ U − ) = ∅ . This contradiction provesthat G is a topological group. (cid:3) The author does not know does counterparts of Proposition 7 hold for T Z , +) endowed with thecofinite topology is a T compact semitopological group which is not a paratopologicalgroup. Let G S be a T sequentially compact paratopological group from Example 5; thesubsemigroup S of the group G S is an open compact subset of G S , but G S is not a topo-logical group. From the other side, Alexander Arhangel’skii, Mitrofan Choban, and PetarKenderov proved in [ACK, Pr. 8.5] that a Hausdorff locally countably compact semi-topological group containing a compact of countable character is a paracompact locallycompact topological group.Example 3 is an extension of Example 5.17 from [Rav3]. Example 3 shows that counter-parts of Proposition 6 and Implication (4 ⇒
1) of Proposition 3 do not hold for countablypracompact paratopological groups. Also Example 3 negatively answers Problem 6.2 from[Tka2] and Problem 8.1 from [ACK].
Example 3.
There exists a functionally Hausdorff second countable paratopological group G such that each power of G is countably pracompact but G is not a topological group.Proof. Let T = { z ∈ C : | z | = 1 } be the unit circle. Denote by τ the standard topologyon the group T . For each ε > B ε (0 T ) = { z ∈ T : | arg z | < ε } . Let s : T → Q be a homomorphism onto. For each a ∈ R put T a = { x ∈ T : s ( x ) > a } and T a = { x ∈ T : s ( x ) ≥ a } . Put B ′ = { T a : a < } , B ′ = { T } , B ′ = {{ T } ∪ T } and B ′ = {{ T } ∪ T a : a > } . It is easy to check that B ′ i is a base at the unit ofa first countable semigroup topology on the group T for every i . Therefore for every i the family B i = { S ∩ B ε (0 T ) : S ∈ B ′ i , ε > } is a base at the unit of a first countablefunctionally Hausdorff semigroup topology on the group T . Denote this topology by τ i and put G i = ( T , τ i ). Clearly, τ ⊂ τ ⊂ τ ⊂ τ ⊂ τ .Choose an element x ∈ T . It is easy to check that the set A = { nx : n ∈ Z , n ≥ } is dense in G . Let κ be an arbitrary cardinal. Then A κ is dense in G κ . Let { y n : n ∈ ω } be an arbitrary sequence of points from A κ . Since the space ( T , τ ) κ is compact, we seethat there is a point y ∈ T κ such that the set { n ∈ ω : y n ∈ U } is infinite for each openin ( T , τ ) κ set U ∋ y .We claim that each open in G κ set U ∋ y has a similar property. Fix such a neigh-borhood U . For each α ∈ κ let π α : T κ → T be the projection onto the α -th coordinate.There exist a finite subset F of κ , a family { U α : α ∈ F } of open in ( T , τ ) sets, and anumber a > V ′ ⊂ U , where V ′ = T α ∈ F π − α (( y α + T a ) ∩ U α ). Decreasingneighborhoods U α , if necessary, we can assume that for each α ∈ F and for every integer0 < n ≤ ( s ( y α ) + a ) /s ( x ) if nx = y α , then nx U α .Put V = T α ∈ F π − ( U α ). Now let z be an arbitrary point of A κ ∩ V . Then z α = y α or s ( z α ) > s ( y α ) for each α ∈ F . Therefore z ∈ V ′ . Hence the set { n ∈ ω : y n ∈ U } ⊃ { n ∈ ω : y n ∈ V ′ } = { n ∈ ω : y n ∈ V } is infinite. Therefore the space G κ is countably compactat its dense subset A κ . Thus the space G κ is countably pracompact. Let i = 1 or i = 2. A set s − (0) is a subgroup of T of countable index and τ i | s − (0) = τ | s − (0). Since the group ( T , τ ) has the countable weight, the group G i has a countablenetwork. Since χ ( G i ) = ω , by [Rav, Pr. 2.3], the group G i is second countable.We claim that the group G has no countable network of closed subsets of G . Assumethe converse. Let N be a countable network such that for each set X ∈ N there existsan element x ∈ G with x + T ⊃ X . We claim that the set X is nowhere dense in( G, τ ). Assume the converse. Since the set A = { nx : n ∈ Z , n < } is dense in ( G, τ ),there exists an element y = nx ∈ int τ X τ such that s ( y ) < s ( x ). Let ε > y + B ε (0 T ) ⊂ X τ . There exists a point z ∈ ( y + B ε (0 T )) ∩ X .Since z ∈ X ⊂ x + T , we see that s ( z ) ≥ s ( x ) > s ( y ). Thus z ∈ y + T . Hence z ∈ y + ( B ε (0 T ) ∩ T ). Therefore y ∈ X τ = X ⊂ x + T . Thus s ( y ) ≥ s ( x ). Thiscontradiction proves that the set X is nowhere dense in ( G, τ ). Then (
G, τ ) = S N is aunion of a countable family of its closed nowhere dense sets. This contradiction provesthat the group G has no countable network of closed subsets of G .The set T \ T is a closed discrete subset of G . Thus e ( G ) = c . The set { ( x, − x ) : x ∈ G } is a closed discrete subset of G × G . Thus e ( G × G ) = c . (cid:3) Moreover, Manuel Sanchis and Mikhail Tkachenko constructed a Hausdorff 2-pseudo-compact Fr´echet-Urysohn paratopological group which is not a topological group [SanTka,Th.1], answering an author’s question from the previous version of the manuscript.Oleg Gutik asked the following question. Let G be a countably compact paratopologicalgroup and H be a subgroup of G . Is H a group? Proposition 8.
Let G be a countably compact paratopological group such that the closureof each cyclic subgroup of G is a group. Then G is a topological group.Proof. Let x be an arbitrary point of G . Put H = h x i . Then H is a separable count-ably compact paratopological group. Since each countably compact left ω -precompactparatopological group is a topological group [Rav3, p. 82], we see that H is a topologicalgroup. Therefore H is a topologically periodic group. Hence G is a topologically periodicgroup. By [BokGur], H is a topological group. (cid:3) The following proposition answers a question of Igor Guran.
Proposition 9.
Each Hausdorff σ -compact pseudocompact paratopological group is acompact topological group.Proof. Let (
G, τ ) be such a group and let (
G, τ r ) be the regularization of the group ( G, τ ).Then (
G, τ r ) is a regular pseudocompact topological group. Let G = S { K n : n ∈ ω } be aunion such that the set K n is compact in ( G, τ ) for each n ∈ ω . Then for each n ∈ ω theset K n is compact in ( G, τ r ). Since the group ( G, τ r ) is Baire there exists a number n ∈ ω such that the set K n has nonempty interior in ( G, τ r ). Since τ r ⊂ τ then the set K n hasnonempty interior in ( G, τ ). Therefore the group (
G, τ ) is locally compact. By [Rav3,Pr.5.5], (
G, τ ) is a topological group. Since (
G, τ ) is precompact, we see that (
G, τ ) iscompact. (cid:3) Cone topologies
In this section G is an abelian group and S ∋ G . In thissection, we consider an interplay between the algebraic properties of the semigroup S andproperties of two semigroup topologies generated by S on the group G . The main aim of SEUDOCOMPACT PARATOPOLOGICAL GROUPS 11 this section is Example 5 of two 2-pseudocompact paratopological groups which are nottopological groups.We need the following technical lemmas.
Lemma 15.
If there is an element a ∈ S such that S ⊂ a − S , then S is a group. (cid:3) Lemma 16.
The following conditions are equivalent:1. For each countable subset C of S − S there is an element a ∈ S such that C ⊂ a − S .2. For each countable subset C of S there is an element a ∈ S such that C ⊂ a − S .3. For each countable infinite subset C of S there are an element a ∈ S and an infinitesubset B of C such that B ⊂ a − S .Proof. (3 ⇒ C be an arbitrary countable subset of S . If C is finite, then S ⊂ ( P c ∈ C c ) − S . Now suppose the set C is infinite. Let C = { c n : n ∈ ω } . Put D = { d n : n ∈ ω } , where d n = P ni =0 c i for each n ∈ ω . If the set D if finite, then S ⊂ ( P d ∈ D d ) − S .Now suppose the set D is infinite. There are an element a ∈ S and an infinite subset B of D such that B ⊂ a − S . Therefore for each number i ∈ ω there exists a number n ∈ ω such that i ≤ n and d n ∈ B . Then a i ∈ d n − S ∈ a − S − S ∈ a − S .(2 ⇒ { c n : n ∈ ω } be a countable subset of the group S . Fix sequences { a n : n ∈ ω } and { b n : n ∈ ω } of S such that c n = b n − a n for each n . There exist anelement a ∈ S such that b n ∈ a − S for each n . Then c n ∈ a − S for each n . (cid:3) Cone topology.
The one-element family { S } satisfies Pontrjagin conditions (see[Rav, Pr. 1]). Therefore there is a semigroup topology on G with the base { S } at thezero. This topology is called the cone topology generated by the semigroup S on G . Denoteby G S . the group G endowed with this topology.It is easy to check that the closure S of the semigroup S is equal to S − S and S = h S i .Moreover, S is clopen. Proposition 10.
The group G S is T if and only if S ∩ ( − S ) = { } . (cid:3) Proposition 11.
The group G S is T if and only if S = { } . (cid:3) Lemma 17.
A subset K of G S is compact if and only if there is a finite subset F of K such that F + S ⊃ K . (cid:3) Lemma 18.
Let A be a subset of G S . Then A = A − S . (cid:3) Proposition 12.
The following conditions are equivalent:1. The group G S is compact2. The group G S is ω -bounded.3. The group G S is totally countably compact.4. The group G S is precompact.5. The semigroup S is a subgroup of G of finite index.Proof. Implications (1 ⇒ ⇒
2) and (2 ⇒
3) are trivial.(4 ⇒ F of G such that F + S = G . Since f + S ⊂ f + S forevery f ∈ F , S ⊂ ( F ∩ S ) + S . Chose a number n and points x , . . . , x n , y , . . . , y n ∈ S such that F ∩ S = { x − y , . . . , x n − y n } . Then S = S x i − y i + S ⊂ S − y i + S . Put y = P y i . Then S ⊂ − y + S . Therefore there is an element z ∈ S such that − y = − y + z .Then z = − y and hence − y i ∈ S for every i . Therefore S = S + S ⊃ S . Consequently S is a group. Since F + S = G , we see that the index | G : S | is finite.(5 ⇒ S is a subgroup of G of finite index, we see that there exists a finitesubset F of G such that F + S = G . For each point f ∈ F find set U f ∈ U such that f ∈ U f . Let f ∈ F be an arbitrary point. Since the set U f is open, f + S ⊂ U f . Therefore { U f : f ∈ F } is a finite subcover of U and S { U f : f ∈ F } = G .(3 ⇒ − S = { } is compact. Therefore there is a finite subset F of S suchthat − F + S ⊃ − S . Let F = { f , . . . , f n } . Put f = P f i . Then − f + S ⊃ F + S ⊃ − S .Moreover, − f + S ⊃ − f + S + S ⊃ S − S . Hence there is an element g ∈ S such that − f = − f + g . Then g = − f and − f + S ⊃ S . Therefore S = S . Suppose the index | G : S | is infinite. Then there is a countable set A ⊂ G that contains at most one pointof each coset of the group S in G . Let B be an arbitrary infinite subset of A . The set B = B − S intersects an infinite number of the cosets. Therefore B = F + S for eachfinite subset F of G . This contradiction proves that the index | G : S | is finite. (cid:3) Proposition 13.
The following conditions are equivalent:1. The group G S is sequentially compact.2. The group G S is countably compact.3. The group G S is 2-pseudocompact.4. The index | G : S | is finite and for each countable subset C of S there is an element a ∈ S such that C ⊂ a − S .Moreover, in this case each power of the space G S is countably compact.Proof. Implications (1 ⇒
2) and (2 ⇒
3) are trivial.(3 ⇒ C be an arbitrary countable infinite subset of S . Let { x n : n ∈ ω } bean enumeration of elements of the set C . For each n ∈ ω put U n = S i ≥ n x i + S . Then { U n : n ∈ ω } is a non-increasing sequence of nonempty open subsets of G S . Since thegroup G S is 2-pseudocompact then there is a point b ∈ G such that ( b + S ) ∩ − U n = ∅ foreach n ∈ ω . Then for each n ∈ ω there is a number i ≥ n such that ( b + S ) ∩ − ( x i − S ) = ∅ and hence − x i ∈ b + S . Let B be the set of all such elements x i . Then B is an infinitesubset of − b − S . Since B ⊂ S , we see that − b ∈ B + S ⊂ S . Now Lemma 16 impliesCondition 4.Suppose the index | G : S | is infinite. Then there is a countable infinite set C ⊂ G thatcontains at most one point of each coset of the group S in G . Let { x n : n ∈ ω } be anenumeration of the elements of the set C . For each n ∈ ω put U n = S i ≥ n x i + S . Then { U n } is a non-increasing sequence of nonempty open subsets of G S . But for each point x ∈ G an open set x + S intersects only finitely many sets of the family {− U n : n ∈ ω } .This contradiction proves that the index | G : S | is finite.(4 ⇒ { c n : n ∈ ω } be a sequence of elements of the group G S . There is a point x ∈ G such that the set I = { n ∈ ω : x n ∈ x + S } is infinite. By Lemma 16, there is apoint a ∈ S such that x − c n ∈ a − S for each n ∈ I . Thus the sequence { c n : n ∈ I } converges to the point x − a .Now let κ be an arbitrary cardinal. Condition 1 and Lemma 16 imply that for eachcountable subset C of S κ there is a point a ∈ S κ such that C ⊂ S κ − a . Consequently,the space S κ is sequentially compact. Since S is a clopen subgroup of G S , we see that thespace G S is homeomorphic to S × D , where D is a finite discrete space. Then a space G κS is homeomorphic to the countably compact space S κ × D κ . (cid:3) The proof of the following lemma is straightforward.
Lemma 19. If X is a countably pracompact space and Y is a compact space then thespace X × Y is countably pracompact. (cid:3) Proposition 14.
The following conditions are equivalent:
SEUDOCOMPACT PARATOPOLOGICAL GROUPS 13
1. The group G S is countably pracompact.2. The group G S is pseudocompact.3. The index | G : S | is finite.Moreover, in this case each power of the space G S is countably pracompact.Proof. Implication (1 ⇒
2) is trivial.(2 ⇒ | G : S | is infinite. Then there is a countable infinite set C ⊂ G that contains at most one point of each coset of the group S in G . Let { x n : n ∈ ω } be an enumeration of the elements of the set C . For each n ∈ ω put U n = S i ≥ n x i + S .Then { U n } is a non-increasing sequence of nonempty open subsets of G S . But for eachpoint x ∈ G an open set x + S intersects only finitely many sets of the family { U n : n ∈ ω } .This contradiction proves that the index | G : S | is finite.(3 ⇒ A ⊂ G such that the intersection of A with each coset ofthe group S in G is a singleton. Put A = A + S . Let x ∈ G be an arbitrary point. Thenthere is an element a ∈ A such that x ∈ a + S . Hence ( x + S ) ∩ ( a + S ) = ∅ . Therefore( x + S ) ∩ A = ∅ . Thus A is dense in G S .Let C be an arbitrary countable infinite subset of A . There is a point a ∈ A suchthat the set B = C ∩ ( a + S ) is infinite. But since C ⊂ A + S , we see that B ⊂ a + S .Therefore a is an accumulation point of the set C .Now let κ be an arbitrary cardinal. Since the set S is dense in S , we see that the set S κ is dense in S κ . Since each neighborhood of the zero of the group S κ contains the set S κ ,the space S κ is countably compact at S κ . Hence the space S k is countably pracompact.Since S is a clopen subgroup of G S , the space G S is homeomorphic to S × D , where D is a finite discrete space. Then a space G κS is homeomorphic to S κ × D κ . By Lemma 19,the space S κ × D κ is countably pracompact. (cid:3) Proposition 15.
The group G S is finally compact if and only if G S is left ω -precompact. (cid:3) Cone* topology.
The family B S = {{ } ∪ ( x + S ) : x ∈ S } satisfies Pontrjaginconditions (see [Rav, Pr. 1]). Therefore there is a semigroup topology on G with the base B S at the zero. This topology is called the a cone* topology generated by the semigroup S and on G . Denote by G ∗ S . the group G endowed with this topology.It is easy to check that the closure S of the semigroup S is equal to S − S and S = h S i .Moreover, S is clopen. Proposition 16.
The group G ∗ S is Hausdorff if and only if S = { } . (cid:3) Proposition 17.
The following conditions are equivalent:1. The group G ∗ S is T .2. The group G ∗ S is T .3. S = { } or S is not a group.4. S = { } or T { x + S : x ∈ S } = ∅ .Proof. Implications (1 ⇒ ⇒ ⇒
3) are trivial. We show that (3 ⇒ y ∈ T { x + S : x ∈ S } . Then y ∈ y + S . Hence there is anelement z ∈ S such that y = 2 y + z . Therefore − y = z ∈ S . Now let x be an arbitraryelement of S . There is an element s ∈ S such that y = x + s . Then − x = − ( x + s ) + s ∈ S .Therefore S is a group. Thus y = 0. (cid:3) Proposition 18.
The following conditions are equivalent:
1. The group G ∗ S is compact.2. The group G ∗ S is ω -bounded.3. The group G ∗ S is totally countably compact.4. The group G ∗ S is sequentially compact.5. The group G ∗ S is countably compact.6. The group G ∗ S is precompact.7. The semigroup S is a subgroup of G of finite index.Proof. Implications (1 ⇒ ⇒ ⇒ ⇒ ⇒
5) and (1 ⇒
6) are trivial.From (7) follows that G ∗ S = G S . Therefore by Proposition 12, (7 ⇒ G S is precompact. Therefore by Proposition 12, (6 ⇒ ⇒ x ∈ S be an arbitrary element.We claim that there is n ≥ − nx ∈ S . Assume the converse. Then the set X = {− nx : n ≥ } is infinite. There is a point b ∈ G such that the set ( b + s + S ) ∩ X is infinite for each s ∈ S . Since X ⊂ S , we see that b ∈ S . Therefore there are elements y, z ∈ S such that b = y − z . But since S ⊃ ( y − z ) + z + S , the set S ∩ X is infinite too.This contradiction proves that there is n ≥ − nx ∈ S . Therefore S is a group.By Proposition 13, the index | G : S | is finite. (cid:3) Proposition 19.
The group G ∗ S is countably pracompact if and only if the index | G : S | is finite and there is a countable subset C of S such that S ⊂ C − S .Moreover, in this case each power of the space G ∗ S is countably pracompact.Proof. The necessity. Since the group G S is countably pracompact, by Proposition 14,the index | G : S | is finite. Let A be a dense subset of S such that the space S iscountably compact at A . Let B be an arbitrary countable infinite subset of A . There areelements x, y ∈ S such that the set ( x − y + s + S ) ∩ B is infinite for each s ∈ S . Since s + S ⊃ ( x − y ) + y + s + S , we see that the set ( s + S ) ∩ B is infinite for each s ∈ S . Thiscondition holds for each countable infinite subset B of A . Therefore the set A \ ( s + S ) isfinite for each s ∈ S . Since A ⊂ S − S , for each element a ∈ A there is an element s a ∈ S such that a + s a ∈ S . Put A ′ = { a + s a : a ∈ A } . Then the set A ′ \ ( s + S ) is finite foreach s ∈ S .Suppose the set A ′ is finite. Put C = A ′ . Since A is dense in S , we see that A ′ isdense in S too. Let s ∈ S be an arbitrary point. Then the set ( s + S ) ∩ C is nonempty.Therefore s ∈ C − S .Suppose the set A ′ is infinite. Let C be an arbitrary countable infinite subset of A ′ .Let s ∈ S be an arbitrary point. There is a point c ∈ ( s + S ) ∩ C . Therefore s ∈ C − S . The sufficiency.
Let C = { c n : n ∈ ω } . Put C ′ = { P ni =0 c i : n ∈ ω } . Then for eachelement s ∈ S and each infinite subset B ′ of C ′ there is a finite subset F of B ′ such that c − S ∋ s for each c ∈ B ′ \ F .Choose a finite set A ⊂ G such that the intersection of A with each coset of thegroup S in G is a singleton. Put A = A + C ′ . Let x ∈ G be an arbitrary point. Thenthere is an element a ∈ A such that x ∈ a + S . Therefore there are elements s, s ′ ∈ S such that x − a = s − s ′ . There is an element c ∈ C ′ such that s ∈ c − S . Then a + c = x + s ′ − s + c ∈ x + s ′ + s − c + c ⊂ x + S . Hence A is dense in G ∗ S .Let A ′ be an arbitrary countable infinite subset of A . There is a point a ∈ A such thatthe set B ′ = A ′ ∩ ( a + S ) is infinite. Let s ∈ S be an arbitrary element. Since B ′ − a is aninfinite subset of C ′ , the set ( s + S ) ∩ ( B ′ − a ) in infinite. Then the set ( a + s + S ) ∩ B ′ is infinite too. Therefore a is an accumulation point of the set B ′ ⊂ A ′ . SEUDOCOMPACT PARATOPOLOGICAL GROUPS 15
Let κ be an arbitrary cardinal. Put A = { } ∪ C ′ . Then A is compact and dense in S .Therefore A κ is compact and dense in S κ . Hence the space S κ is countably compact atits dense subset. Consequently S κ is countably pracompact. Since S is a clopen subgroupof G ∗ S , we see that the space G ∗ S is homeomorphic to S × D , where D is a finite discretespace. Then a space G ∗ κS is homeomorphic to S κ × D κ . by Lemma 19, the space S κ × D κ is countably pracompact. (cid:3) Proposition 20.
The group G ∗ S is 2-pseudocompact if and only if the index | G : S | isfinite and for each countable subset C of S there is an element a ∈ S such that C ⊂ a − S .Moreover, in this case each finite power of the group G ∗ S is 2-pseudocompact and eachpower of the group S is 2-pseudocompact.Proof. The necessity. If the group G ∗ S is 2-pseudocompact then the group G S is 2-pseudocompact too. Thus Proposition 13 implies the necessity. The sufficiency.
Let { U n : n ∈ ω } be a non-increasing sequence of nonempty opensubsets of G . For each n choose a point z n ∈ U n such that z n + S ⊂ U n . Since theindex | G : S | if finite, there are an infinite subset I of ω and an element z ∈ G such that z n ∈ z + S for each n ∈ I . For each n ∈ I fix elements x n , y n ∈ S such that z n − z = x n − y n .There is an element a ∈ S such that x n ∈ a − S for each n ∈ I . Let n ∈ I be an arbitrarynumber. Then z n − z = x n − y n ∈ a − S . Therefore U n ∋ z n + S ∋ a + z . Since the family { U n } is non-increasing, we see that a + z ∈ T { U n : n ∈ ω } .Now let κ be an arbitrary cardinal. Let { U n : n ∈ ω } be an arbitrary non-increasingsequence of nonempty open subsets of S κ . For each n choose a point z n ∈ U n such that z n + S κ ⊂ U n and fix elements x n , y n ∈ S κ such that z n = x n − y n . There is an element a ∈ S κ such that x n ∈ a − S κ for each n ∈ ω . Thus U n ∋ z n + S κ ∋ a . (cid:3) Corollary 1.
The group G ∗ S is compact if and only if the group G ∗ S is countably pracom-pact and 2-pseudocompact.Proof. The proof is implied from Proposition 20, Proposition 19, and Lemma 15. (cid:3)
Proposition 21.
The group G ∗ S is pseudocompact if and only if the index | G : S | is finite.Proof. The necessity. If the group G ∗ S is pseudocompact then the group G S is pseudo-compact too. Thus Proposition 14 implies the necessity. The sufficiency.
Let { U n : n ∈ ω } be a family of nonempty open subsets of G . Foreach n choose a point x n ∈ U n such that x n + S ⊂ U n . Since the index | G : S | if finite,there are a point b ∈ G and an infinite set I ⊂ ω such that x n ∈ b + S for each n ∈ I .Let n ∈ I be an arbitrary number. Then ( x n + S ) ∩ ( b + s + S ) = ∅ for each s ∈ S .Therefore b ∈ x n + S ⊂ U n . Hence the family { U n } is not locally finite. Thus the space G ∗ S is pseudocompact. (cid:3) The counterpart of Proposition 15 does not hold for the group G ∗ S . Example 4.
There are an abelian group G and a subsemigroup S of the group G suchthat G ∗ S is ω -precompact and not finally compact.Proof. Let G = R and S = [0 , ∞ ). Then G ∗ S is ω -precompact. Consider an open cover U = { x ∩ S : x < } of the group G ∗ S . Then U has no subcover of cardinality less than c .By Proposition 19, the group G ∗ S is countably pracompact. By Proposition 20, the group G ∗ S is not 2-pseudocompact. (cid:3) Proposition 22.
If the group G ∗ S is Baire then S is a group or there is no countablesubset C of S such that S ⊂ C − S . Proof.
Suppose that there is a countable subset C of S such that S ⊂ C − S . Then S ⊂ S c ∈ C c − S . Since the set − S is closed in G ∗ S , there are elements c ∈ S and x ∈ S such that c − S ⊃ x + S . It follows easily that − S ⊃ S . Thus S is a group. (cid:3) Corollary 2.
The countably pracompact group G ∗ S is Baire if and only if S is a group. (cid:3) Corollary 3.
The countable group G ∗ S is 2-pseudocompact if and only if the semigroup S is a subgroup of G of finite index.Proof. If the semigroup S is a subgroup of G of finite index then from Proposition 18 itfollows that G ∗ S is compact. If the group G ∗ S is 2-pseudocompact then from Lemma 5 itfollows that G ∗ S is Baire. By Proposition 22, S is a group. Proposition 20 implies thatthe index | G : S | = | G : S | is finite. (cid:3) A space X is called a P -space if every G δ subset of X is open. Proposition 23.
The group G ∗ S is a P-space if and only if for each countable subset C of S there is an element a ∈ S such that C ⊂ a − S .Proof. The necessity. Let C be an arbitrary countable subset of G . Then the set { } ∪ T { c + S : c ∈ C } is a neighborhood of the zero. Therefore there is a point b ∈ S suchthat b + S ⊂ { } ∪ T { c + S : c ∈ C } . If b = 0 then C ⊂ b − S . Suppose b = 0. If S = { } then C ⊂ S = 0 − S . If there is an element a ∈ S \{ } then a ∈ T { c + S : c ∈ C } and C ⊂ a − S . The sufficiency.
Let { U n : n ∈ ω } be an arbitrary sequence of nonempty open subsetsof G , U = T { U n : n ∈ ω } and x ∈ U . For each n choose a point c n ∈ S such that x + ( { } ∪ ( c n + S )) ⊂ U n . There exists an element a ∈ S such that { c n : n ∈ ω } ⊂ a − S .Then a + S ⊂ c n + S for each n ∈ ω . Hence x + ( { } ∪ ( a + S )) ⊂ U . Thus the set U isopen. (cid:3) Corollary 4.
The group G ∗ S is 2-pseudocompact if and only if the group G ∗ S is a pseu-docompact P-space.Proof. The proof is implied from Proposition 20, Proposition 21, and Proposition 23. (cid:3)
Corollary 5.
The group G ∗ S is compact if and only if the group G ∗ S is a countably pra-compact P-space.Proof. The proof is implied from Corollary 1 and Corollary 4. (cid:3)
Example 5.
There are an abelian group G and a subsemigroup S of the group G such thatthe paratopological group G S is T sequentially compact, not totally countably compact,not precompact, and not a topological group and the paratopological group G ∗ S is T Let G = L α ∈ ω Z be the direct sum of the groups Z . Let S = { } ∪ { ( x α ) ∈ G : ( ∃ β ∈ ω )(( ∀ α > β )( x α = 0)&( x β > } . Since S ∩ ( − S ) = { } then by Proposition 10, the group G S is T and by Proposition 17,the group G ∗ S is T . Since G = S − S and for each countable infinite subset C of S thereis an element a ∈ S such that C ⊂ a − S , we see that by Proposition 13, the group G S is sequentially compact and by Proposition 20, the group G ∗ S is 2-pseudocompact. ByProposition 19, the group G ∗ S is not countably pracompact. By Proposition 12, the group G S is not totally countably compact and not precompact. Therefore the group G ∗ S is notprecompact too. (cid:3) SEUDOCOMPACT PARATOPOLOGICAL GROUPS 17
In [San], Iv´an S´anchez proved that the group G ∗ S from Example 5 is a P-space.From Proposition 3 in [RavRez] it follows that each T paratopological group G suchthat G × G is countably compact is a topological group. Example 1 shows that in generalthis proposition cannot be generalized for T countably pracompact groups. Example 5shows that the proposition cannot be generalized for T sequentially compact groups.5. Products of pseudocompact paratopological groups
The product of an arbitrary nonempty family of compact spaces is compact by Ty-chonoff Theorem. But there are Tychonoff countably compact spaces X and Y such that X × Y is not pseudocompact [Eng, 3.10.19]. The product of an arbitrary nonempty fam-ily of Tychonoff pseudocompact topological groups is pseudocompact [ComRos, Th. 1.4].Under Martin Axiom van Douwen obtained the first example of countably compact topo-logical groups G and G such that the product G × G is not countably compact. Later,under M A countable
Hart and van Mill proved that there is a countably compact topologicalgroup G such that G × G is not countably compact. As far as the author knows, there areno ZFC-example of a family { G α : α ∈ A } of countably compact Tychonoff topologicalgroups such that the product Q { G α : α ∈ A } is not countably compact, see the surveys[Com] and [ComHofRem].The next proposition generalizes the result of Comfort and Ross [ComRos, Th. 1.4] forparatopological groups. Proposition 24.
The product of an arbitrary nonempty family of pseudocompact paratopo-logical groups is pseudocompact.
To prove Proposition 24, we need the following lemmas.The proof of the following technical lemma is straightforward.
Lemma 20.
Let ( X, τ ) be the product of a family { ( X α , τ α ) : α ∈ A } of topological spaces.Then ( X, τ r ) = Q α ∈ A ( X α , τ αr ) . (cid:3) Let (
G, τ ) be a paratopological group and H ⊂ G be a normal subgroup of G . Thenthe quotient group G/H endowed with the quotient topology is a paratopological group,see [Rav]. If H = T { U ∩ U − : U ∈ B G } then we denote the quotient group G/H by T G .It is easy to check that the group T G is a T space for each paratopological group G . Lemma 21.
A paratopological group G is pseudocompact if and only if the paratopologicalgroup T G is pseudocompact.Proof. If the group G is pseudocompact then the group T G is a continuous image ofa pseudocompact space. Thus T G is pseudocompact. Now suppose the group T G ispseudocompact. Let U be a locally finite family of open subsets of the space G . Put H = T { U ∩ U − : U ∈ B G } . Then U H = H for each open subset U of the space G .Let π : G → T G be the quotient homomorphism. Therefore the family { π ( U ) : U ∈ U } is locally finite too. Since the space T is pseudocompact, the family { π ( U ) : U ∈ U } isfinite. Thus the family U is finite too. (cid:3) Proof. (Of Proposition 24). Let (
G, τ ) be the product of a nonempty family { ( G α , τ α ) : α ∈ A } of pseudocompact paratopological groups. Let B r be an open base of the topology τ r at the unit e of the paratopological group ( G, τ r ). Put H = T { U ∩ U − : U ∈ B r } . ¿Lemma 20 implies that ( G, τ r ) = Q α ∈ A ( G α , τ αr ). Let π : G → G/H be the quotient map.
Let α ∈ A be an arbitrary element, B αr an open base of the topology τ αr at the unit e α of the paratopological group ( G α , τ αr ), H α = T { U ∩ U − : U ∈ B αr } , π α : ( G α , τ αr ) → ( G α , τ αr ) /H α the quotient map, and p α : ( G, τ r ) → ( G α , τ αr ) the projection.Let i : Q α ∈ A ( G α , τ αr ) → Q α ∈ A ( G α , τ αr ) /H α be the product of the family { π α : α ∈ A } . Proposition 2.3.29 from [Eng] implies that the map i is open. Since H = { e } τ r ∩ ( { e } τ r ) − and H α = { e α } τ αr ∩ ( { e } τ αr ) − for each α ∈ A , we see that H = Q α ∈ A H α = T α ∈ A p − α ( H α ) = ker i . Let i ′ : ( G, τ r ) /H → Q α ∈ A ( G α , τ αr ) /H α be a map such that i ′ ( xH ) = i ( x ) for each x ∈ G . Theorem on continuous epimorphism [Rav, p. 42] impliesthat the map i ′ is well-defined and a topological isomorphism.Let α ∈ A be an arbitrary element. Then ( G α , τ αr ) is a T pseudocompact paratopolog-ical group (see the remark after Lemma 1). So ( G α , τ αr ) is a pseudocompact topologicalgroup by Lemma 9. Then the group ( G α , τ αr ) /H α is a T pseudocompact topologicalgroup. Therefore by Theorem 1.4 from [ComRos], the product Q α ∈ A ( G α , τ αr ) /H α is pseu-docompact. Hence the group ( G, τ r ) /H = T ( G, τ r ) is pseudocompact. By Lemma 21,the group ( G, τ r ) is pseudocompact. Thus by Lemma 3, the group ( G, τ ) is pseudocom-pact. (cid:3)
Problem 4.
Is the product of two 2-pseudocompact paratopological groups 2-pseudocompact?Of a countable nonempty family? Of an arbitrary nonempty family?Is the product of two countably pracompact paratopological groups countably pracompact?Of a countable nonempty family? Of an arbitrary nonempty family?The product of a nonempty family of 2-pseudocompact topological groups is 2-pseudocompact.Are there other nontrivial conditions sufficient for 2-pseudocompactness of the product? Isthe product of two countably compact paratopological groups 2-pseudocompact? Of a count-able nonempty family? Of an arbitrary nonempty family? In particular, by [RavRez, Pr.3] , the square ( G, σ ) × ( G, σ ) of a group ( G, σ ) from Example 2 is not countably compact.By Proposition 24, ( G, σ ) × ( G, σ ) is pseudocompact. Is ( G, σ ) × ( G, σ ) Extensions of pseudocompact paratopological groups
In this section we shall use the next results and notions to prove Proposition 25, whichpositively solves Problem 5.3 and quasiregular case of Problem 5.4 from [Tka2].
Proposition 25.
Let N be a normal pseudocompact subgroup of a paratopological group G such that both N and the quotient paratopological group G/N are pseudocompact. Then G is pseudocompact. Lemma 22.
Let ( G, τ ) and ( ˆ G, σ ) be paratopological groups such that G ≤ ˆ G , σ | G ⊃ τ , ( G, τ ) is a T -space and the set G is dense in the space ( ˆ G, σ ) . Then σ r | G ⊃ τ . Inparticular, if σ | G = τ then σ r | G = τ .Proof. Let U be an arbitrary neighborhood of the unit in the topology τ . Since thetopology τ is T there exists an open neighborhood V of the unit in the topology τ suchthat V τ ⊂ U . Since σ | G ⊃ τ , there exists an open neighborhood V ′ of the unit in thetopology σ such that V ′ ∩ G = V . Since the group G is dense in the space ( ˆ G, σ ) we seethat V ′ σ = V ′ ∩ G σ = V σ . Then V ′ σ ∩ G = V σ ∩ G ⊂ V τ ⊂ U . (cid:3) Example 6.
Let ( G, τ ) and ( ˆ G, σ ) be paratopological groups such that G ≤ ˆ G . We try todefine a topology τ σ on the group ˆ G as follows. SEUDOCOMPACT PARATOPOLOGICAL GROUPS 19
In this example as · we denote the closure with respect to the topology σ .Let N be the family of all sequences U = { U n : n ∈ ω } of neighborhoods of the unit inthe topology τ . For every sequence U = { U n : n ∈ ω } ∈ N put U = S n ∈ ω U · · · U n . Put B = {U : U ∈ N } .Now we shall check that the family B satisfies Pontrjagin conditions (see [Rav, Pr. 1] ).Condition 1. It is satisfied because for each pair U = { U n : n ∈ ω } , V = { V n : n ∈ ω } of members of the family N we have U ∩ V ⊂ U ∩ V , where
U ∩ V = { U n ∩ V n : n ∈ ω } .Condition 2. Let U = { U n : n ∈ ω } be an arbitrary member of the family N . Byinduction we can construct a sequence V = { V n : n ∈ ω } ∈ N such that V ⊂ U and V n +1 ⊂ V n for each n ∈ ω . The last condition implies that V · · · V n ⊂ V · · · V n − V n − ⊂ V · · · V n − V n − ⊂ · · · ⊂ V V for each n ∈ ω . Since the multiplication on the group ( ˆ G , σ ) is continuous, we have that V · · · V n ⊂ V · · · V n ⊂ V V for each n ∈ ω . Then V = S n ∈ ω V · · · V n ⊂ V V and V · V ⊂ V V · V V ⊂ V ⊂ U ⊂ U . Hence Condition 2is satisfied.Condition 3. Let U = { U n : n ∈ ω } be an arbitrary member of the family N and x ∈ U . Then there exists a number m ∈ ω such that x ∈ U · · · U m . For each n ∈ ω put V n = U m + n +1 and put V = { V n : n ∈ ω } . Then V ∈ N , V ∈ B and x V ⊂ U · · · U m [ n ∈ ω V · · · V n ! = U · · · U m [ n ∈ ω U m +1 · · · U n + m +1 ! = [ n ∈ ω U · · · U m · U m +1 · · · U n + m +1 ⊂ [ k ∈ ω U · · · U k = U . Hence Condition 3 is satisfied.Condition 4. We claim that this condition is satisfied provided the groups ( G, τ ) and ( ˆ G, σ ) satisfy one of the following conditions.(C1) The group G is contained in the center of the group ˆ G . Indeed, in this casewe have the following. Let U = { U n : n ∈ ω } be an arbitrary member of the fam-ily N and x ∈ ˆ G . Then x − U x = x − (cid:0)S n ∈ ω U · · · U n (cid:1) x = S n ∈ ω x − U · · · U n x = S n ∈ ω ( x − U x ) · · · ( x − U n x ) = S n ∈ ω ( x − U x ) · · · ( x − U n x ) = S n ∈ ω U · · · U n = U . Hencein this case Condition 4 is satisfied.(C2) The set G is dense in ( ˆ G, σ ♯ ) and ( G, τ ) is a paratopological SIN-group. (Recallthat a paratopological group G is called a paratopological SIN-group provided G possessesa neighborhood base B at the unit such that y − U y = U for any U ∈ B and y ∈ G (asexpected, SIN is abbreviated from Small Invariant Neighborhoods)). Indeed, in this casewe have the following. Let U = { U n : n ∈ ω } be an arbitrary member of the family N and x ∈ ˆ G . For each n ∈ ω there exists a neighborhood V n of the unit in the topology τ such that y − V n y ⊂ U n for each element y ∈ G . Suppose that there exists an element z ∈ x − V n x \ U n . Then z = x − vx for some element x ∈ V n . Since the multiplication onthe group ( ˆ G, σ ) is continuous, there exist a neighborhood U of the unit in the topology τ such that U x − vxU ⊂ ˆ G \ U n . Since the set G is dense in the group ( ˆ G, σ ♯ ) , there existsan element y ∈ G ∩ x ( U ∩ U − ) . Then y − vy ⊂ U x − vxU ⊂ ˆ G \ U n , a contradiction.Therefore x − V n x ⊂ U n and x − V n x ⊂ U n . Put V = { V n : n ∈ ω } . Then V ∈ N , V ∈ B and x − V x = x − [ n ∈ ω V · · · V n ! x = [ n ∈ ω x − V · · · V n x = [ n ∈ ω ( x − V x ) · · · ( x − V n x ) ⊂ [ n ∈ ω U · · · U n = U . Hence in this case Condition 4 is satisfied.(C3) The set G is dense in a topological group ( ˆ G, σ ) , and σ | G ⊃ τ . Indeed, in thiscase we have the following.Let U = { U n : n ∈ ω } be an arbitrary member of the family N and x ∈ ˆ G . For each n ∈ ω there exists a neighborhood V n of the unit in the topology τ such that V n ⊂ U n . Theset V n ∩ V − n is a neighborhood of the unit of the topological group ( G, τ ♯ ) . Since the set G is dense in ( ˆ G, σ ) and σ | G ⊃ τ ♯ , we see that the set V n ∩ V − n has the nonempty interiorin the group ( ˆ G, σ ) . Therefore there exists a point x n ∈ xV n ∩ V − n ∩ G .Since ( G, τ ) is a paratopological group, the set x n V n x − n is a neighborhood of the unitof the group ( G, τ ) . Moreover, x − x n V n x − n x = x − x n V n x − n x ⊂ x − xV n ∩ V − n · V n · ( xV n ∩ V − n ) − x = V n ∩ V − n · V n · V n ∩ V − n ⊂ V n · V n · V n ⊂ V n ⊂ U n . Put V = { x n V n x − n : n ∈ ω } . Then V ∈ N , V ∈ B and x − V x = x − [ n ∈ ω x V x − · · · x n V n x − n ! x = [ n ∈ ω x − x V x − · · · x n V n x − n x = [ n ∈ ω x − x V x − x · · · x − x n V n x − n x ⊂ [ n ∈ ω U · · · U n = U . Hence Condition 4 is satisfied.
From here till the end of the example we shall assume that Condition 4 is satisfied .Since Conditions 1-4 are satisfied, the family { x U : x ∈ ˆ G, U ∈ N } is a base of asemigroup topology on the group ˆ G (see, for instance, [Rav2, p.93]). Denote this topologyas τ σ .The construction of the topology τ σ implies τ σ | G ⊂ τ .If σ | G ⊃ τ then τ σ | G ⊃ τ r . Indeed, let U r be an arbitrary neighborhood of the unitin the topology τ r . Since the topology τ r is T (see the paragraph after Lemma 1), thereexists an open neighborhood V r of the unit in the topology τ r such that V rτ r ⊂ U r . Bydefinition of the topology τ r , there exists a canonically open neighborhood V of the unitin the topology τ such that V ⊂ V r . Canonical openness of the neighborhood V means V = int τ V τ , where the interior is taken with the respect to the topology τ . By inductionwe can build a sequence { V n : n ∈ ω } of open neighborhoods of the unit in the topology τ such that V ⊂ V and V n +1 ⊂ V n for each n ∈ ω . Then V = { V n : n ∈ ω } ∈ N . Similarlyto Condition 2 we can show that V ⊂ V . Then V ∩ G ⊂ V ∩ G ⊂ V τ , because σ | G ⊃ τ .Then V ∩ G ⊂ V τ ⊂ V τr ⊂ V rτ r ⊂ U r .If σ | G ⊃ τ and the set G is dense in the space ( ˆ G, σ ) then τ σ ⊂ σ . Indeed, eachneighborhood U of the unit in the topology τ contains the closure U in the topology σ of aneighborhood U of the unit in the topology τ . Since σ | G ⊃ τ , there exists a neighborhood V of the unit in the topology σ such that V ∩ G ⊂ U . Since the set G is dense in the space ( ˆ G, σ ) , we see that V ⊂ V = V ∩ G ⊂ U .If σ | G ⊂ τ then τ σ ⊃ σ r . Indeed, let U r be an arbitrary neighborhood of the unitin the topology σ r . Since the topology σ r is T there exists an open neighborhood V r ofthe unit in the topology σ r such that V rσ r ⊂ U r . By definition of the topology σ r , thereexists a canonically open neighborhood V of the unit in the topology σ such that V ⊂ V r .Canonical openness of the neighborhood V means V = int σ V , where the interior is takenwith the respect to the topology σ . Since σ | G ⊂ τ , by induction we can build a sequence SEUDOCOMPACT PARATOPOLOGICAL GROUPS 21 { V n : n ∈ ω } of open neighborhoods of the unit in the topology τ such that V ⊂ V and V n +1 ⊂ V n for each n ∈ ω . Then V = { V n : n ∈ ω } ∈ N . Similarly to Condition 2 we canshow that V ⊂ V . Then V ⊂ V ⊂ V r ⊂ V rσ r ⊂ U r .Hence, if the group ( G, τ ) is T and σ | G ⊃ τ then τ σ | G = τ r = τ . Moreover, if theset G is dense in the space ( ˆ G, σ ) then then τ σ ⊂ σ and the set G is dense in the space ( ˆ G, τ σ ) too. Then, by Lemma 22, τ σr | G = τ . Lemma 23.
Let G be a T paratopological group and N be a compact normal subgroup ofthe group G . Then a quotient paratopological group G/N is a T space too.Proof. Let π : G → G/N be the quotient map, ˜ F be an arbitrary closed subset of theparatopological group G/H and ˜ x ∈ G/H \ ˜ F be an arbitrary point. Take an arbitrarypoint x ∈ π − (˜ x ). Then x π − ( ˜ F ). Proposition 1.8 from [Rav] imply that there existson open neighborhood U of the unit of the group G such that U N x ∩ π − ( ˜ F ) = ∅ . Sincethe group G is T , there exists exists on open neighborhood U of the unit of the group G such that V ⊂ U . Propositions 1.12 and 1.13 from [Rav] imply that the map π is openand closed. Then ˜ x ∈ π ( V x ) ⊂ π ( V x ) ⊂ π ( V x ) ⊂ π ( U N x ) ⊂ ( G/N ) \ ˜ F . Thus the space G/N is T . (cid:3) Corollary 6. If G is a T paratopological group then the quotient paratopological group T G is regular.Proof. Clearly, the group H = T { U ∩ U − : U ∈ B G } is compact. Therefore the space T G is T . By [Eng, Ex. 1.5.D], each T + T space is regular. (cid:3) Lemma 24. [Rav, Cor.1.2]
Let N and L be normal subgroups of a paratopological group G and L ⊂ N . Then G/N ≃ ( G/L ) / ( N/L ) . (cid:3) Lemma 25.
Let N be a normal pseudocompact subgroup of a topological group G suchthat both N and the quotient topological group G/N are pseudocompact. Then G is pseu-docompact.Proof. The closure N of the group N is a normal pseudocompact subgroup of a topologi-cal group G . The space N is pseudocompact, because it contains a dense pseudocompactspace N . By Lemma 24, G/N ≃ ( G/N ) / ( N /N ). Therefore the space
G/N is pseudo-compact as an image of a pseudocompact space
G/N . Since the pseudocompactness isa three space property in topological groups (see the paragraph before Problem 5.3 in[Tka2]), the group G is pseudocompact too. (cid:3) Given a paratopological group G let τ ♯ be the weakest group topology on G , strongerthan the topology of G . The topological group G ♯ = ( G, τ ♯ ) is called the group co-reflexion of G . According to [Rav], a neighborhood base of the unit of the group co-reflexion G ♯ ofa paratopological group G consists of the sets U ∩ U − where U runs over neighborhoodsof the unit in G .At last we are able to prove Proposition 25. Proof.
At first we suppose that the paratopological group (
G, τ ) is regular. Then thegroup co-reflexion G ♯ is a T -space and hence a Tychonoff space. Let ( c G ♯ , b τ ♯ ) be a Raikovcompletion of a topological group G ♯ .In the proof of this proposition as · we denote the closure with respect to the topology b τ ♯ . By Proposition 3, a pseudocompact regular paratopological group (
N, τ | N ) is a topo-logical group. Therefore τ ♯ | N ⊂ τ | N and a space ( N, τ ♯ | N ) is pseudocompact. The space( N , b τ ♯ | N ) is pseudocompact because it has a dense pseudocompact subspace ( N, τ ♯ | N ). ByTheorem 3.7.2 from [ArkTka], the topological group ( N , b τ ♯ | N ) is precompact. Since N is aclosed precompact subset of a Raikov complete topological group ( c G ♯ , b τ ♯ ), by Proposition3.7.9 from [ArkTka], the space ( N , b τ ♯ | N ) is compact.Since the group G ♯ is dense in its completion, which is a topological group ( c G ♯ , b τ ♯ )and b τ ♯ | G = τ ♯ ⊃ τ , we see that, by Example 6 and its Condition C3, on the group c G ♯ exists the topology τ b τ ♯ , which we denote as τ . Moreover, by Example 6, τ ⊂ b τ ♯ and τ r | G = τ | G = τ r = τ .Now we use the following Theorem on isomorphism [Rav, p. 42 in corrected and sub-stituted notations]. Lemma 26.
Let ( c G ♯ , τ r ) be a paratopological group, G be a subgroup of c G ♯ and N be anormal subgroup of c G ♯ . Then GN is a subgroup of c G ♯ , N = N ∩ G is a normal subgroup of G and the map σ : G/N → ( GN ) /N defined as σ ( gN ) = gN is a homomorphic bijectivecontinuous map. (cid:3) Thus the space (
GN , τ r | GN ) /N is pseudocompact as a continuous image of a pseu-docompact space ( G, τ r | G ) /N = ( G, τ ) /N . We recall that ( GN , τ r | GN ) is T paratopo-logical group, the space ( N , b τ ♯ | N ) is compact and τ r ⊂ b τ ♯ . Thus the space ( N , τ r | N ) iscompact too. Now Lemma 23 implies that the space ( GN , τ r | GN ) /N is T . By Propo-sition 3 ( GN , τ r | GN ) /N is a pseudocompact topological group. Since ( N , τ r | N ) is acompact paratopological group, it is a topological group too (for instance, by Proposi-tion 1). Then by [Rav4, 1.3] or [Rav3, Pr. 5.3], ( GN , τ r | GN ) is a topological group too.Then, by Lemma 25, the group ( GN , τ r | GN ) is pseudocompact. Since this group is densein the group ( c G ♯ , τ r ), the group ( c G ♯ , τ r ) is pseudocompact too. Since the group ( c G ♯ , τ r )is T , by Proposition 3 ( c G ♯ , τ r ) is a topological group. Since τ r | G = τ , ( G, τ ) is a topolog-ical group as a subgroup of a topological group ( c G ♯ , τ r ). Then the groups ( N, τ | N ) and( G, τ ) /N are topological too. Then, by Lemma 25, the group ( G, τ ) is pseudocompact.Now we consider the case when the group (
G, τ ) is not necessarily regular. Since τ r ⊂ τ then τ r | N ⊂ τ | N and the quotient topology of the group ( G, τ r ) /N is weakerthan the quotient topology of the group ( G, τ ) /N . Hence both groups ( N, τ r ) and( G, τ r ) /N are pseudocompact as images of pseudocompact spaces. Let H = T { U ∩ U − : U ∈ B ( G,τ r ) } . Since the topology τ r is T (see the paragraph after Lemma 1),by Corollary 6, the group T ( G, τ r ) = ( G, τ r ) /H is regular. Since both N and H are normal subgroups of the group G , a set N H is a normal subgroup of the group G . By Lemma 24, (( G, τ r ) /N ) / (( N H, τ r | N H ) /N ) ≃ ( G, τ r ) /N H . Therefore the group( G, τ r ) /N H is pseudocompact as an image of a pseudocompact group ( G, τ r ) /N . ByLemma 24 again, (( G, τ r ) /H ) / (( N H, τ r | N H ) /H ) ≃ ( G, τ r ) /N H . By Theorem on iso-morphism [Rav, p. 42], the group ( N H, τ r | N H ) /H is a continuous bijective image ofthe group ( N, τ r | N ) / ( H ∩ N ), which is a continuous image of a pseudocompact group( N, τ r | N ). Therefore the group ( N H, τ r | N H ) /H is pseudocompact. Since the group( G, τ r ) /H is regular, the groups ( G, τ r ) /N H and ( N H, τ r | N H ) /H are pseudocompact,and (( G, τ r ) /H ) / (( N H, τ r | N H ) /H ) ≃ ( G, τ r ) /N H , by the already proved proposition forthe case of the regular group, the group ( G, τ r ) /H = T ( G, τ r ) is pseudocompact. Then SEUDOCOMPACT PARATOPOLOGICAL GROUPS 23 by Lemma 21, the group (
G, τ r ) is pseudocompact too. Finally, by Lemma 3, the group( G, τ ) is pseudocompact. (cid:3) Acknowledgements
The author would like to express his thanks to Artur Tomita and to Manuel Sanchiswho provided him their articles, to Taras Banakh who turned his attention to the classof countably pracompact spaces, intermediate between the classes of countably compactand pseudocompact spaces and to Mikhail Tkachenko who have inspired him to proveProposition 2.
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