aa r X i v : . [ m a t h . G R ] M a r ON H-CLOSED PARATOPOLOGICAL GROUPS
Oleksandr RAVSKY
Ivan Franko National University of Lviv, 1 Universitetska str. 79000 Lviv, Ukraine
A Hausdorff paratopological group is it H-closed if it is closed in every Hausdorffparatopological group containing it as a paratopological subgroup. Obtained a criterionwhen abelian topological group is H-closed and for some classes of abelian paratopolog-ical groups are obtained simple criteria of H-closedness.Key words: paratopological group, minimal topological group, absolutely closed topo-logical group.
All topological spaces considered in this paper are Hausdorff, if the opposite is notstated. We shall use the following notations. Let A be a subset of a group and n bean integer. Put A n = { a a · · · a n : a i ∈ A } and nA = { a n : a ∈ A } . For a grouptopology τ the closure of set A we define as A τ and the base of the unit as B τ .A topological space X of a class C of topological spaces is C-closed provided X isclosed in any space Y of the class C containing X as a subspace. It is well known thatwhen C is the class of Tychonoff spaces than C -closedness coincides with compactness.For the class of Hausdorff spaces the following conditions for a space X are equivalent[1, 3.12.5](1) The space X is H-closed.(2) If V is a centered family of open subsets of X then T { V : V ∈ V} 6 = ∅ .(3) Every ultrafilter in the family of all open subsets of X is convergent.(4) Every cover U of the space X contains a finite subfamily V such that S { V : V ∈ V} = X .The group G with topology τ is called a paratopological group if the multiplicationon the group G is continuous. If the inversion on the group G is continuous then ( G, τ )is a topological group . A group (
G, τ ) is paratopological if and only if the followingconditions (known as Pontrjagin conditions) are satisfied for base B at unit e of G [4,5].1. T { U U − : U ∈ B} = { e } .2. ( ∀ U, V ∈ B )( ∃ W ∈ B ) : W ⊂ U ∩ V .3. ( ∀ U ∈ B )( ∃ V ∈ B ) : V ⊂ U .4. ( ∀ U ∈ B )( ∀ u ∈ U )( ∃ V ∈ B ) : uV ⊂ U .5. ( ∀ U ∈ B )( ∀ g ∈ G )( ∃ V ∈ B ) : g − V g ⊂ U .The paratopological group G is a topological group if and only if6. ( ∀ U ∈ B )( ∃ V ∈ B ) : V − ⊂ U .A topological group is absolutely closed if it is closed in every Hausdorff topologicalgroup containing it as a topological subgroup. A topological group G is H-closed ifand only if it is Rajkov-complete , that is complete with respect to the upper uniformitywhich is defined as the least upper bound
L ∨ R of the left and the right uniformitieson 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 c (cid:13) Ravsky Oleksandr, 2002 OLEKSANDR RAVSKY right uniformity R , the condition x − y ∈ U is replaced by yx − ∈ U . The Rajkovcompletion ˆ G of a topological group G is the completion of G with respect to theupper uniformity L ∨ R . For every topological group G the space ˆ G has a naturalstructure of a topological group. The group ˆ G can be defined as a unique (up to anisomorphism) Rajkov complete group containing G as a dense subgroup.A paratopological group is H-closed if it is closed in every Hausdorff paratopologicalgroup containing it as a subgroup. In the present section we shall consider H-closedparatopological groups.
Question
Let G be a regular paratopological group which is closed in every regularparatopological group containing it as a subgroup. Is G H-closed?
1. Lemma.
Let ( G, τ ) be a paratopological group. If there exists a paratopology σ onthe group G × Z such that σ | G ⊂ τ and e ∈ ( G, σ then ( G, τ ) is not H-closed. Proof.
We shall build the paratopology ρ on the group G × Z such that ρ | G = τ and G ρ = G . Determine the base of unit B ρ as follows. Let S = { ( x, n ) : x ∈ G, n > } .For every neighborhoods U ∈ τ , U ∈ σ such that U ⊂ U put ( U , U ) = U ∪ ( U ∩ S ). Put B ρ = { ( U , U ) : U ∈ B τ , U ∈ B σ } . Verify that B ρ satisfies the Pontrjaginconditions.1. It is satisfied since ( U , U ) ⊂ U .2. It is satisfied since ( U ∩ V , U ∩ V ) ⊂ ( U , U ) ∩ ( V , V ).3. Select V ∈ B σ and V ∈ B τ such that V ⊂ U , V ⊂ U and V ⊂ V . Let y , y ∈ ( V , V ). The following cases are possibleA. y , y ∈ V . Then y y ∈ V ∈ ( U , U ).B. y ∈ V , y ∈ V ∩ S . Then y y ∈ V ∈ U . Since y ∈ G and y ∈ S then y y ∈ S and hence y y ∈ U ∩ S .C. y ∈ V ∩ S, y ∈ V is similar to the case B.D. y , y ∈ V ∩ S . Since S is a semigroup then y y ∈ U ∩ S .4. Let y ∈ ( U , U ). There exist V ∈ B σ and V ∈ B τ such that yV ⊂ U and V ⊂ V . The following cases are possibleA. y ∈ U . We may suppose that yV ⊂ U . Since y ∈ G then y ( V ∩ S ) ⊂ U ∩ S .B. y ∈ U ∩ S . Since V ⊂ G then yV ∈ U ∩ S . Since S is a semigroup and y ∈ S then y ( V ∩ S ) ⊂ U ∩ S . Therefore y ( V , V ) ⊂ ( U , U ).5. Let ( g, n ) ∈ G × Z . There exist V ∈ B σ and V ∈ B τ such that V ⊂ V , g − V g ⊂ U and g − V g ⊂ U . Then ( g, n ) − ( V , V )( g, n ) = g − ( V , V ) g = g − ( V ∪ ( V ∩ S ) g ⊂ U ∪ ( U ∩ S ) = ( U , U ).Therefore ( H, ρ ) is a paratopological group. Since ( U , U ) ∩ G = U then ρ | G = τ .Since e ∈ ( G, σ then for every U ∈ B σ there exists g ∈ G such that ( g, ∈ U .Then g ∈ ( e, − U ∩ S ) and therefore ( e, − ∈ G ρ . (cid:3) A group topology τ on the group G is called complementable if there exist anondiscrete group topology τ on G and neighborhoods U i ∈ τ i such that U ∩ U = { e } . In this case we say that τ is a complement to τ . Proposition 1.4 from [1] impliesthat in this case a topology τ ∧ τ is Hausdorff.A Banach measure is a real function µ defined on the family of all subsets of agroup G which satisfies the following conditions:(a) µ ( G ) = 1.(b) if A, B ⊂ G and A ∩ B = ∅ then µ ( A ∪ B ) = µ ( A ) + µ ( B ).(c) µ ( gA ) = µ ( A ) for every element g ∈ G and for every subset A ⊂ G . N H-CLOSED PARATOPOLOGICAL GROUPS 3
2. Lemma. [3, p.37] . Let G be an abelian group and let µ be a Banach measureon G . Let τ be a group topology on G . Suppose that the set nG is U -unboundedfor some natural number n and for some neighborhood U of zero in ( G, τ ) . Then µ ( { x ∈ G : nx ∈ gW } ) = 0 for every element g ∈ G and for every neighborhood W ofzero satisfying W W − ⊂ U . Let U be a neighborhood of zero in a topological group ( G, τ ). We say that asubset A ⊂ G is U -unbounded if A KU for every finite subset K ⊂ G .Given any elements a , a , . . . , a n of an abelian group G put Y ( a , a , . . . , a n ) = { a x a x · · · a x n n : 0 x i i + 1 , i n, X x i > } ,X ( a , a , . . . , a n ) = { a x a x · · · a x n n : − ( i + 1) x i i + 1 , i n } . Then X ( a , a , . . . , a n ) = Y ( a , a , . . . , a n ) Y ( a , a , . . . , a n ) − .
3. Lemma.
Let ( G, τ ) be an abelian paratopological group of the infinite exponent.If there exists a neighborhood U ∈ B τ such that a group nG is U U − -unbounded forevery natural number n then the paratopological group ( G, τ ) is not H-closed. Proof.
Define a seminorm | · | on the group G such that for all x, y ∈ G holds | xy | | x | + | y | . Suppose that there exists a non periodic element x ∈ G . Determinea map φ : h x i → Z putting φ ( x n ) = n . Since Q is a divisible group then themap φ can be extended to a homomorphism φ : G → Q . Put | x | = | φ ( x ) | for everyelement x ∈ G . If G is periodic then put | e | = 0 and | x | = [ln ord ( x )] + 1, where ord ( x ) denotes the order of the element x .Fix a neighborhood V ∈ B τ such that V ⊂ U and put W = V V − . We shallconstruct a sequence { a n } such that(a) | a n | > n .(b) W ∩ X ( a , a , . . . , a n ) = { e } .(c) Y ( a , a , . . . , a n ) e .(d) if − n k n, k = 0 then a kn X ( a , a , . . . , a n − ).Take any element a W . Suppose that the elements a , . . . , a n have been chosensatisfying conditions (a) and (b). Put B n = { x ∈ G : ( ∀ g ∈ X ( a , a , . . . , a n − ))( ∀ k ∈ Z \{ } : − e n +1 k e n +1 ) : kx gW } . If the group G is periodic then | x | > n for every element x ∈ B n . Lemma 2 impliesthat µ ( B n ) = 1. If the group G is not periodic then the construction of the seminorm | · | implies that µ ( { x ∈ G : | x | n } ) = µ ( φ − [ − n ; n ]) = 0. In both cases thereexists an element a n ∈ B n such that | a n | > n . Then W ∩ X ( a , a , . . . , a n ) = ∅ .Considering a subsequence and applying condition (a) we can satisfy conditions (c)and (d) also.Define a base B τ { a n } at the unit of group topology τ { a n } on the group G × Z asfollows. Put A + n = { ( e, } ∩ { ( a k ,
1) : k > n } . For every increasing sequence { n k } put A [ n k ] = S l ∈ N A + n · · · A + n l . Put B τ { a n } = { A [ n k ] } . We claim that ( G × Z, τ { a n } ) isa zero dimensional paratopological group.Put F = S n ∈ ω X ( a , a , . . . , a n ). Let A [ n k ] ∈ B τ { a n } , ( x, n x ) A [ n k ]. If x F then( x, n x ) A [ n ] ∩ A [ n k ] = ∅ . Let x ∈ X ( a , a , . . . , a m ). Put m k = m + k . Suppose that OLEKSANDR RAVSKY ( x, n x ) A [ m k ] ∩ A [ n k ] = ∅ . Select the minimal k such that ( x, n x )( A + m · · · A + m k ) ∩ A [ n k ] = ∅ . Let( ∗ ) ( x, n x )( a l , · · · ( a l k ,
1) = ( a l ′ , · · · ( a l ′ k ′ , i, i ′ holds m i l i l i +1 , n ′ i l ′ i ′ l ′ i ′ +1 . Remark that a member a q occurs in each part of the equality ( ∗∗ ) no more than q times. If l k > l ′ k ′ then if wemove all members which are not equal to ( a l k ,
1) from the left side of the equality(*) to the right one, we obtain contradiction to condition ( d ). The case l k < l ′ k ′ isconsidered similarly. Therefore l k = l ′ k ′ , a contradiction to that k is the minimalnumber such that the equality (*) holds. It is showed similarly that if x = e and m k = m + k + 1 then ( x, n x ) A [ m k ]. If x = e and n x = 0 then condition (c) impliesthat A [ n ] ( x, n x ). Hence Pontrjagin condition 1 for B τ { a n } is satisfied. Since A [ n k ] ⊂ A [ n k ], Pontrjagin condition 3 is satisfied. All other Pontrjagin conditionsare obvious.Condition (b) implies that A [ n ] A [ n ] − ∩ V V − = { ( e, } . Therefore the topology τ { a n } g is a complement to the topology ( τ × { } ) g , where τ × { } is the producttopology on the group ( G, τ ) × Z . Therefore the topology σ = τ { a n } ( τ × { } ) isHausdorff. Since ( e, ∈ ( G, τ { a n } ⊂ ( G, σ then ( G, τ ) is not H-closed. (cid:3)
We shall need the following lemma.
4. Lemma.
Let G be a paratopological group and H be a normal subgroup of thegroup G . If H and G/H are topological groups then G is a topological group. Proof.
Let U be an arbitrary neighborhood of the unit. There exist neighborhoods V, W of the unit such that V ⊂ U , ( V − ) ∩ H ⊂ U and W ⊂ V , W − ⊂ V H .If x ∈ W − then there exist elements v ∈ V, h ∈ H such that x = vh . Then h = v − x ∈ V − W − ∩ H ⊂ U . Therefore x ∈ V U ⊂ U . Hence G is a topologicalgroup. (cid:3)
5. Theorem.
An abelian topological group ( G, τ ) is H-closed if and only if ( G, τ ) isRajkov complete and for every group topology σ ⊂ τ on G the quotient group ˆ G/G isperiodic, where ˆ G is the Rajkov completion of the group ( G, σ ) . Proof.
Suppose that there exists a group topology σ ⊂ τ on G such that thequotient group ˆ G/G is not periodic, where ˆ G is the Rajkov completion of the group( G, σ ). Select a non periodic element x ∈ ˆ G such that h x i ∩ G = { e } . Then G × h x i is naturally isomorphic to a group G × Z and Lemma 1 implies that the group ( G, τ )is not H-closed.Let a paratopological group (
H, τ ′ ) contains ( G, τ ) as non closed subgroup. Since G is abelian then G is an abelian semigroup. Choose an arbitrary element x ∈ G \ G .Then a group hull F = h G, x i with a topology τ ′ | F is an abelian paratopologicalgroup. Then the group G is dense in ( F, τ ′ g ). Since the Rajkov completion ˆ F ofthe topological group ( F, τ ′ | F g ) is periodic then there exists a natural number n suchthat x n ∈ G . Therefore F n ⊂ G . Lemma 4 implies that F is a topological group andtherefore G is closed in ( F, τ ′ g ), a contradiction. (cid:3)
6. Corollary.
A Rajkov completion of a isomorphic condensation of H-closed abeliantopological group is H-closed.
N H-CLOSED PARATOPOLOGICAL GROUPS 5
7. Proposition.
Let G be a Rajkov complete topological group, H be H-closedparatopological subgroup of the group G . If a group G/H has finite exponent then G is an H-closed paratopological group. Proof.
Select a number n such that g n ∈ H for every element g ∈ G . Let F ⊃ G be a paratopological group. Since H is closed in F then for every element g ∈ G holds g n ∈ H . Denote continuous maps φ : G → G as φ ( g ) = g n − and ψ : G → H as ψ ( g ) = ( g n ) − . Then for every element g ∈ G holds g − = φ ( g ) ψ ( g ) and thereforethe inversion on the group G is continuous. Since G is a topological group and G isRajkov complete then G = G . (cid:3)
8. Proposition.
Let G be a paratopological group and K be a compact normalsubgroup of the group G . If a group G/K is H-closed then the group G is H-closed. Proof.
Suppose that there exists a paratopological group F containing the group G such that G = G . Since K is compact then F/K is a Hausdorff paratopologicalgroup by Proposition 1.13 from [4]. Let π : F → F/K be the standard map. Then
G/K ⊃ π ( π − ( G/K )) ⊃ π ( G ) = π ( G ) = G/K . This implies that the group
G/K isnot H-closed, a contradiction. (cid:3)
Let G be a topological group, N be a closed normal subgroup of the group G .Then if N and G/N are Rajkov complete so is the group G [5]. This suggests thefollowing
9. Question.
Let G be a paratopological group, N be a closed normal subgroup ofthe group G and the groups N and G/N are H-closed. Is the group G H-closed?Let (
G, τ ) be a paratopological group. Then there exists the finest group topology τ g coarser than τ (see [2]), which is called the group reflection of the topology τ .
10. Proposition.
Let ( G, τ ) be an abelian paratopological group. If ( G, τ g ) is H-closed then ( G, τ ) is H-closed. If ( G, τ ) is H-closed and ( G, τ g ) is Rajkov completethen ( G, τ g ) is H-closed. Proof.
Suppose that the group (
G, τ g ) is H-closed and ( G, τ ) is not. Let a paratopo-logical ( H, ˆ τ ) contains ( G, τ ) as non closed subgroup. Without loss of generality wemay suppose that there exists an element x ∈ H \ G such that H = h G, x i and thegroup H is abelian. Let ˆ τ g be the group reflection of the topology ˆ τ . Since ˆ τ g | G ⊂ τ g then Theorem 5 implies that the group H/G is periodic. Without loss of generalitywe may suppose that x p ∈ G for some prime p .Denote the family of neighborhoods at unit in the topology τ as B ˆ τ . Let U ∈ B ˆ τ .If U ∩ xG = ∅ then there exists a neighborhood V of unit such that V p ⊂ U and thus V ⊂ G and G is open in ( H, ˆ τ ). Therefore a set F = { x − ( xG ∩ U ) : U ∈ B τ } is afilter. Let U ∈ B ˆ τ . There exists V ∈ B ˆ τ such that V p ⊂ U . Then ( xG ∩ V ) p ⊂ U . Let xg ∈ ( xG ∩ V ). Then x − ( xG ∩ V ) ⊂ x − (( xg ) − p ( xG ∩ V ) p ) ∩ G ⊂ x − p g − p ( U ∩ G )and hence F is a Cauchy filter in the group ( G, τ g ). Let h ∈ G be a limit of the filter F on the group ( G, τ g ). But then for every neighborhood of the unit U in the topologyˆ τ g holds U ∩ xhU ⊃ U ∩ xh ( U ∩ G ) = ∅ and therefore ( H, ˆ τ g ) is not Hausdorff, acontradiction.Let ( G, τ g ) is Rajkov complete and ( G, τ g ) is not H-closed. Then Theorem 5 impliesthat there exists a group topology σ ⊂ τ on G such that the quotient group ˆ G/G ofthe Rajkov completion ˆ G of the group ( G, σ ) is not periodic. Then Lemma 1 impliesthat a group (
G, τ ) is not H-closed. (cid:3)
OLEKSANDR RAVSKY
11. Lemma.
Let topological group ( H, σ H ) be a closed subgroup of an abelian topo-logical group ( G, τ ) and σ H ⊂ τ | H . Then there exists a group topology σ ⊂ τ on thegroup G such that σ | H = σ H . Proof
Let B τ and B σH be bases of unit of ( G, τ ) and (
H, σ H ) respectively.Put B σ = { U U : U ∈ B τ , U ∈ B σH } . Verify that the family B σ satisfies thePontrjagin conditions.2. It is satisfied since ( U ∩ V )( U ∩ V ) ⊂ U U ∩ V V .3. Select V ∈ B σH and V ∈ B τ such that V ⊂ U , V ⊂ U . Then ( V V ) ⊂ U U .4. Let y ∈ U U . Then there exist points y ∈ U and y ∈ U such that y = y y .Therefore there exist a neighborhoods V ∈ B τ and V ∈ B σH such that y i V i ⊂ U i .Then yV V ⊂ U U .5. It is satisfied since G is abelian.6. ( U − U − ) − ⊂ U U .1. Since all others Pontrjagin conditions are satisfied then it suffice to show that T B σ = { e } . Let x ∈ G and x = e . If x ∈ H then there exists U ∈ B σH such that U x and U ∈ B σ such that U ∩ H ⊂ U . Then U U ∩ { x } = U U ∩ { x } ∩ H ⊂ U ∩ { x } = ∅ . If x H then ( G \ xH ) H x .Therefore ( G, σ ) is a topological group. Since U U ∩ H = ( U ∩ H ) U then σ | H = σ H . (cid:3)
12. Proposition.
A closed subgroup of an H-closed abelian group is H-closed.
Proof.
Let H be a closed subgroup of an H-closed abelian group ( G, τ ). Then G and H are Rajkov complete. Let σ H ⊂ τ | H be a group topology on the group H .Lemma 11 implies that there exists a group topology σ on the group G such that σ | H = σ H . Let ( ˆ G, ˆ σ ) be the Rajkov completion of the group ( G, σ ). Then a closure H ˆ σ of the group H in the group ( ˆ G, ˆ σ ) is a Rajkov completion of the group ( H, σ H ).Let x ∈ H ˆ σ . Theorem 5 implies that there exists n > x n ∈ G . Since H ˆ σ ∩ G = H then x n ∈ H . Therefore Theorem 5 implies that H is H-closed. (cid:3)
13. Proposition.
Let G be a H-closed abelian topological group. Then K = T n ∈ N nG is compact and for each neighborhood U of zero in G there exists a nat-ural n with nG ⊂ KU . Proof.
Let Φ be a filter on G with a base { nG : n ∈ N } , and Ψ be an arbitraryultrafilter on G with Ψ ⊃ Φ. Let U be a closed neighborhood of the unit in G .Lemma 2 implies that there exists a number n such that the set nG is U -bounded.Since nG ∈ Φ and Ψ is an ultrafilter, there exists g ∈ G with gU ∈ Ψ. HenceΨ is a Cauchy filter on G . By the completeness of G , Ψ is convergent. Thereforeeach ultrafilter Ψ on G with Ψ ⊃ Φ converges. In particular each ultrafilter on K isconvergent, and since K is closed, K is compact.To show that there exists a number n with nG ⊂ KU it suffices to prove that KU ∈ Φ. Assume that KU Φ. Then there exists an ultrafilter Ψ ⊃ Φ with G \ KU ∈ Ψ. As we have proved, Ψ is convergent. Clearly lim Ψ ∈ K . Therefore KU ∈ Ψ which is a contradiction. Hence KU ∈ Φ, and this completes the proof. (cid:3)
14. Corollary.
A divisible abelian H-closed topological group is compact. (cid:3)
N H-CLOSED PARATOPOLOGICAL GROUPS 7
15. Proposition.
Every H-closed abelian topological group is a union of compactgroups.
Proof.
Let G be such a group. It suffice to show that every element x ∈ G is contained in a compact subgroup. Let X be the smallest closed subgroup of G containing the element x . Then X = S nk =0 ( kx + nX ) for every natural n . Let U bean arbitrary neighborhood of the zero. By Lemma 15 there exists a natural number n such that nG is U -bounded. Then X is also U -bounded. Hence X is a precompactgroup. Since X is Rajkov complete then X is compact. (cid:3)
16. Conjecture.
An abelian topological group G is H-closed if and only if G isRajkov complete and nG is precompact for some natural n .
17. Proposition.
The Conjecture 16 is true provided the group ( G, τ ) satisfies thefollowing two conditions:(1) There exists a σ -compact subgroup L of G such that G/L is periodic.(2) There exists a group topology τ ′ ⊂ τ such that the Rajkov completion ˆ G of thegroup ( G, τ ′ ) is Baire. Proof.
Let G be such a group and L = S k ∈ N L k be a union of compact subsets L k .Put G ( n, k ) = { x ∈ ˆ G : nx ∈ L k } for every natural n and k . Then every set G ( n, k )is closed. By Theorem 5 holds ˆ G = S n,k ∈ N G ( n, k ). Since ˆ G is Baire then there existnatural numbers n and k such that int G ( n, k ) = ∅ . Then F = G ( n, k ) − G ( n, k )is a neighborhood of the zero. By Corollary 6 the group ˆ G is H-closed. Put K = T n ∈ N n ˆ G . By Proposition 13 there exists a natural m such that m ˆ G ⊂ F + K . Then mnG ⊂ mn ˆ G ⊂ L k − L k + K and hence the group mnG is precompact. (cid:3) ——————————1. Engelking R.
General topology. – Monografie Matematyczne, Vol. 60, Polish Sci-entific Publ. – Warsaw, 1977.2.
Graev M.I.
Theory of topological groups // UMN, 1950 (in Russian).3.
Protasov I., Zelenyuk E. , Topologies on Groups Determined by Sequences. – VNTLPublishers, 1999.4.
Ravsky O.V. // Paratopological groups I // Matematychni Studii. – 2001. –Vol. 16. – 1. – P.37-48.5.