Bounded Topological Groups

BOUNDED TOPOLOGICAL GR OUPS KAZEM HAGHNEJAD AZAR Abstract. In this note for a topological group G , we introduce a bounded sub set of G and we find some relationships of this de finition with other topological properties of G . 1. prelim inaries and I n tro duction Suppo se that G is a topolo gical gr oup and E ⊆ G . In this pap er, w e wan t to know when E is bounded or un bounded subset of G and if G is metrizablity , we show that E ⊆ G is b ounded with resp ect to top ology if and only if it is bo unded with resp ect to metric. Let E ⊆ G be bounded and closed. Then E is compact subset of G . Conv ersely if E is a comp onent o f e and compa c t, then E is b ounded. W e inv estigated s ome topolog ical prop erty for b ounded subset o f G . Now w e in tro duce some notations a nd definitions that we used throughout this pap er. F or topolog ical group G , e is iden tity element of G a nd for E ⊆ G , E − is clo sure of E and for every n ∈ N , E n = { x 1 x 2 x 3 ...x n : x i ∈ E , 1 ≤ i ≤ n } . A to po lo gical spac e X is O − dimensi onal if the family of all sets that are bo th o pen and closed is op en basis fo r the top ology . 2. Bounded T op ological Groups Definition 2-1. Let G b e top ologic a l group and E ⊆ G . W e say that E is b ounded subset of G , if for every neighborho o d V of e , there is natura l n um b er n such that E ⊆ V n . It is clear that if E is b ounded subset of G a nd H is subgroup of G , then E /H is b ounded subset o f G/H . Theorem 2-2. Let G b e top ologic a l group and metriza ble with res pect to a le ft inv ar iant metr ic d . Then G is bounded with resp ect to topolog y if and only if G is bo unded with resp ect to metric d . 2000 Mathematics Subje ct Classific ation. 46L06; 46L07; 46L10; 47L25. Key wor ds and phr ases. T opological Group, Bounded T op ological Groups, Group. 1 2 Pr o of. Let G b e a bo unded top olog ic a l g r oup a nd ε > 0. T ake d ([0 , ε )) = U × V where U and V are neighbor ho o ds of e . Suppo se that W is symmetric neighbor ho o d of e such that W ⊆ U ∩ V . Then there is natural num ber n such that W n = G . Since d ( W × W ) < ε , we s how that d ( W 2 × W 2 ) < 2 ε , and so d ( W n × W n ) < nε . Assume that x, y , x ′ , y ′ ∈ W . Then we hav e d ( xy , x ′ y ′ ) ≤ d ( xy , e ) + d ( e, x ′ y ′ ) = d ( y , x − 1 ) + d ( x ′ − 1 , y ′ ) < 2 ε. Then d ( G × G ) = d ( W n × W n ) < n ε . Conv ersely , supp ose that G is b ounded with resp ect to metric d . Then there is M > 0 such that d ( G × G ) < M . Let U b e a neighborho o d of e . Cho ose ε > 0 such that d − 1 ([0 , ε )) ⊆ U × U . Tha k natural num b er n such that nε > M . Then we hav e G × G = d − 1 ([0 , M )) = d − 1 ([0 , nε )) ⊆ V n × V n . It follows that G = V n , a nd so that G is b ounded.  Theorem 2 - 3. Let G b e top ological group and let H b e a norma l subgr o up of G . If H a nd G/H are b o unded, then G is b ounded. Pr o of. Let U be a neighbor ho o d of e . P ut V = U ∩ H . Then there a re natural nu mbers m and n such that ( U / H ) n = G/H and V m = H . W e s how that U n + m = G . Let x ∈ G . Then if x ∈ H , we hav e x ∈ V m ⊂ U m ⊂ U n + m . Now let x / ∈ H . The n xH ∈ ( U /H ) n . Assume tha t x 1 , x 2 , ..., x n ∈ U such that xH = x 1 x 2 ...x n H. Consequently there is h ∈ H such that xh ∈ U n , and so x ∈ U n H ⊂ U n V m ⊂ U n U m = U n + m . W e conclude that U n + m = G , and s o G is b ounded.  Theorem 2 - 4. If G is a lo cally compac t O- dimensional to po logical gr oup, then G is un b ounded. Pr o of. Let U b e a neighborho o d of e such that U − is co mpa ct and U − 6 = G . Since G is a O-dimensional top ologica l group, U contains an open and clos e d neig h b o rho o d as V . Then V is a compact neigh b orho o d of e . By apply [1, Theor em 4 .10] to obtain a neighbor ho o d W of e such tha t W V ⊂ V . T ake W 0 = W ∩ V . Then W 0 2 ⊂ W V ⊂ V ⊂ U − . By finite induction, we have W 0 n ⊂ W 0 W 0 n − 1 ⊂ W V ⊂ V ⊂ U − , for every natural num ber n . It follows that W 0 n $ G for every natural num ber n , and so G is unbounded.  Theorem 2-5. Supp ose that G is a locally co mpact, Hausdorff, and to tally discon- nected top olo gical gro up. Then G is unbounded. Pr o of. By using [1, The o rem 3.5 ] and Theor em 2-4, pro o f is hold.  Theorem 2-6. Let G b e top olo g ical group. Then we hav e the following as sertions. 3 (1) If E ⊆ G is b ounded, then E − is b ounded subset o f G . (2) If G is b ounded, then G is connected and moreov er G has no pr op er op en subgroups. Pr o of. 1) Let U be a neigh bo rho o d of e and suppose that V is a neighborho o d of e such that V − ⊂ U . Since E is bounded subset of G , there is na tural num ber n such that E ⊂ V n . Then E − ⊂ ( V n ) − ⊂ ( V − ) n ⊂ U n . It fo llows that E − is a b ounded subset o f G . 2) Since G is b ounded, there is a natural num ber n such that G = V n where V is neighborho o d of e . By us ing [1, Co rollary 7 .9], pro of is hold.  Corollary 2 -7. Assume that G is a lo ca lly compact top olo g ical gr oup. Then e very bo unded and closed subset of G is compact, moreov er if E ⊆ G is b ounded, then E − is compact. Every bo unded top o logical gro up G , in gene r al, is no t c ompact, for example R / Z is b ounded, but is not compact. Theorem 2-8. Let G be top olog ical group a nd supp o se tha t E ⊆ G is the c ompo nen t of e . If E is compa ct, then E is b ounded. Pr o of. Since E is the c ompo nen t of e , by using [1, Theor em 7 .4], for every neig h b or- ho o d U of e , we hav e E ⊆ S ∞ k =1 U k . Since E is compact there is natural n um b er n such that E ⊆ U n . Then E is b ounded s ubset of G .  In gener al, every compa ct subset E of a top ologica l gr oup G is not bounded and in ab ove Theo rem, it is necessar y that E m ust b e a c o mpo nen t o f e . F or example Z n = { ¯ 0 , ¯ 1 , ¯ 2 , ..., ¯ n } for e very n ≥ 1, with dis c rete top ology is not b ounded, but it is compact. Corollary 2 -9. If G is a lo cally compact top olog ical gro up, then the comp onent of e is b ounded. Theorem 2 -10. Let G and G ′ be top olo gical group and supp ose that π : G → G ′ is gro up isomorphism. If π is contin uous and E ⊆ G is a b ounded subset o f G , then π ( E ) is bo unded subset of G ′ . Pr o of. Let V ′ be a neig h b orho o d of e ′ ∈ G ′ . Then π − 1 ( V ′ ) is a neighborho o d of e . Since E is a b ounded subset of G , there is a natura l num ber n such that E ⊆ ( π − 1 ) n ( V ′ ) ⊆ π − 1 ( V ′ n ) implies tha t π ( E ) ⊆ V ′ n . Thus π ( E ) is a b ounded subset of G ′ .  Definition 2-11. Let G and G ′ be top olog ical g roup. W e say that the mapping π : G → G ′ is co mpact, if for ev ery b ounded subset E ⊆ G , π ( E ) is r elatively com- pact. 4 Theorem 2-12. L et G a nd G ′ be top olog ical group and suppo se that π : G → G ′ is contin uo us and gro up isomo rphism. Then if G ′ is lo cally compact, then π is compa ct. Pr o of. Let E ⊆ G b e bounded. By using Theorem 2.1 0, π ( E ) is bounded subset of G ′ and by using The o rem 2.6 , π ( E ) − is co mpact, and so that π is compact.  References 1. E. Hewitt, K. A. R oss, Abstr act harmonic analysis , Springer, Berlin, V ol I 1963.