User - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-24T16:34:05Zhttp://mathoverflow.net/feeds/user/7188http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/29950/countable-open-subgroupCountable open subgroupunknown (google)2010-06-29T18:36:47Z2013-04-12T02:45:00Z
<p>In a Hausdorff topological group, how can I show that every infinite topological group has a countable open subgroup?</p>
http://mathoverflow.net/questions/30092/discrete-cyclic-subgroupDiscrete cyclic subgroup.unknown (google)2010-06-30T18:46:55Z2010-08-02T05:59:07Z
<p>Let T is a hausdorff group topology and (G,T) is locally compact abelian group.If (G,T) has no open compact subgroups then can we say G has an infinite discrete cyclic subgroup?</p>
http://mathoverflow.net/questions/31511/totally-bounded-topology-and-indiscrete-topologyTotally bounded topology and indiscrete topologyunknown (google)2010-07-12T07:22:40Z2010-07-12T07:22:40Z
<p>Is this true?
A refinement of a totally bounded topology and an indiscrete topolgy is an indiscrete topology.</p>
http://mathoverflow.net/questions/30094/finitely-generated-groupFinitely generated groupunknown (google)2010-06-30T18:56:18Z2010-06-30T19:40:41Z
<p>If G is finitely generated how can we say G/nG is finite for every natural number n?</p>
http://mathoverflow.net/questions/30093/totally-bounded-subgroupTotally bounded subgroup.unknown (google)2010-06-30T18:51:29Z2010-06-30T18:51:29Z
<p>Is it true to say every totally bounded subgroup of a complete group is finite?</p>
http://mathoverflow.net/questions/30092/discrete-cyclic-subgroupComment by 2010-07-01T06:23:15Z2010-07-01T06:23:15Zhere is the article <a href="http://www.4shared.com/document/ih8d9nMA/29-Complemented_Topologies_on_.html" rel="nofollow">4shared.com/document/ih8d9nMA/…</a> and please see theorem 3.2http://mathoverflow.net/questions/29950/countable-open-subgroupComment by 2010-06-30T18:34:58Z2010-06-30T18:34:58ZThanks to all for your attentions.
here is the theorem.<a href="http://666kb.com/i/bkj3854twsrqna1hf.bmp" rel="nofollow">666kb.com/i/bkj3854twsrqna1hf.bmp</a> and also requirment definition:<a href="http://666kb.com/i/bkj3c6i3ushyjzchv.bmp" rel="nofollow">666kb.com/i/bkj3c6i3ushyjzchv.bmp</a> .And if you need Article here it is:<a href="http://www.4shared.com/document/ih8d9nMA/29-Complemented_Topologies_on_.html" rel="nofollow">4shared.com/document/ih8d9nMA/…</a>http://mathoverflow.net/questions/29950/countable-open-subgroupComment by 2010-06-29T20:52:27Z2010-06-29T20:52:27ZSome requirement Definition. A sequence (an) of elements of a group G is a T-sequence if there exists
a topology on G making this sequence vanish.
.A group G furnished with a maximal topology in which
a given T-sequence (an) vanishes we say topology on a group
G is determined by T-sequence (an).http://mathoverflow.net/questions/29950/countable-open-subgroupComment by 2010-06-29T20:42:05Z2010-06-29T20:42:05ZIn Article "Complemented topologies on abelian groups" All groups are abelian and hausdorff.
In this article 3.6.Theorem followed as below:
"Let Let T1 be a topology on an infinite group G which is determined by some T-sequence.
There exists a complement T2 of T1 on G which is determined by a T-sequence.
Proof. Note that (G, T1) has a countable open subgroup. Now, apply Theorems 1.6 and 3.5 and
Lemma 2.3."
My question is:
Why (G, T1) has a countable open subgroup?