characterization of Alexandroff groups
Topological group is called Alexandroff if is an Alexandroff space as a topological space. For example every finite topological group is Alexandroff. We wish to characterize them. First recall, that if is a subset of a topological space, then denotes an intersection of all open neighbourhoods of .
Lemma. Let be an Alexandroff space, be a continuous map and such that . Then , where .
Proof. Let . Of course is open (because is Alexandroff). Therefore is open in . Thus (from the definition of product topology and continuous map), there are open subsetes such that each is an open neighbourhood of and
Now let . Of course , so is nonempty and is open. Furthermore and thus
On the other hand and is open neighbourhood of . Thus , because is minimal open neighbourhood of . Therefore
which completes the proof.
Proof. Let be an intersection of all open neighbourhoods of the identity . Let be an open subset of . If , then is an open neighbourhood of . Thus and therefore . Thus
To complete the proof we need to show that is normal subgroup of . Consider the following mappings:
This shows that is a normal subgroup, which completes the proof.
Corollary. Let be a topological group such that is finite and simple. Then is either discrete or antidiscrete.
Proof. Of course finite topological groups are Alexandroff. Since is simple, then there are only two normal subgroups of , namely the trivial group and entire . Therfore (due to proposition) the topology on is ,,generated” by either the trivial group or entire . In the first case we gain the discrete topology and in the second the antidiscrete topology.
|Title||characterization of Alexandroff groups|
|Date of creation||2013-03-22 18:45:43|
|Last modified on||2013-03-22 18:45:43|
|Last modified by||joking (16130)|