example of an Alexandroff space which cannot be turned into a topological group
Let denote the set of real numbers and . One can easily verify that is an Alexandroff space.
Proof. Assume that is a topological group. It is well known that this implies that there is which is open, normal subgroup of . This subgroup ,,generates” the topology (see the parent object for more details). Thus because is not antidiscrete. Let such that (and thus ). Then is again open (because the mapping is a homeomorphism). But since both and are open, then . Indeed, every two open subsets in have nonempty intersection. Contradiction, because diffrent cosets are disjoint.
|Title||example of an Alexandroff space which cannot be turned into a topological group|
|Date of creation||2013-03-22 18:45:46|
|Last modified on||2013-03-22 18:45:46|
|Last modified by||joking (16130)|