example of an Alexandroff space which cannot be turned into a topological group
Let ℝ denote the set of real numbers and τ={[a,∞)|a∈ℝ}∪{(b,∞)|b∈ℝ}. One can easily verify that (ℝ,τ) is an Alexandroff space.
Proposition. The Alexandroff space (ℝ,τ) cannot be turned into a topological group
.
Proof. Assume that ℝ=(ℝ,τ,∘) is a topological group. It is well known that this implies that there is H⊆ℝ which is open, normal subgroup of ℝ. This subgroup
,,generates” the topology
(see the parent object for more details). Thus H≠ℝ because τ is not antidiscrete. Let g∈ℝ such that g∉H (and thus gH∩H=∅). Then gH is again open (because the mapping f(x)=g∘x is a homeomorphism). But since both H and gH are open, then gH∩H≠∅. Indeed, every two open subsets in τ have nonempty intersection
. Contradiction
, because diffrent cosets are disjoint. □
Title | example of an Alexandroff space which cannot be turned into a topological group |
---|---|
Canonical name | ExampleOfAnAlexandroffSpaceWhichCannotBeTurnedIntoATopologicalGroup |
Date of creation | 2013-03-22 18:45:46 |
Last modified on | 2013-03-22 18:45:46 |
Owner | joking (16130) |
Last modified by | joking (16130) |
Numerical id | 4 |
Author | joking (16130) |
Entry type | Example |
Classification | msc 22A05 |