example of an Alexandroff space which cannot be turned into a topological group

Let $ℝ$ denote the set of real numbers and $\tau =\{[a,\mathrm{\infty })|a\in ℝ\}\cup \{(b,\mathrm{\infty })|b\in ℝ\}$. One can easily verify that $(ℝ,\tau )$ is an Alexandroff space.

Proposition. The Alexandroff space $(ℝ,\tau )$ cannot be turned into a topological group.

Proof. Assume that $ℝ=(ℝ,\tau ,\circ )$ is a topological group. It is well known that this implies that there is $H\subseteq ℝ$ which is open, normal subgroup of $ℝ$. This subgroup ,,generates” the topology (see the parent object for more details). Thus $H\ne ℝ$ because $\tau$ is not antidiscrete. Let $g\in ℝ$ such that $g\notin H$ (and thus $gH\cap H=\mathrm{\varnothing }$). Then $gH$ is again open (because the mapping $f(x)=g\circ x$ is a homeomorphism). But since both $H$ and $gH$ are open, then $gH\cap H\ne \mathrm{\varnothing }$. Indeed, every two open subsets in $\tau$ have nonempty intersection. Contradiction, because diffrent cosets are disjoint. $\mathrm{\square }$

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