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

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

Proof. Assume that $\mathbb{R}=(\mathbb{R},\tau, \circ)$ is a topological group. It is well known that this implies that there is $H\subseteq\mathbb{R}$ which is open, normal subgroup of $\mathbb{R}$ This subgroup ,,generates'' the topology (see the parent object for more details). Thus $H\neq\mathbb{R}$ because $\tau$ is not antidiscrete. Let $g\in\mathbb{R}$ such that $g\not\in H$ (and thus $gH\cap H=\emptyset$ . 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 $gH\cap H\neq\emptyset$ Indeed, every two open subsets in $\tau$ have nonempty intersection. Contradiction, because diffrent cosets are disjoint. $\square$