# A paratopological group $T_1$ that is not $T_2$?

A group $G$ endowed with a topology is called a paratopological group if the multiplication $G\times G\to G$ is continuous.

It is known every $T_1$ topological group is completely regular, and I've seen mentioned in several places that in the class of paratopological groups $T_1$ does not even imply the Hausdorff separation axiom. But I don't know examples. Can anyone help me to find some example about this?

• Maybe I will make some comments on things that I have tried so that over people will not try them. $\mathbb{Z}$ with the "cofinite" topology does not work since the group operation is not continuous. $\mathbb{Z}$ with the topology generated by $[a, \infty)$ does not work since the space is not $T_1$. Mar 9 '17 at 3:57

take $G= (\mathbb{Z},+)$ as the group, and a basic neighbourhood of $0$ looks like $U_n(0) = \{0\} \cup \{k \in \mathbb{Z}: k \ge n\}$, $n \in \mathbb{N}$. At $p \in \mathbb{Z}$ we of course get the shifted neighbourhoods $U_n(p) = \{p\} \cup \{k+p: k \ge n\}$ base of neighbourhoods.
This is a $T_1$ topology, as can easily be checked, but any two non-empty open sets intersect, so $G$ is anti-Hausdorff even. And one checks that the addition map is continuous as $+[U_n(p) \times U_n(q)] \subseteq U_n(p+q)$, so $+$ is continuous at $(p,q) \in G \times G$.