If $\quad(\sigma_{\lambda})_{\lambda\in \Lambda}\quad $ is a family of topologies defined on a group $E$, then its intersection $\quad t=\bigcap_{\lambda\in\Lambda}\sigma_{\lambda}$ forms a topology on $E$ that is coarser than each one of the topologies $\sigma_{\lambda}$. As for the union of these topologies, it was mentioned in Professor Hazi's book (titled Key Principles in Topological Concepts) that we do not have a similar result in general unless these topologies are comparable with each other. My question is about whether there are other conditions that make this union a topology on $E$ without them being comparable.
Note: We say two topologies are comparable if one of them is finer than the other, which is defined as: $t_1$ is finer than $t_2$ if one of the following equivalent claims is true:
- Each open set in $t_2$ is open in $t_1$;
- Each closed set in $t_2$ is closed with respect to $t_1$;
Where $t_1$ and $t_2$ are topologies defined on the same set $E$.
