Let $X$ be a set. Let $L$ be the set of all topologies on $X$ . We may order $L$ by inclusion. When $\mathcal{T}_1\subseteq \mathcal{T}_2$ , we say that $\mathcal{T}_2$ is finer than $\mathcal{T}_1$ , or that $\mathcal{T}_2$ refines $\mathcal{T}_1$ .

Proof. Clearly $L$ is a partially ordered set when ordered by $\subseteq$ . Furthermore, given any family of topologies $\mathcal{T}_i$ on $X$ , their intersection $\bigcap \mathcal{T}_i$ also defines a topology on $X$ . Finally, let $\mathcal{B}_i$ 's be the corresponding subbases for the $\mathcal{T}_i$ 's and let $\mathcal{B}=\bigcup \mathcal{B}_i$ . Then $\mathcal{T}$ generated by $\mathcal{B}$ is easily seen to be the supremum of the $\mathcal{T}_i$ 's.

Let $L$ be the lattice of topologies on $X$ . Given $\mathcal{T}_i\in L$ , $\mathcal{T}:=\bigvee \mathcal{T}_i$ is called the common refinement of $\mathcal{T}_i$ . By the proof above, this is the coarsest topology that is finer than each $\mathcal{T}_i$ .

If $X$ is non-empty with more than one element, $L$ is also an atomic lattice. Each atom is a topology generated by one non-trivial subset of $X$ (non-trivial being non-empty and not $X$ ). The atom has the form $\lbrace \varnothing, A, X\rbrace$ , where $\varnothing \subset A\subset X$ .

Remark. In general, a lattice of topologies on a set $X$ is a sublattice of the lattice of topologies $L$ (mentioned above) on $X$ .