lattice of topologies

Let $X$ be a set. Let $L$ be the set of all topologies on $X$. We may order $L$ by inclusion. When ${𝒯}_1\subseteq {𝒯}_2$, we say that ${𝒯}_2$ is finer (http://planetmath.org/Finer) than ${𝒯}_1$, or that ${𝒯}_2$ refines ${𝒯}_1$.

Theorem 1.

$L$, ordered by inclusion, is a complete lattice.

Proof.

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

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

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 $\{\mathrm{\varnothing },A,X\}$, where $\mathrm{\varnothing }\subset A\subset X$.

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