lattice of topologiesLatticeOfTopologies2007-04-09 16:09:142007-04-10 16:48:03Definitioncommon refinement\usepackage{amssymb,amscd}
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\newcommand{\real}{\mathbb{R}}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 \PMlinkname{finer}{Finer} than $\mathcal{T}_1$, or that $\mathcal{T}_2$ refines $\mathcal{T}_1$.
\begin{thm} $L$, ordered by inclusion, is a complete lattice. \end{thm}
\begin{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.
\end{proof}
Let $L$ be the lattice of topologies on $X$. Given $\mathcal{T}_i\in L$, $\mathcal{T}:=\bigvee \mathcal{T}_i$ is called the \emph{common refinement} of $\mathcal{T}_i$. By the proof above, this is the coarsest topology that is \PMlinkescapetext{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 $\varnothing \subset A\subset X$.
\textbf{Remark}. In general, a \emph{lattice of topologies} on a set $X$ is a sublattice of \emph{the} lattice of topologies $L$ (mentioned above) on $X$.