54A10-LatticeOfTopologies.tex

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\pmtitle{lattice of topologies}
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\begin{document}
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 $X$.
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