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'lattice of topologies'
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| Title of object: |
lattice of topologies |
| Canonical Name: |
LatticeOfTopologies |
| Type: |
Definition |
| Created on: |
2007-04-09 16:09:14 |
| Modified on: |
2007-04-09 17:25:25 |
| Classification: |
msc:54A10 |
| Defines: |
common refinement |
Preamble:
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Content:
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 $\mathcal{T}_1$.
\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 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 $X$). |
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