# 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 ${\mathcal{T}}_{1}\subseteq {\mathcal{T}}_{2}$, we say that ${\mathcal{T}}_{2}$ is finer (http://planetmath.org/Finer) than ${\mathcal{T}}_{1}$, or that ${\mathcal{T}}_{2}$ refines ${\mathcal{T}}_{1}$.

###### 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 ${\mathcal{T}}_{i}$ 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.
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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 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 $\{\mathrm{\varnothing},A,X\}$, 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 $X$.

Title | lattice of topologies |
---|---|

Canonical name | LatticeOfTopologies |

Date of creation | 2013-03-22 16:54:42 |

Last modified on | 2013-03-22 16:54:42 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 8 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 54A10 |

Related topic | Coarser^{} |

Defines | common refinement |