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Homelattice of topologies

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# 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 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. ∎

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

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

## Mathematics Subject Classification

54A10*no label found*

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