Let $latex U$ is a set. A filter (on $latex U$) $latex \mathcal{F}$ is by definition a non-empty set of subsets of $latex U$ such that $latex A,B\in\mathcal{F} \Leftrightarrow A\cap B\in\mathcal{F}$. Note that unlike some other authors I do not require $latex \varnothing\notin\mathcal{F}$.

For greater clarity I will use *filter objects* instead of filters. Below I will describe the properties of filter objects without exact definition and the proofs. You can look here for the formalistic behind.

I will denote the set of all filters objects on a set $latex U$ as $latex \mathfrak{F}$. Filter objects are bijectively related with filters by the bijection “$latex \mathrm{up}$” from the set of filter objects to the set of filters. A filter object corresponding to principal filter generated by a set $latex A$ is equal to $latex A$. (Thus the set of subsets of $latex U$ is a subset of $latex \mathfrak{F}$.)

Formal definition of filter objects in the framework of ZF is given here. We will not need the exact definition of filter objects, but only the facts that “$latex \mathrm{up}$” is a bijection from filter objects to filters and that a filter object corresponding to principal filter generated by a set $latex A$ is equal to $latex A$.

I will define the order on the set of filter objects by the formula $latex \mathcal{A}\subseteq\mathcal{B} \Leftrightarrow \mathrm{up} \mathcal{A} \supseteq \mathrm{up} \mathcal{B}$ for every filter objects $latex \mathcal{A}$ and $latex \mathcal{B}$. This order well-agrees with the order of sets on $latex U$.

$latex \mathfrak{F}$ with the above defined order is a complete lattice. (See this draft article for a proof.)

## 10 thoughts on “Filter objects”