Revision as of 21:14, 13 October 2019 by Mrmxs (talk | contribs) (Link)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

A filter on a set $X$ is a structure of subsets of $X$.


Let $\mathcal{F}$ be a set of subsets of $X$. We say that $\mathcal{F}$ is a filter on $X$ if and only if each of the following conditions hold:

  • The empty set is not an element of $\mathcal{F}$.
  • If $A$ and $B$ are subsets of $X$, $A$ is a subset of $B$, and $A$ is an element of $\mathcal{F}$, then $B$ is an element of $\mathcal{F}$.
  • The intersection of two elements of $\mathcal{F}$ is an element of $\mathcal{F}$.

It follows from the definition that the intersection of any finite family of elements of $\mathcal{F}$ is also an element of $\mathcal{F}$. Also, if $A$ is an element of $\mathcal{F}$, then its complement is not.

More generally, one can define a filter on any Partially ordered set $(P,\leq)$: Let $F$ be a subset of $P$. We say $F$ is a filter if and only if

  • $F\neq\emptyset$.
  • For all $x,y\in F$, there exists $z\in F$ such that $z\leq x$ and $z\leq y$.
  • If $x\in F$ and $x\leq y\in P$, then $y\in F$.

A filter on a set $S$ is a filter on the poset $(\mathcal{P}(S),\subseteq)$.


Let $Y$ be a subset of $X$. Then the set of subsets of $X$ containing $Y$ constitute a filter on $X$.

If $X$ is an infinite set, then the subsets of $X$ with finite complements constitute a filter on $X$. This is called the cofinite filter, or Fréchet filter.

See also

This article is a stub. Help us out by expanding it.