In order theory ideals are generally defined for partially ordered sets. Preorders (aka quasiorders) need not be anti-symmetric. They're reflexive and transitive.

Can I extend the notion of an ideal to preordered sets? That is, can I define a subset S of a preordered set P such that S satisfies the axioms of an ideal on a poset (i.e. a non-empty, directed, and lower set) and call S an ideal?


1 Answer 1


You can, but you won't really get anything new.

For each preorder $Q$, there is a univeral poset $\tilde Q$ and a map of preorders $Q\to\tilde Q$. Concretely, the elements of $\tilde Q$ are equivalence classes of elements in $Q$, where two elements $p,q$ are equivalent if and only if $p\leq q\leq p$.

Given an ideal $J\subset \tilde Q$, you can pull it back along $Q\to\tilde Q$ (i.e. you take the preimage) to get an ideal $J$ in $Q$. This gives an isomorphism of posets $$\{\text{ideals of }\tilde Q\}\xrightarrow{\cong} \{\text{ideals of }Q\}.$$ So from an ideal-theoretic point of view there is really nothing about $Q$ which is not already known by the poset $\tilde Q$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .