What is a partial order? Are $<$ and $\le$ partial orders?

Question

According to Jech's set theory, a partial ordering (of $P$) is:

A binary relation $<\subseteq P\times P$ where we have:

1. $\forall p\in P[(p,p)\notin <]$ (which we may say as: $\forall p\in P[p\not<p$])
2. $\forall p,q,r\in P[p<q\wedge q<r\implies p<r]$

It then goes on to say if $<$ is a partial order then we can define a relation $\le$ where $p\le q\iff[p=q\vee p<q]$ and that is a partial order.

However this violates point 1 of the definition of partial orders.

It notes that

"$<$ is sometimes called a strict order"

I have a theorem that shows there is a 1:1 correspondance between strict and "non-strict" orderings, so perhaps this is more of a "we may as well call it a partial, for it is induced by and induces exactly one partial order"

Answer 1

Usually, the term partial order, when unqualified by any other adjectives, refers to a reflexive, antisymmetric and transitive relation (i.e. the $\le$ you mention). A strict (partial) order is an irreflexive, transitive relation (i.e. the $<$ you mention).

As you mention, given a set $X$, there is a correspondence between partial orders and strict partial orders on $X$;

• Given a partial order $\le$, there is a strict partial order $<$ defined by $$x < y \quad \Leftrightarrow \quad x \le y \text{ and } x \ne y$$
• Given a strict partial order $<$, there is a partial order $\le$ defined by $$x \le y \quad \Leftrightarrow \quad x < y \text{ or } x = y$$

So really, referring to strict partial orders simply as "partial orders" is a (forgivable) abuse of notation.