Have ordering relation the same domain as codomain?

Question

The question is fairly straight foreward:

$R$ being an ordering relation means that $R \subseteq (a \times b)$ for some sets $a$ and $b$. Is it necessary that $a=b$?

My expected answer is no.

However, how would one then refer to an ordered set $(a, R)$ if $\texttt{Dom}(R) \neq \texttt{CoDom}(R)$. Is there a terminology that I am not aware of?

Answer 1

While you can certainly make sense of order relation with a different domain and co-domain, you are correct that in the prevalent definition of order relations $R$ we always have, by definition, that $\operatorname{Dom}(R) = \operatorname{Co-Dom}(R)$.

Moreover, if we define order relations $R \subseteq a \times b$ with possibly $a \neq b$, as transitive and antisymmetric relations, we can define a new order relation $R^* \subseteq (a \cup b) \times (a \cup b)$ s.t.

• $R^* \restriction a \times b = R$ and
• for $x,y \in (a \cup b) \times (a \cup b) \setminus a \times b \colon$ $$ R^{*}(x,y) \iff x = y $$ that naturally extends $R$ and again is transitive and antisymmetric. Therefore I argue that there is no real need for the definition of 'generalized order relations' on $a \times b$.

Also note that for a relation $R \subseteq a \times b$ to be reflexive, we necessarily need that $a = b$ since for $x \in a \Delta b \colon (x,x) \not \in a \times b$ and thus trivially $\neg R(x,x)$.