How can we express that a partial order is more complete than another one?

Question

Suppose we have two partial orders $R$ and $I$ on $\mathbb{C}$ (conplex numbers) such that:

• $R$ is a total order on $\mathbb{R}$ (real numbers).
• $I$ is also a total order on $\mathbb{R}$ and, additionally, on $\mathbb{I}$ (imaginary numbers).

Is there a technical term fore expressing that $I$ is more complete than $R$ in the sense that it has more pairs in its domain?

Would there be a different terminology for the following cases?

• The case where the hierarchy defined by $I$ for $\mathbb{R}$ is the same as $R$'s.
• The case where the hierarchy defined by $I$ for $\mathbb{R}$ is different from $R$'s.

Answer 1

From Partially ordered set - Wikipedia:

A partial order $\leq^*$ on a set $X$ is an extension of another partial order $\leq$ on $X$ provided that for all elements $x$ and $y$ of $X,$ whenever $x \leq y,$ it is also the case that $x \leq^* y.$ A linear extension is an extension that is also a linear (i.e., total) order.

Apart from Wikipedia, it seems to be surprisingly hard to find a reference for this general use of the term extension, except in the special case of a linear extension. Even Wikipedia's definition occurs under the latter heading. Order theory aside, however, the term extension is used fairly widely in mathematics, with a meaning that is at least vaguely similar to the one given in Wikipedia. I can't imagine anyone objecting to it. We'll see!

Fortunately, in the case you're interested in (at least, in the first of the two cases you listed), $I$ is a linear extension of $R.$ So there is a well-established technical term that says exactly what you want.

It's not clear to me what would be meant by $I$ having (in your words) more pairs than $R$ when it is not a superset of $R,$ as seems to be implied by this sentence in the question:

The case where the hierarchy defined by $I$ for $\mathbb{R}$ is different [from] $R$'s.