I'm confused about the definition of poset.

Question

The definition of poset :
$\qquad$A set with a partial ordering.

Partial ordering is a binary relation $\preceq$ over a set ($P$). If I understood the definition of relation correctly, then ${\preceq}\subseteq P^2$ holds which also means that partial ordering $\preceq$ is a set.
Then, poset is a set with a set. What is this mean?

Is it simply a partial ordering (i.e. $(P,\preceq)={\preceq}$)?

Answer 1

Many concepts in mathematics can be captured by the idea of a "set with structure" - there is some underlying set $S$ of objects we're interested in, and then we endow or equip $S$ with a "structure" (a vague word including, e.g., operations, topologies, orderings, etc.) by doing this:

• We form the ordered pair $(S,\,\mathsf{structure})$.
• By an abuse of language, this ordered pair is also referred to as $S$.

The purpose of doing this is that we want to implement our mathematical ideas using sets in a way such that $S$ with one structure (e.g., an ordering $\preceq$) is, in the literal sense, not equal to the same set $S$ with a different structure (e.g., a different ordering $\Subset$). By the definition of what it means for two ordered pairs to be equal, this artifice with ordered pairs serves the purpose.

Let me try to illustrate this idea with a brief digression.

For example, we often want to think of $\mathbb{Z}$ (the integers) as more than just a set; we want to think of $\mathbb{Z}$ as a ring (Wikipedia), under its binary operations ${+}:\mathbb{Z}^2\to\mathbb{Z}$ and ${\times}:\mathbb{Z}^2\to\mathbb{Z}$.

Now, let's say someone else wants to think about the set $\mathbb{Z}$ as a ring, under the ordinary addition operation, but where they use a strange multiplication $\otimes$ defined by $a\otimes b=0$ for every $a$ and $b$. How are we going to distinguish our ring $\mathbb{Z}$ from this other person's ring $\mathbb{Z}$? We can't just say $\mathbb{Z}\neq\mathbb{Z}$, because of course $\mathbb{Z}=\mathbb{Z}$.

We equip the set $\mathbb{Z}$ with the different operations, so that our ring $\mathbb{Z}$ is really the ordered triple $(\mathbb{Z},{+},{\times})$, and their ring $\mathbb{Z}$ is really the ordered triple $(\mathbb{Z},{+},{\otimes})$, and voila, they're no longer equal.

So, in the case of an ordered set, since the same underlying set $P$ might have multiple orderings on it that we're interested in, we define (again, really as an artifice) a partially ordered set to be an ordered pair $(P,\preceq)$, where ${\preceq}\subseteq P^2$ is a partial ordering on the set $P$.

Answer 2

No. We have a set $P$ and a partial order on $P$, $\preceq$. A poset is the ordered pair, $(P,\preceq)$.

Note that an ordered pair can also be thought of as a set. But that's not the point. The point is that a poset is a fixed set with a fixed partial order.

Answer 3

The relation that is the partial ordering is a subset of ordered pairs of $P^2$.

The poset is the elements of the set $P$ together with this relation, the ordered pair $(P, \leq)$.