Category of Ordered Sets has Enough Constants

Theorem
Let $\mathbf{Pos}$ be the category of posets.

Then $\mathbf{Pos}$ has enough constants.

Proof
By Singleton Poset is Terminal Object, we have that any poset with a singleton as underlying set is terminal in $\mathbf{Pos}$.

Let $1$ be such a singleton poset.

To show that $\mathbf{Pos}$ has enough constants, it is to be shown that if:


 * $f: P \to Q \ne g: P \to Q$

then there exists an $x: 1 \to P$ such that $f \circ x \ne g \circ x$.

Now by definition of $\mathbf{Pos}$, $f$ and $g$ are increasing mappings.

In particular, they are mappings, and by Equality of Mappings, there must be a $p \in P$ such that:


 * $f \left({p}\right) \ne g \left({p}\right)$

Define $\bar p: 1 \to P$ by $\bar p \left({*}\right) = p$ (where $*$ is the unique element of $1$).

If this is a morphism in $\mathbf{Pos}$, we are evidently done.

This follows from Constant Mapping both Increasing and Decreasing.

Hence $\mathbf{Pos}$ has enough constants.