Understanding the dual of a partially ordered set

Question

I would like to check my understanding regarding the dual of a partially ordered set $P$.

The dual of $(P,\leq)$ is defined to be $(P^*,\geq)$ which satisfies the property $$x\leq_{P} y\Longleftrightarrow x\geq_{P^*} y$$

I see that $(P^*,\geq)$ is also a partially ordered set. Looking at a lattice it is clear that the least upper bound of $P$ is the greatest lower bound of $P^*$, but proving it has got me confused.

Let $u$ be the least upper bound of $P$, then $$p\leq_{P} u\Longleftrightarrow p\geq_{P^*} u$$ This shows that $u$ is a lower bound of $P^*$. If $\alpha$ is any lower bound of $P^*$, how can we get that $\alpha\geq_{P^*}u$? Is the answer because $$\alpha\geq_{P^*}u\Longleftrightarrow \alpha\leq_{P}u$$ and $u$ is the least upper bound of $P$?

This leads into a bigger question that if $(P,\leq)$ has the supremum property wouldn't we be able to see $(P^*,\geq)$ also having the supremum property? (I believe my answer above will prove this result).

Thank you for looking!

Answer 1

A simple calculation suffices,

   ⊔ is the lub in (P, ⊑)
≡⟨ definition ⟩
   ∀ x, y • ∀ z • x ⊑ z ∧ y ⊑ z ≡ x ⊔ y ⊑ z
≡⟨ Dual order ⟩
   ∀ x, y • ∀ z • z ⊒ x ∧ z ⊒ y ≡ z ⊒ x ⊔ y
≡⟨ definition ⟩
   ⊔ is the glb in (P, ⊒)   

;-)