# Partially ordered set with no maximal element

• Mar 8th 2010, 12:20 PM
Mimi89
Partially ordered set with no maximal element

Let $\displaystyle P$ be a non-empty partially ordered set and assume that no element of $\displaystyle P$ is maximal. Use the Axiom of Choice to show that there exists a function $\displaystyle f: \omega \rightarrow P$ such that $\displaystyle f(n) < f(n^+)$ $\displaystyle \forall n \in \omega$
• Mar 8th 2010, 12:34 PM
arsenicbear
Here's my idea. We're going to construct a relation that you can apply Axiom of Choice to for obtaining the function you desire.

Pick some arbitrary $\displaystyle y_{0} \in P$

Construct a set of ordered pairs R s.t. it has two properties.

(1) $\displaystyle <0,y_{0}> \in R.$

(2) if for some $\displaystyle x \in \mathbb{N}$ and $\displaystyle y \in P \ \exists <x,y> \in P$ $\displaystyle \rightarrow$ <x^{+}, y'> s.t. y P_{r} y'

You know that there will always exist greater y's because P has no maximal unit.

Then from axiom of choice you can create a function from this relation and I believe it should have all the properties you want.
• Mar 8th 2010, 11:52 PM
Mimi89
Thank you a lot for your help; could you please just edit the equations, as they're given as "Latex Error: Syntax Error"'s now?

Thank a lot,

Mimi
• Mar 9th 2010, 12:37 AM
arsenicbear
I've edited my eqns. Also P_{r} represents the partial order relationship that is assumed in P.