# Question about proof of "every finite poset has a maximal element".

I have seen various proofs (see, for example, here: https://math.stackexchange.com/a/436074/264885) where we're saying if $a$ is not a maximal element, then there necessarily is some $b$ such that $a < b$. My question is, why is this true? How can we assume $a$ and $b$ are comparable because by my understanding of the poset axioms any two elements need not be comparable.

Also, I'm using the following definition of a maximal element: "Let $(P,\leq)$ be a partially ordered set and $S\subset P$. Then $m\in S$ is a maximal element of $S$ if for all $s\in S, m \leq s$ implies $m = s$."

• If $m$ is not maximal, you can simply negate the definition of maximal to get $(\exists s \in S)\ m \leq s \wedge m \neq s$.
– user133249
Nov 12 '15 at 11:38
• An element of a poset is by definition a maximal element if it is not smaller than some other element. So if it is not maximal then it will be smaller that some other element. In order to be smaller it also must be comparable with that element. Nov 12 '15 at 11:39

This is just the negation of the definition of maximal. Not being maximal, means that it is not true that for all $s\in S$, if $m\leq s$ then $m=s$. Which is to say that there exists $s\in S$ such that $m\leq s$ and $m\neq s$.
In other words, if $a$ is not maximal, then there is some $b$ such that $a<b$.