Show that there is exactly one maximal element in a poset with a greatest element? This is true, any idea how to say it in proof form? 
I would guess: 
In a poset with one maximal element, then that element has no other elements above it and has elements below it. If its the only one with no other elements above it, then it has to be the greatest element.
 A: Suppose that $R$ is a partial order on $A$ with $B\subseteq A$. Suppose that $g$ is the greatest element of $B$.
Suppose $x\in B$ and $(g,x)\in R$. Since $g$ is the greatest element, $(x,g)\in R$. But since $R$ is antisymmetric, $g=x$ which proves $g$ is a maximal element of B.
Suppose $m$ is also a maximal element of $B$. Then either $(m,g)\not\in R$ or $m=g$. But since $g$ is the greatest element, $(m,g)\in R$. Thus, $m=g$ which proves that $g$ is the only maximal element of $B$.
A: First, for the poset $P$, if $g \in P$ is the greatest element it implies that $g$ is related to every other element in $P$ (this is really the key, since in general a maximal element need not be related to every other element in $P$) since then for any other element $s \in P$ we must have $s \leq g$. So $s$ cannot be maximal. Then $g$ must be the only maximal element in $P$.
A: Is a matter of definition actually.
An element $g$ is the greatest element if $g\geq s$ for all $s$ in the poset while it is a maximal element if there is not an element $s$ in the poset such that $g\lt s$.
Do you see why the first imply the second?
