Partially ordered sets that has maximal element but no last element

Question

I'm trying to understand the idea behind partially ordered sets. I'm trying to think of a simple example of a partially ordered set that has a one maximum element but no last element. Is it possible?

Answer 1

If you have $A=\{1,2,4,8,10\}$ with ordinary divisibility $\mid$ then $8$ is maximal element in $A$ but not last element. Also $10$ is maximal element. If we have $B=\{1,2,4,8\}$ then 8 is maximal and last element. Last element is always the maximal element (if exist) but not vice versa. Last element is therefore unique and maximal are not necessary unique.

Answer 2

Trying to guess at what idea you have in mind... consider the set whose elements are the natural numbers together with an extra element called $*$. Extend the usual ordering on the natural numbers by:

• $* \leq *$ is true
• $* \leq n$ and $n \leq *$ are both false for every natural number $n$

This partial order has two features:

• $*$ is the only maximal element, but it is not a maximum
• The collection of natural numbers is an increasing chain without an upper bound