# Thread: A partial order set with a non-unique maximal element

1. ## A partial order set with a non-unique maximal element

Find an example of a partial order set with a maximal element that is not unique.

My question here is, since a maximal element is define as:

m is a max of a set X if $m \leq x$, then x = m.

Not unique means that there are at least two, but how can I find two such elements? If we have two different max, say m and n, then either $m \leq n$ or $n \leq m$ since $m \neq n$, but then by definition of a maximal element, m is n...

Thank you!

2. let $x\le y$ if $x^2\le y^2$. And to this on the interval $[-a,a]$ for $a>0$.

Find an example of a partial order set with a maximal element that is not unique.
Remember this is a partial order not a total ordering.
Consider $\{1,2,3,4,5,6,7,8,9\}$ ordered by divides, $
x \prec y \Leftrightarrow x|y$
.

4. Or just make up something like {a, b, c, d} with a< b< d, a< c< d with b and c "non-comparable". ("Non-comparable meaning none of b< c, c< b, or b= c are true. d is maximal (in fact, the maximum) because every other member is less than it. That would not be possible in a linearly ordered set but is in a partially ordered set.)

If you want something a little more concrete use sets. Supose A= {a}, B= {a, b}, C= {a, c}, D= {a, b, c} with " $\le$" defined by "inclusion": [tex]X\le Y[tex] if and only if X is a subset of Y. That is partially ordered because neither B nor C is a subset of the other. $B\le C$ is false because B is not a subset of C. $C\le B$ is not true because C is not a subset of B. Clearly B= C is not true. So this is a partially ordered set in which D is the maximum (and so "maximal").

Yet another example, where D is maximal but not "the" maximum, is {A, B, C, D} where A= {a}, B= {a,b}, C= {c}, D= {c, d}. Now both B and D are maximal.