In poset maximal number is defined as: An element 'a' belongs to 'A 'is called a maximal number if there is no element 'c' in 'A' such that a is less than c.
but it again says that there can be more than one maximal number. How can it be possible?
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2 Answers
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Yes, it is possible for a poset to have more than one maximal element.
For example, let $R$ be the divides relation on the set $A = \{1,2,3,5\}$. Then $2$ is a maximal element of the poset $(A, R)$ because none of the other elements is a multiple of $2$. If you draw the Hasse diagram of this poset, you'll find that there are no elements "above" $2$ in this diagram, and so $2$ is a maximal element. Similarly,$3$ and $5$ are also maximal elements. These three elements are pairwise incomparable (similar to how apples and oranges are incomparable).
The terminology partially-ordered set is justified because the elements can't be totally ordered by being placed on a straight line, whereas the set of all integers or the set of all real numbers are totally ordered under the $\le$ relation.
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Consider for example the poset of all subsets of $\{1,2,3\}$ of size at most $2$ ordered according to inclusion. The maximal elements in this poset are $\{1,2\},\{1,3\},\{2,3\}$.
For an even more striking example, every antichain (set in which no two elements are comparable) is a poset in which all elements are maximal and minimal. For example, consider the poset of all subsets of $\{1,2,3\}$ of size exactly $2$ ordered according to inclusion. There are three elements $\{1,2\},\{1,3\},\{2,3\}$, and all are maximal and minimal.
It is always possible to add a maximal element to a poset so that it contains a unique maximal element. See if you can figure out how.
answered Jan 21, 2015 at 15:09