# Can maximal number in poset be more than one?

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?

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.
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.