The wikipedia says:
Maximal chain. A chain in a poset to which no element can be added without losing the property of being totally ordered. This is stronger than being a saturated chain, as it also excludes the existence of elements either less than all elements of the chain or greater than all its elements. A finite saturated chain is maximal if and only if it contains both a minimal and a maximal element of the poset.
Q:
But the minimal element may be incomparable since its definition only ensures "not greater than any other", then why does one finite saturated chain need to contain both a minimal and a maximal element to become maximal?
Edited:
After reading some comments, I have questions about
A finite saturated chain is maximal if and only if it contains both a minimal and a maximal element of the poset.
As the comment says, for one finite saturated chain $0 < 1$ based on the relation $<$ on the subset $\{x\in\mathbb{N}\vert -N\le x\le N\}$ of the set $\mathbb{N}$, after adding "both a minimal and a maximal element", i.e. $\{-N,N\}$, the chain $-N < 0 < 1 < N$ is not maximal. Then it seems that one half of the statement "A finite saturated chain is maximal if it contains both a minimal and a maximal element of the poset." is wrong.
Could someone give one proof of this or offer some hints?