The way I understand minimal and maximal is the following.
Consider the following collection of sets:
{{1,2},{2,4},{1,6,7},{1,2,4,5,6},{8}}
The minimal members of this collection of sets is: {1,2},{2,4},{1,6,7},{8}
Meaning there are no other sets in the collection that are proper subsets of the set.
Minimal Set Example 1: in the set {1,2} this case holds true because the only proper subsets possible are {1} and {2}. Those sets {{1},{2}) are nowhere to be found in the collection. Therefore we include {1,2} as a part of the minimal collection of sets.
Minimal Set Example 2: in the set {1,6,7} this case holds true because the only proper subsets possible are {1},{6},{7},{1,6},{1,7}, and {6,7}. Those proper subsets ({1},{6},{7},{1,6},{1,7}, and {6,7}) are nowhere to be found in the collection. Therefore we include {1,6,7} as a part of the minimal collection of sets.
The maximal members of this collection of sets is: {1,6,7},{1,2,4,5,6},{8}
Meaning none of these sets are proper subsets of other sets in the collection. NOTICE: The maximal collection is not mutually exclusive of the minimum collection. See {1,6,7}.
Maximal Set Example 1: in the set {8} this case holds true because this set is not found to be a subset of any of the other sets in the collection. Meaning 8 is not found in any of the other sets in the collection. Therefore we include {8} as a part of the maximal collection of sets.
Maximal Set Example 2: in the set {1,6,7} this case holds true because this set is not found to be a subset of any of the other sets in the collection. Meaning all of the numbers of the set {1,6,7} are not found together as a part of another set in the collection. Therefore we include {1,6,7} as a part of the maximal collection of sets.
First: let's review definitions, keeping in mind that "$\leq$" simply represents "any" partial order on a given set $P$. (In your case, the order relation of "$\mid$" replaces the relation $\leq$, and "is a multiple of" replaces $\geq$.) Can you figure out what these definitions tell you in terms of your given relation?
Given a set $P$ and an order relation $\leq$:
Greatest element and least element:
An element $x$ in a partially ordered set $P$ is a greatest element if for every element $a \in P, a \leq x.$ An element $y\in P$ is a least element if for every element $a \in P, a \geq y$.
- That is, in your case, $x\in P$ is a greatest element if for every $a$ in $P$, $a\mid x$. And $y\in P$ is a least element if for every $a\in P,\;y\mid a$.
- So in your example above, there is neither a greatest nor a least element in the set, given the relation "|".
A poset can only have one greatest or least element.
Maximal elements and minimal elements:
An element $x \in P$ is a maximal element if there is no element $a \in P$ such that $a > x$. Similarly, an element $y \in P$ is a minimal element if there is no element $a \in P$ such that $a < y$. If a poset has a greatest element, it must be the unique maximal element, but otherwise there can be more than one maximal element, and similarly for least elements and minimal elements.
- This addresses your first question: "If you have more than one extremal element, is it possible to have a greatest or least element?" No. If an element $x$ is the greatest element of the set $P$, meaning that every element in $P$, including $x$, divides $x$, it is the unique maximal element. Likewise for a least element: in your case, that would mean if there were any element, like, e.g., $1$ in the set that it divides every element, including itself, of the set $P$, then it would be the unique minimal element. So if you have more than one maximal (minimal) element in the set, there cannot be a greatest (least) element in the set.
- In terms of comparability. You are on target with your take on whether or not elements are comparable under a particular ordering relation comparability, and how that impacts the determination of extrema.
Upper and lower bounds:
For a subset $A \subset P$, an element $x \in P$ is an upper bound of $A$ if $a \leq x$, for each element $a \in A$. In particular, $x$ need not be in $A$ to be an upper bound of $A$. Similarly, an element $x \in P$ is a lower bound of $A$ if $a \geq x$, for each element $a \in A$. A greatest element of $P$ is an upper bound of $P$ itself, and a least element is a lower bound of $P$.
- Given your edit, and the corresponding answers, it seems you have a good grasp of this.
Comparability
Any two elements $x$ and $y$ of a set $P$ that is partially ordered by a binary relation $\leq$ are comparable when either $x \leq y$ or $y \leq x$. If it is not the case that $x$ and $y$ are comparable, then they are called incomparable. A totally ordered set is exactly a partially ordered set in which every pair of elements is comparable.
- In your partially ordered set call it $P$, under "|", for any $x, y\in P$, $x$ and $y$ are comparable if and only if either $x|y$ or $y|x$, as you seem to understand.
Best Answer
Good up to the last line, but lacking justifications.
You should observe that $\emptyset\in B$, so it is obviously the smallest element and therefore also the greatest lower bound.
The five-element sets are indeed maximal, but you should add a reason why they are. If $|X|=5$ and $Y\in B$, with $X\subseteq Y$, then $Y=X$ because…
There is no largest element, because there are at least two distinct maximal elements (show them).
The least upper bound exists. Of course it's not in $B$, but that's not a requirement.
Every subset of $\mathscr{P}(\mathbb{N})$ has a least upper bound, by the way.