This is an exercise from Velleman's "How To Prove It":
- In each case find all minimal and maximal elements of $B$. Also find, if they exist, the largest and smallest elements of $B$, and the least upper bound and greatest lower bound of $B$.
c. $R = \{ (x,y) \in \mathscr{P}(\mathbb{N}) \times \mathscr{P}(\mathbb{N})| x \subseteq y \}$ and $B = \{ x \in \mathscr{P}(\mathbb{N}) | \text{ $x$ has at most 5 elements.}\}$
My solution:
Minimal element: $\emptyset$, since $\emptyset$ will be a subset of every set of natural numbers.
Maximal element: The set of all maximal elements can be given by $\{X \in \mathscr{P}(\mathbb{N}) | \text{ $X$ has exactly 5 elements}\}$
Smallest element: $\emptyset$. This is also the greatest lower bound of $B$.
Largest element: None, since there is no $X \in B$ such that $\forall Y \in B (Y \subseteq X)$. There is also no least upper bound.
Is this solution correct?
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.