Find all maximal elements of B. Also find if they exist, the largest element of B, and the least upper bound of B, where
$R = \{(x, y) \in 2^\mathbb{N}\times2^\mathbb{N}\mid x\subseteq y\}, B = \{x \in 2^\mathbb{N} \mid \text{x has at most 5 elements}\}.$
This exercise is from Velleman's book "How to prove, it", in the book is only basic theory about ordering.
- maximal element: is the the element $x \in B$ such that $\lnot\exists x(bRx \wedge x\neq b)$.
B has several maximal elements, namely all $A \in 2^\mathbb{N}$ such that $|A| = 5$.
- largest element: is the element $x \in B$ such that $\forall x(xRb)$.
B has no largest element, because for any sets $A, B \in 2^\mathbb{N}, A \neq B$ such that $|A| = |B| = 5$ we have $A \not\subseteq B$ or $B \not\subseteq A$.
- greatest lower bound: is the largest element of the set of all lower bound.
There is only one lower bound, so it is greatest lower bound. It is the set $\mathbb{N}$, since for all every member $x \in B$ we have $x \subseteq \mathbb{N}$.
In the book is not answer to the exercise. I found on the internet solution, but I think that the solution is not correct, but I am not sure. Thank you for check my solution.
A set with less than five elements can be extended to one with five, because $\mathbb{N}$ is infinite. Thus no such set can be maximal in $B$. Conversely, each five element set is maximal, because it cannot be contained in a larger set in $B$.
Since $B$ has more than one maximal element, for instance $\{0,1,2,3,4\}$ and $\{1,2,3,4,5\}$, it cannot have a greatest element.
Since every natural number $x$ belongs to at least one set with at most five elements, namely the singleton $\{x\}\in B$, the least upper bound is $\mathbb{N}$.
The greatest lower bound is the empty set, because it is the unique lower bound.