Let $A=\{1,2,3,4\}$. Let $K$ be the set of all transitive relations over $A$ except for the empty relation which is not an element of $K$. As we know $\subseteq$ relation is a partially ordered relation therefore $\subseteq$ is a relation over $K$, which means that $K$ is a partially ordered set. Prove that $K$ has a greatest element and find it. Find minimal element of $K$ and prove that it's minimal.
In order to avoid confusion the definitions of greatest and minimal elements are according to Wikipedia.
Let $G$ be: $$ \bigg\{(1,1), (2,2), (3,3), (4,4),(1,2),(2,1), (1,3), (3,1),\\ (1,4), (4,1), (2,3), (3,2), (2,4), (4,2), (3,4), (4,3)\bigg\} $$ There're $16$ unique ordered pairs in $K$ and $G$ contains all of them. Also any subset of $K$ is a combination of the $16$ unique pairs which are all contained in $G$. Therefore $G$ is the greatest element of $K$.
I think there're $16$ minimal elements in $K$ as each unique ordered pair doesn't contain any other subsets. We're given that the empty set is not part of $K$ and the $16$ pairs are unique, therefore cannot contain each other. Therefore they're minimal members.
There's no least element because it only the empty set is the subset of any set but it's given that it's not part of $K$.
I'm not sure whether my proof is formal enough and I really prove the greatest and minimal elements.
You are correct. The set $G$ of all possible pairs is always transitive, and is maximal since if $R$ is any relation it must be a subset of $G$. Since they took out the empty relation, which is also transitive, you do get 16 minimal elements, since any one element set is transitive whether or not it is a pair of distinct or the same elements.