Directed Set has Strict Successors iff Unbounded Above

Theorem
Let $\left({S, \le}\right)$ be a directed set.

Then every element of $S$ has a strict successor in $S$ iff $S$ has no upper bound in $S$.

Proof
Suppose $S$ has no upper bound in $S$.

Let $x \in S$.

Since $S$ has no upper bound, $x$ is not an upper bound of $S$.

Thus for some $p \in S$, $p \not\le x$.

By the definition of a directed set, there is a $y \in S$ such that $p \le y$ and $x \le y$.

Suppose for the sake of contradiction that $x = y$.

Then since $p \le y$ we would conclude that $p \le x$, a contradiction.

Thus $x \ne y$.

Since we already know $x \le y$, in fact $x < y$.

Thus $y$ is a strict successor of $x$.

Suppose, on the other hand, that each element of $S$ has a strict successor in $S$.

If $x$ is any element of $S$, then $x$ has a strict successor, so $x$ is not an upper bound of $S$.