Directed Set has Strict Successors iff Unbounded Above

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

Suppose that $S$ has no upper bound.

Let $x \in S$.

Then there exists $y \in S$ such that $x < y$.

Proof
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 $x < y$.