Directed Set has Strict Successors iff Unbounded Above

Theorem

Let $\struct {S, \le}$ be a directed set.

Then every element of $S$ has a strict successor in $S$ if and only if $S$ has no upper bound in $S$.

Suppose that each element of $S$ has a strict successor in $S$.

If $x$ is any element of $S$, then $x$ has a strict successor.

Therefore, $x$ is not an upper bound of $S$.

$\Box$

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 $x \le y$.

Since $p \not\le x$, it follows that $x \ne y$.

Hence, since we already know that $x \le y$, in fact $x < y$.

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

$\blacksquare$