Definition 1   Let $P$ be a poset and $A$ a subset of $P$ . The upper set of $A$ is defined to be the set $$ \up A = \lbrace b \in P \mid (\exists \, a \in A) \, a \le b \rbrace $$

Theorem 1   $\up \emptyset = \emptyset$

Proof. Any statement of the form ``$(\exists \, a \in \emptyset) P(a)$ '' is identically false no matter what the predicate $P$ (i.e. it is an antitautology) and the set of objects satsfying an identically false condition is empty, so $\up \emptyset = \emptyset$ .

Theorem 2   $A\subseteq \up A$

Proof. This follows from reflexivity -- for every $a \in A$ , one has $a \le a$ , hence $a \in \up A$ .

Theorem 3   $\uparrow \up A=\up A$

Proof. By the previous result, $\up A \subseteq \uparrow \up A$ . Hence, it only remains to show that $\uparrow \up A \subseteq \up A$ . This follows from transitivity. In order for some $a$ to be an element of $\uparrow \up A$ , there must exist $b$ and $c$ such that $a \ge b \ge C$ and $C \in A$ . By transitivity, $A \ge C$ , so $a \in \up A$ , hence $\uparrow \up A \subseteq \up A$ as well.

Theorem 4   If $A$ and $B$ are subsets of a partially ordered set, then $$ \up (A \cup B) = (\up A) \cup (\up B) $$

Proof. On the one hand, if $a \in \up (A \cup B)$ , then $a \ge b$ for some $b \in A \cup B$ . It then follows that either $b \in A$ or $b \in B$ . In the former case, $a \in \up A$ , in the latter case, $a \in \up B$ so, either way $a \in (\up A) \cup (\up B)$ . Hence $\up (A \cup B) \subseteq (\up A) \cup (\up B)$ .

On the other hand, if $a \in (\up A) \cup (\up B)$ , then either $a \in (\up A)$ or $a \in (\up B)$ . In the former case, there exists $b$ such that $a \ge b$ and $b \in A$ . Since $A \subseteq A \cup B$ , we also have $b \in A \cup B$ , hence $a \in \up (A \cup B)$ . Likewise, in the second case, we also conclude that $a \in \up (A \cup B)$ . Therefore, we have $(\up A) \cup (\up B) \subseteq \up (A \cup B)$ .

Theorem 5   $\up P = P$

Theorem 6   $A\subseteq B$ , $\up A\subseteq \up B$