Let $A$ be a set and $\subseteq $ is a partial order on $\mathcal{P}(A)$. Let $C$ and $D$ be subsets of A. Prove that the supremum (least upper bound) of $\left \{ C, D \right \}$ is $C \vee D$ and the infimum (greatest lower bound) is $C \wedge D$.
First, we will prove that $ C \vee D \subseteq E$ where $E$ is the supremum of $\left \{ C,D \right \}$.Then $C \subseteq E$ and $D \subseteq E$. Hence, if there is an element in $C$ or $D$, then that element is also in $E$. Therefore, $C \vee D \subseteq E$. Now we must prove that $E \subseteq C \vee D$. Let $x \in E$. Then $x \in C \vee D$ or $x \notin C \vee D$. Suppose $x\notin C \vee D$. Then $E - \left \{ x \right \} \subseteq E$ and $C \vee D \subseteq E - \left \{ x \right \}$. Then E is not supremum. But it was assumed so. Hence, $ x \in C \vee D$. Therefore, $C \vee D = E$ and $C \vee D$ is the supremum.
We proceed similarly. First, we will prove that $F \subseteq C \wedge D$ where $F$ is the infimum of $\left \{ C,D \right \}$. Then $F \subseteq C$ and $F \subseteq D$. Hence, every element in $F$ is also and element in $C$ and $D$. Then, $F \subseteq C \wedge D$. Now, we must prove that $C \wedge D \subseteq F$. Let $x \in C \wedge D$, then either $x \in F$ or $x \notin F$. Suppose $x \notin F$, then $F \subseteq F + \left \{ x \right \}$ and $F + \left \{x\right \} \subseteq C \wedge D$. Hence, $F$ is not the infimum. But it was assumed so. Therefore, $x \in F$. Therefore, $ C\wedge D = F$ and $C \wedge D$ is the infimum.
Verification? I'm a little uneasy about the part where we get $E - \left \{ x \right \}$ and $F + \left \{ x \right \}$.
$\vee$ and $\wedge$ mean "or" and "and".
The are logical connectivies between statements.
$\cup$ and $\cap$ mean "union" and "intersection".
They are binary operations on sets.
Forget the clumsy {x} stuff.
Let A and B be two sets.
As A subset A $\cup$ B and B subset A $\cup$ B,
A $\cup$ B is an upper bound of A and B.
Let D be an upper bound of A and B.
Since A subset D and B subset D, A $\cup$ B subset D.
As A $\cup$ B is an upper bound of A and B
and A $\cup$ B is a subset of every upper bound of A and B
A $\cup$ B is the least upper bound of A and B.
The reverse order dual problem showing A $\cap$ B
is the greatest lower bound of A and B is similar.
Just reverse the order throughout the proof,