What needs to be proved to show a partially ordered set is a lattice? I have recently been introduced to lattices, but I am not clear on how to prove a partially ordered set is a lattice.
For example, let $B$ be any set and denote its power set by $2^{B}$. Let the partial order on $2^{B}$ be defined by the $\subseteq$ relation.
To prove that $2^{B}$ is a lattice I know I have to show that for any two sets $B_{1}, B_{2} \in 2^{B}$ there is a unique $A_{sup} = B_{1} \vee B_{2}$ and $A_{inf} = B_{1} \wedge B_{2}$ (is there anything else to show?).
I know from set theory that the smallest superset of $B_{1}, B_{2}$ will be $B_{1} \cup B_{2}$ and the largest subset will be $B_{1} \cap B_{2}$. It is also clear that these set will be in the powerset. 
So what is there to show? The proof here seems to have different steps. 
 A: To show that a partial order is a lattice, you have to show that it has well-defined "meet" and "join" operations. Because the partial order here is $\subseteq$ and the underlying set is the full powerset, meet and join are $\cap$ and $\cup$ respectively. 
To show that a set of sets $S$ is a lattice with respect to $\subseteq$, you have to show that it's closed under those operations. When "meet" and "join" are actual intersection and union, this amounts to showing: 
$\forall X, Y \in S$, both $X \cap Y \in S$ and $X \cup Y \in S$. When $S = \mathcal{P}(B) = 2^B$ is the powerset of a set $B$, there isn't a lot to actually "prove". You have to show that $X \cap Y$ is actually the largest set contained ($\subseteq$) in both $X$ and $Y$, and that $X \cup Y$ is actually the smallest set containing ($\subseteq$) in both $X$ and $Y$... and that both are actually subsets of $B$ :)
Note, however, that even when the underlying set is a set of sets, and  the partial order is $\subseteq$, "meet' and 'join' may not actually be set-theoretic $\cap$ and $\cup$. Example: a four-element diamond lattice $\{\emptyset, \{0, 1\}, evenInts, \mathbb{N}\}$, where $evenInts$ is... the set of even integers. The "meet" of $\{0, 1\}$ and $evenInts$ is the bottom of the diamond, $\emptyset$, but of course their intersection is nonempty; similarly, the "join" of $\{0, 1\}$ and $evenInts$ is the top of the diamond, $\mathbb{N}$, but their union doesn't equal  $\mathbb{N}$.
