Suppose I have an antichain $\{a,b\}$ and I form this into a poset $P = \{a,b,\top,\bot\}$ so that $a \le \top$, $b \le \top$, $\bot \le a$ and $\bot \le b$. I believe this forms a lattice.
So far so good. Now, it is known that every finite lattice is complete, which means that every subset of that lattice is also a lattice. Suppose I form the subset $S = \{a,b\}$. My formula doesn't work anymore because $\top \notin S$ and $\bot \notin S$, so it seems like my sublattice doesn't meet the requirement, which makes me think my poset P isn't a lattice either. What am I misunderstanding?
- Is it OK for $a \vee b = \top$ when $\top \notin S$?
- Or have I misunderstood the definition of lattice, finite lattice, or complete lattice; or is it not the case that every finite lattice is complete?
I agree that $P$ is a lattice. And a finite lattice is (trivially) complete.
What $P$ complete means here is that every subset of $P$ has a meet and a join in $P$. This certainly holds for your subset $S$, which has meet $\top$ and join $\bot$.
A subset $S \subseteq P$ is a sublattice if $\forall x,y \in S: x \land y \in S, x \lor y \in S$. $S$ should be closed under the lattice operations meet and join (like subgroups in groups, subspaces in vector spaces etc.). Also, $\top \in S$ and $\bot \in S$ must hold (neutral elements must lie in the substructure too).
So your $S$ is certainly not a sublattice of $P$. And complete does not imply that every subset is a sublattice. It's not even true that a sublattice of a complete lattice is itself a complete lattice (linear orders already suffice to show this, consider $\mathbb{Q}$ inside $\mathbb{R}$).