Show that B has infimum

Question

Let $(A,\leq)$ is a lattice. if $B\subseteq A$ and $|B|=3$ then show that there is $\inf(B)$

If A is a lattice we know for every $x,y\in A$ there are $\inf\{x,y\}=x\wedge y $ and $\sup\{x,y\}=x\vee y$ since it satisfy all $x,y\in B$, $B$ is also a lattice. how do we use $|B|=3$?

Answer 1

Let $B=\{x,y,z\}$. By definition of lattice each of the pairs in this set have both an infimum and a supremum. The number of such pairs is $\binom{3}{2}=6$. Now look into the infimum among these and that will be the infimum of $B$.

Answer 2

As long as $B$ is finite, we can compute its meet and join.

Indeed, meet and join are well-defined binary operations on the lattice. Furthermore, they are symmetric and associative and as such we can quantity over them like $\prod,\sum$ for $\cdot,+$ in arithmetic.

More precicely, if $B = \{b_1 , b_2, \ldots, b_n \} \subseteq A$, for some $n \in \mathbb{N}$, then $$\inf B = \bigwedge B = \bigwedge_{i = 1}^n b_i = b_1 \land b_2 \land \cdots \land b_n$$

Hope that helps! :)

(Perhaps a good exercise is to prove the symmetry and associativity of meet $\land$).