Completeness of product of two lattices

Question

As I know, a lattice is complete if both sup and inf do exist for that. Now consider this question and its proof.

I'm gonna use this sort of argument for general lattices (as posets for which meet and join are defined for every two elements.)

The OP has shown that each arbitrary subset of $X$ and $Y$ has a sup. And then he concluded that each of them is complete without addressing the existence of inf for them!

1- How is this case justified in the proof?!

2- Is the argument reversible? I mean, can one use the same argument to prove that "if $X$ and $Y$ are complete then their product is complete"?

Answer 1

1. If, in a poset $A$ every subset has a supremum, then also every subset $X\subseteq A$ will have an infimum, namely it must be: $$\inf X = \sup\{a\in A\mid \forall x\in X: a\le x\}$$
2. Yes, this direction is not a big surprise: consider $U\subseteq A\times B$ and its projection in $A$, $\pi_A(U)=\{a\mid\exists b:(a,b)\in U\}$. This has a supremum $x$ in $A$.
   Then take $y:=\sup\{b\mid (x,b)\in U\}$.
   Finally, verify that $(x,y)=\sup U$ with the lexicographic order.