A Lattice that is not a Complete Lattice

The set $\mathbb Q$ of all rational numbers, with the usual linear order, is an infinite distributive lattice which is not complete. For example, $\mathbb Q$ itself has neither a least upper bound nor a greatest lower bound in $\mathbb Q$; neither does the bounded nonempty set $\{x\in\mathbb Q:x^2\lt2\}.$

Consider the lattice of finite subsets of $\Bbb N$. Since the union and intersection of two finite subsets is again finite, this is indeed a lattice. But the set $A=\{\{0\},\{0,1\},\{0,1,2\},\dots\}$ has no upper bound in the lattice. Of course $\Bbb N$ is an upper bound of $A$ in $\mathcal P(\Bbb N)$, the powerset of $\Bbb N$.