Michael: An arbitrary poset \\(\langle A, \le \rangle\\) need not contain an element like \\(\varnothing\\); consider \\(\\{\mathrm{false}, \mathrm{true}\\}\\) with the order given in the book. Are you concerned specifically with the powerset order \\(\langle \mathcal{P}(A), \subseteq \rangle\\)? In that case, \\(a \vee \varnothing = a\\), since \\(\varnothing \subseteq a\\) for all \\(a\\), and hence \\(a\\) is already the least upper bound of \\(\\{a, \varnothing\\}\\). Similarly, \\(a \wedge \varnothing = \varnothing\\), since \\(\varnothing\\) is already the greatest lower bound of the two.