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.