$(3.\Rightarrow 2.)$ If $A=\varnothing$ , then $\bigvee A=0$ by definition. So assume $A$ be a non-empty subset of $L$ . Let $A^{\prime}$ be the set of all finite subsets of $A$ and $B=\lbrace \bigvee F\mid F\in A^{\prime}\rbrace$ . By assumption, $B$ is well-defined and $A\subseteq B$ . Next, let $B^{\prime}$ be the set of all directed subsets of $B$ , and $C=\lbrace \bigvee D\mid D\in B^{\prime}\rbrace$ . By assumption again, $C$ is well-defined and $B\subseteq C$ . Now, every chain in $C$ has a maximal element in $C$ (since a chain is a directed set), $C$ itself has a maximal element $d$ by Zorn's Lemma. We will show that $d$ is the least upper bound of elments of $A$ . It is clear that each $a\in A$ is bounded above by $d$ ($A\subseteq B\subseteq C$ ). If $t$ is an upper bound of elements of $A$ , then it is an upper bound of elements of $B$ , and hence an upper bound of elements of $C$ , which means $d\le t$ .

$(2.\Rightarrow 1.)$ By assumption $\bigvee \varnothing$ exists ($=0$ ), so that $\bigwedge L=0$ . Now suppose $A$ is a proper subset of $L$ . We want to show that $\bigwedge A$ exists. If $A=\varnothing$ , then $\bigwedge A=\bigvee L=1$ by definition of an arbitrary meet over the empty set. So assume $A\neq \varnothing$ . Let $A^{\prime}$ be the set of lower bounds of $A$ : $A^{\prime}=\lbrace x\in L\mid x\le a\mbox{ for all }a\in A\rbrace$ and let $b=\bigvee A^{\prime}$ , the least upper bound of $A^{\prime}$ . $b$ exists by assumption. Since $A$ is a set of upper bounds of $A^{\prime}$ , $b\le a$ for all $a\in A$ . This means that $b$ is a lower bound of elements of $A$ , or $b\in A^{\prime}$ . If $x$ is any lower bound of elements of $A$ , then $x\le b$ , since $x$ is bounded above by $b$ ($b=\bigvee A^{\prime}$ ). This shows that $\bigwedge A$ exists and is equal to $b$ .