PlanetMath: criteria for a poset to be a complete lattice

Proof. Implications $1.\Rightarrow 2.\Rightarrow 3.$ are clear. We will show $3.\Rightarrow 2.\Rightarrow 1.$

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

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