criteria for a poset to be a complete lattice
Proposition^{}. Let $L$ be a poset. Then the following are equivalent^{}.

1.
$L$ is a complete lattice^{}.

2.
for every subset $A$ of $L$, $\bigvee A$ exists.

3.
for every finite subset $F$ of $L$ and every directed set^{} $D$ of $L$, $\bigvee F$ and $\bigvee D$ exist.
Proof.
Implications^{} $1.\Rightarrow 2.\Rightarrow 3.$ are clear. We will show $3.\Rightarrow 2.\Rightarrow 1.$
$(3.\Rightarrow 2.)$ If $A=\mathrm{\varnothing}$, then $\bigvee A=0$ by definition. So assume $A$ be a nonempty subset of $L$. Let ${A}^{\prime}$ be the set of all finite subsets of $A$ and $B=\{\bigvee F\mid F\in {A}^{\prime}\}$. By assumption^{}, $B$ is welldefined and $A\subseteq B$. Next, let ${B}^{\prime}$ be the set of all directed subsets of $B$, and $C=\{\bigvee D\mid D\in {B}^{\prime}\}$. By assumption again, $C$ is welldefined 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 $\bigvee \mathrm{\varnothing}$ 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 $A=\mathrm{\varnothing}$, then $\bigwedge A=\bigvee L=1$ by definition of an arbitrary meet over the empty set^{}. So assume $A\ne \mathrm{\varnothing}$. Let ${A}^{\prime}$ be the set of lower bounds of $A$: ${A}^{\prime}=\{x\in L\mid x\le a\text{for all}a\in A\}$ 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 $b$. ∎
Remarks.

•
Dually, a poset is a complete lattice iff every subset has an infimum^{} iff infimum exists for every finite subset and every directed subset.

•
The above proposition shows, for example, that every closure system is a complete lattice.
Title  criteria for a poset to be a complete lattice 

Canonical name  CriteriaForAPosetToBeACompleteLattice 
Date of creation  20130322 16:37:53 
Last modified on  20130322 16:37:53 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  7 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 06B23 
Classification  msc 03G10 
Related topic  MeetContinuous 
Related topic  IntersectionStructure 