properties of arbitrary joins and meets
In this entry, we list and prove some of the basic properties of arbitrary joins and meets. Some of the properties work in general posets, while others work only in lattices, and sometimes only in Boolean algebras^{}.
Let $P$ be a poset and $B$ and $C$ are subsets of $P$ such that $\bigvee B$ and $\bigvee C$ exist.

1.
$b\le \bigvee B$ for any $b\in B$. More generally, if $A\subseteq B$ and $\bigvee A$ exists, then $\bigvee A\le \bigvee B$.

2.
if $b\le a$ for every $b\in B$, then $\bigvee B\le a$.

3.
If $B=\bigcup \{{B}_{i}\mid i\in I\}$, and each ${b}_{i}:=\bigvee {B}_{i}$ exists, then $\bigvee \{{b}_{i}\mid i\in I\}$ exists and is equal to $\bigvee B$. Conversely, if we drop the assumption^{} that $\bigvee B$ exists, but assume instead that $\bigvee \{{b}_{i}\mid i\in I\}$ exists, then $\bigvee B$ exists and is equal to $\bigvee \{{b}_{i}\mid i\in I\}$.
Proof.
Let $b=\bigvee B$. For each $i\in I$, and each $c\in {B}_{i}\subseteq B$, we clearly have that $c\le b$. So ${b}_{i}\le b$, or that $b$ is an upper bound^{} of the collection^{} $D:=\{{b}_{i}\mid i\in I\}$. If $d$ is any upper bound of $D$, then ${b}_{i}\le d$. For any $c\in B$, $c\in {B}_{i}$ for some $i\in I$, so that $c\le {b}_{i}$ and hence $c\le d$. This shows that $b\le d$, or that $b$ is the least upper bound of $D$.
Conversely, suppose $D:=\bigvee \{{b}_{i}\mid i\in I\}$ exists and is equal to $d$. Then for any $b\in B$, $b\in {B}_{i}$ for some $i\in I$, so that $b\le {b}_{i}$, and hence $b\le d$. This shows that $d$ is an upper bound of $B$. If $f$ is any upper bound of $B$, then $f$ is an upper bound of ${B}_{i}$ in particular, so ${b}_{i}\le f$. Since $i$ is arbitray, $d\le f$, or that $d$ is the least upper bound of $B$. ∎

4.
If $\bigvee \{a\vee b\mid b\in B\}$ exists, then it is equal to $a\vee \bigvee B$.
Proof.
Let $c=\bigvee B$ and $d=\bigvee \{a\vee b\mid b\in B\}$. We want to show that $a\vee c=d$. Since $b\le c$ for all $b\in B$, we have that $a\vee b\le a\vee c$, and so $d\le a\vee c$ as $d$ is the least upper bound of $\{a\vee b\mid b\in B\}$. On the other hand $a\vee b\le d$, so that $a\le d$ and $b\le d$, for all $b\in B$, the last inequality^{} means that $c\le d$ as well. Therefore $a\vee c\le d$, and we are done. ∎

5.
If $P$ is a Boolean algebra then the following hold:

(a)
$\bigwedge ({B}^{\prime})$ exists, where ${B}^{\prime}:=\{{b}^{\prime}\mid b\in B\}$, and is equal to ${(\bigvee B)}^{\prime}$.
Proof.
Let $c=\bigvee B$. Then $b\le c$ for any $b\in B$, so that ${c}^{\prime}\le {b}^{\prime}$, or ${c}^{\prime}$ is a lower bound for ${B}^{\prime}$. If $d$ is any lower bound of ${B}^{\prime}$, then $d\le {b}^{\prime}$ for every $b\in B$, so that $b\le {d}^{\prime}$, which implies $c\le {d}^{\prime}$, or $d\le {c}^{\prime}$. This means that ${c}^{\prime}$ is the greatest lower bound^{} of ${B}^{\prime}$, or that ${c}^{\prime}=\bigwedge ({B}^{\prime})$. ∎

(b)
$\bigvee \{a\wedge b\mid b\in B\}$ exists and is equal to $a\wedge \bigvee B$ for any $a\in A$.
Proof.
Let $c=\bigvee B$. Then $b\le c$ for any $b\in B$ and so $a\wedge b\le a\wedge c$. Therefore $a\wedge c$ is an upper bound of $\{a\wedge b\mid b\in B\}$. Now, if $d$ is an upper bound of $\{a\wedge b\mid b\in B\}$, then $a\wedge b\le d$ for every $b\in B$. So $b=({a}^{\prime}\vee a)\wedge b=({a}^{\prime}\wedge b)\vee (a\wedge b)\le ({a}^{\prime}\wedge b)\vee d\le {a}^{\prime}\vee d$. This means that ${a}^{\prime}\wedge d$ is an upper bound of $B$, so $c\le {a}^{\prime}\vee d$. Therefore, $a\wedge c\le a\wedge ({a}^{\prime}\vee d)=(a\wedge {a}^{\prime})\vee (a\wedge d)=a\wedge d$. Hence, $a\wedge c$ is the least upper bound of $\{a\wedge b\mid b\in B\}$. ∎

(c)
Define $B\wedge C:=\{b\wedge c\mid b\in B\text{and}c\in C\}.$ Then $\bigvee (B\wedge C)$ exists and is equal to $\bigvee B\wedge \bigvee C$.
Proof.
Let $d=\bigvee B$ and $e=\bigvee C$. Then $\bigvee B\wedge \bigvee C=d\wedge \bigvee C=\bigvee \{d\wedge c\mid c\in C\}$ by 4.b above. Now, $d\wedge c=\bigvee B\wedge c=\bigvee \{b\wedge c\mid b\in B\}$ again by 4.b. For each $c\in C$, set ${B}_{c}:=\{b\wedge c\mid b\in B\}$. Then $\bigvee {B}_{c}=d\wedge c$ and $B\wedge C=\bigcup \{{B}_{c}\mid c\in C\}$. Therefore, by (3), $\bigvee (B\wedge C)$ exists and is equal to $\bigvee \{\bigvee {B}_{c}\mid c\in C\}=\bigvee \{d\wedge c\mid c\in C\}=\bigvee B\wedge \bigvee C$. ∎

(a)
Remarks.

•
All of the properties above can be dualized: assume that $B$ and $C$ are subsets of a poset $P$ such that $\bigwedge B$ and $\bigwedge C$ exist, then:

(a)
if $A\subseteq B$ and $\bigwedge A$ exists, then $\bigwedge B\le \bigwedge A$.

(b)
if $a\le b$ for every $b\in B$, then $a\le \bigwedge B$.

(c)
if $B=\bigcup \{{B}_{i}\mid i\in I\}$, and each ${b}_{i}:=\bigwedge {B}_{i}$ exists, then $\bigwedge \{{b}_{i}\mid i\in I\}$ exists iff $\bigwedge B$ does, and they are equal when one exists.

(d)
if $\bigwedge \{a\wedge b\mid b\in B\}$ exists, then it is equal to $a\wedge \bigwedge B$.

(e)
If $P$ is a Boolean algebra, then

i.
$\bigvee ({B}^{\prime})$ exists, where ${B}^{\prime}:=\{{b}^{\prime}\mid b\in B\}$, and is equal to ${(\bigwedge B)}^{\prime}$.

ii.
$\bigwedge \{a\vee b\mid b\in B\}$ exists and is equal to $a\vee \bigwedge B$ for any $a\in A$.

iii.
Define $B\vee C:=\{b\vee c\mid b\in B\text{and}c\in C\}.$ Then $\bigwedge (B\vee C)$ exists and is equal to $\bigwedge B\vee \bigwedge C$.

i.

(a)

•
Notice that for property 5 above, the condition that $P$ be Boolean can not be dropped. For example, consider the set $P$ of nonnegative integers. For any two elements $a,b\in P$, define $a\le b$ by the divisibility relation^{} $ab$. It is easy to see that $P$ is a bounded distributive lattice^{}, with top element $0$ and bottom element $1$. However, it is not complemented (suppose ${2}^{\prime}$ is a complement of $2$, then ${2}^{\prime}\wedge 2=1$, so that ${2}^{\prime}$ must be odd, but then ${2}^{\prime}\vee 2=2\cdot {2}^{\prime}\ne 0$, a contradiction^{}).
More generally, for any subset $A$ of $P$, define $\bigvee A$ to be the smallest nonnegative integer $c$ such that $ac$ for all $a\in A$, while $\bigwedge A$ is the largest nonnegative integer $d$ such that $da$ for all $a\in A$. If $A=\mathrm{\varnothing}$, define $\bigvee A=1$ and $\bigwedge A=0$. Then it is not hard to see that $P$ is in addition a complete lattice^{}. However, if we take $A$ to be the set of all odd prime numbers, then $\bigvee A=0$, so that for any $x\in P$, $x\wedge \bigvee A=0$. But if $x$ is any element in $A$, then $\bigvee \{x\wedge a\mid a\in A\}=x\ne 0$.
Title  properties of arbitrary joins and meets 

Canonical name  PropertiesOfArbitraryJoinsAndMeets 
Date of creation  20130322 17:52:38 
Last modified on  20130322 17:52:38 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  12 
Author  CWoo (3771) 
Entry type  Derivation 
Classification  msc 06A06 