distributive inequalities

Let $L$ be a lattice. Then for $a,b,c\in L$, we have the following inequalities:

1. $a\vee (b\wedge c)\le (a\vee b)\wedge (a\vee c)$,

2. $(a\wedge b)\vee (a\wedge c)\le a\wedge (b\vee c)$.

Proof.

Since $a\le a\vee b$ and $a\le a\vee c$, $a\le (a\vee b)\wedge (a\vee c)$. Similarly, $b\wedge c\le b\le a\vee b$ and $b\wedge c\le c\le a\vee c$ imply $b\wedge c\le (a\vee b)\wedge (a\vee c)$. Together, we have $a\vee (b\wedge c)\le (a\vee b)\wedge (a\vee c)$.

The second inequality is the dual of the first one. ∎

The two inequalities above are called the distributive inequalities.

Proposition A lattice $L$ is a distributive lattice if one of the following inequalities holds:

1. $(a\vee b)\wedge (a\vee c)\le a\vee (b\wedge c)$,

2. $a\wedge (b\vee c)\le (a\wedge b)\vee (a\wedge c)$.

Proof.

By the distributive inequalities, all we need to show is that 1. implies 2. (that 2. implies 1. is just the dual statement). So suppose 1. holds. Then

$(a\wedge b)\vee (a\wedge c)$ $\ge ((a\wedge b)\vee a)\wedge ((a\wedge b)\vee c)$ $\mathrm{\hspace{1em}\hspace{1em}}\text{by assumption}$

$=a\wedge ((a\wedge b)\vee c)$ $\mathrm{\hspace{1em}\hspace{1em}}\text{by absorption}$

$\ge a\wedge ((c\vee a)\wedge (c\vee b))$ $\mathrm{\hspace{1em}\hspace{1em}}\text{by assumption}$

$=(a\wedge (c\vee a))\wedge (c\vee b)$ $\mathrm{\hspace{1em}\hspace{1em}}\text{meet associativity}$

$=a\wedge (c\vee b).$ $\mathrm{\hspace{1em}\hspace{1em}}\text{by absorption}$

∎

Title	distributive inequalities
Canonical name	DistributiveInequalities
Date of creation	2013-03-22 16:37:48
Last modified on	2013-03-22 16:37:48
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	5
Author	CWoo (3771)
Entry type	Derivation
Classification	msc 06D99
Related topic	ModularInequality