 2.1. ORDER THEORY

19

1

.

d

S

S

=

d

X

S

d

X

;

2

.

d

S

S

=

d

X

S

d

X

.

Proof.

We will prove only the first as the second is dual.

By definition of joins, it is enough to prove

y

w

d

S

S

y

w

d

X

S

d

X

.

Really,

y

w

l

[

S

x

[

S

:

y

w

x

X

S

x

X

:

y

w

x

X

S

:

y

w

l

X

y

w

l

X

S

l

X.

Definition

69

.

A

sublattice

of a lattice is it subset closed regarding

t

and

u

.

Obvious

70

.

Sublattice with induced order is also a lattice.

2.1.6. Distributivity of lattices.

Definition

71

.

A

distributive

lattice is such lattice

A

that for every

x, y, z

A

1

.

x

u

(

y

t

z

) = (

x

u

y

)

t

(

x

u

z

);

2

.

x

t

(

y

u

z

) = (

x

t

y

)

u

(

x

t

z

).

Theorem

72

.

For a lattice to be distributive it is enough just one of the

conditions:

1

.

x

u

(

y

t

z

) = (

x

u

y

)

t

(

x

u

z

);

2

.

x

t

(

y

u

z

) = (

x

t

y

)

u

(

x

t

z

).

Proof.

(

x

t

y

)

u

(

x

t

z

) =

((

x

t

y

)

u

x

)

t

((

x

t

y

)

u

z

) =

x

t

((

x

u

z

)

t

(

y

u

z

)) =

(

x

t

(

x

u

z

))

t

(

y

u

z

) =

x

t

(

y

u

z

)

(applied

x

u

(

y

t

z

) = (

x

u

y

)

t

(

x

u

z

) twice).

2.1.7. Difference and complement.

Definition

73

.

Let

A

be a distributive lattice with least element

. The

difference

(denoted

a

\

b

) of elements

a

and

b

is such

c

A

that

b

u

c

=

and

a

t

b

=

b

t

c

. I will call

b

substractive

from

a

when

a

\

b

exists.

Theorem

74

.

If

A

is a distributive lattice with least element

, there exists

no more than one difference of elements

a

,

b

.

Proof.

Let

c

and

d

be both differences

a

\

b

. Then

b

u

c

=

b

u

d

=

and

a

t

b

=

b

t

c

=

b

t

d

. So

c

=

c

u

(

b

t

c

) =

c

u

(

b

t

d

) = (

c

u

b

)

t

(

c

u

d

) =

⊥ t

(

c

u

d

) =

c

u

d.

Similarly

d

=

d

u

c

. Consequently

c

=

c

u

d

=

d

u

c

=

d

.