product of posets

product of posets

Cartesian Ordering

Let

(

P

1

,

≤

1

)

and

(

P

2

,

≤

2

)

be posets. Let

P

=

P

1

×

P

2

, the

Cartesian product

of the underlying sets. Next, define a

binary relation

≤

on

P

, given by

(

a

,

b

)

≤

(

c

,

d

)

iff

a

≤

1

c

⁢

and

⁢

b

≤

2

d

.

Then

≤

is a

partial order

on

P

.

(

P

,

≤

)

is called the

product of posets

(

P

1

,

≤

1

)

and

(

P

2

,

≤

2

)

. The

ordering

≤

is called the

Cartesian ordering

. As it is customary, we write

P

to mean

(

P

,

≤

)

.

If

P

1

and

P

2

are

antichains

, their

product

is also an antichain. If they are both join semilattices, then their product

P

is a

join semilattice

as well. The join of

(

a

,

b

)

and

(

c

,

d

)

are given by

(

a

,

b

)

∨

(

c

,

d

)

=

(

a

∨

1

c

,

b

∨

2

d

)

.

Conversely

, if the product

P

of two posets

P

1

and

P

2

is a join semilattice, then

P

1

and

P

2

are both join semilattices. If

(

a

,

b

)

∨

(

c

,

d

)

=

(

e

,

f

)

, then

e

is the

upper bound

of

a

and

c

. If

g

≤

e

is an upper bound of

a

and

c

, then

(

g

,

f

)

is an upper bound of

(

a

,

b

)

and

(

c

,

d

)

, whence

(

g

,

f

)

=

(

e

,

f

)

, or

g

=

e

. So

g

=

a

∨

1

c

. Similarly,

f

=

b

∨

2

d

. Dually,

P

=

P

1

×

P

2

is a

meet semilattice

(and consequently, a

lattice

) iff both

P

1

and

P

2

are. Equivalently, the product of (semi)lattices can be defined purely algebraically (using

∨

and

∧

only).

Another simple fact about the product of posets is the following: the product is never a chain unless one of the posets is

trivial

(a singleton). To see this, let

P

=

P

1

×

P

2

and

(

a

,

b

)

,

(

c

,

d

)

∈

P

. Then

(

a

,

b

)

and

(

c

,

d

)

are

comparable

, say

(

a

,

b

)

≤

(

c

,

d

)

, which implies

a

≤

1

c

and

b

≤

2

d

. Also,

(

c

,

b

)

and

(

a

,

d

)

are comparable. But since

b

≤

2

d

, we must have

(

c

,

b

)

≤

(

a

,

d

)

, which means

c

≤

1

a

, showing

a

=

c

, or

P

1

=

{

a

}

.

Remark

. The product of two posets can be readily extended to any finite product,

countably infinite

product, or even arbitrary product of posets. The definition is

similar

to the one given above and will not be repeated here.

An example of a product of posets is the lattice in

ℝ

n

(

http://planetmath.org/LatticeInMathbbRn

), which is defined as the

free abelian group

over

ℤ

in

n

generators

. But from a poset perspective, it can be viewed as a product of

n

chains, each

order isomorphic

to

ℤ

. As we have just seen earlier, this product is a lattice, and hence the name “lattice” in

ℝ

n

.

Lexicographic Ordering

Again, let

P

1

and

P

2

be posets. Form the Cartesian product of

P

1

and

P

2

and call it

P

. There is another way to partial order

P

, called the

lexicographic order

. Specifically,

(

a

,

b

)

≤

(

c

,

d

)

iff

{

a

≤

c

⁢

, or

a

=

c

⁢

and

⁢

b

≤

d

.

More generally, if

{

P

i

∣

i

∈

I

}

is a

collection

of posets

indexed by

a set

I

that is

linearly ordered

, then the Cartesian product

P

:=

∏

P

i

also has the lexicographic order:

(

a

i

)

≤

(

b

i

)

iff there is some

k

∈

I

such that

a

j

=

b

j

for all

j

<

k

and

a

k

≤

b

k

.

We show that this is indeed a partial order on

P

:

Proof.

The three things we need to verify are

•

(

Reflexivity

). Clearly,

(

a

i

)

≤

(

a

i

)

, since

a

i

≤

a

i

for any

i

∈

I

.

•

(

Transitivity

). If

(

a

i

)

≤

(

b

i

)

and

(

b

i

)

≤

(

c

i

)

, then for some

k

,

ℓ

∈

I

we have that

(a)

a

j

=

b

j

for all

j

<

k

and

a

k

≤

b

k

, and

(b)

b

j

=

c

j

for all

j

<

ℓ

and

b

ℓ

≤

c

ℓ

.

Since

I

is a

total order

,

k

and

ℓ

are comparable, say

k

≤

ℓ

, so that

a

j

=

b

j

=

c

j

for all

k

<

ℓ

and

a

k

≤

b

k

≤

c

k

. Since

P

k

is partially ordered,

a

k

≤

c

k

as well. Therefore

(

a

i

)

≤

(

c

i

)

.

•

(Antisymmetry). Finally, suppose

(

a

i

)

≤

(

b

i

)

and

(

b

i

)

≤

(

a

i

)

. If

(

a

i

)

≠

(

b

i

)

, then

(

a

i

)

≤

(

b

i

)

implies that we can find

k

∈

I

such that

a

j

=

b

j

for all

j

<

k

and

a

k

<

b

k

. By the same token,

(

b

i

)

≤

(

a

i

)

implies the

existence

of

ℓ

∈

I

with

b

j

=

a

j

for all

j

<

ℓ

and

b

ℓ

<

a

ℓ

. Since

I

is linearly ordered, we can again assume that

k

≤

ℓ

. But then this means that either

k

<

ℓ

, in which case

b

k

=

a

k

, a

contradiction

, or

k

=

ℓ

, in which case we have that

a

k

<

b

k

=

b

ℓ

<

a

ℓ

=

a

k

, another contradiction. Therefore

(

a

i

)

=

(

b

i

)

.

This

completes

the proof.
∎

Title

product of posets

Canonical name

ProductOfPosets

Date of creation

2013-03-22 16:33:25

Last modified on

2013-03-22 16:33:25

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

14

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06A06

Classification

msc 06A99

Related topic

LatticeInMathbbRn

Defines

product of lattices

Defines

Cartesian ordering

Defines

lexicographic ordering

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