way below

way below

Let

P

be a poset and

a

,

b

∈

P

.

a

is said to be

way below

b

, written

a

≪

b

, if for any

directed set

D

⊆

P

such that

⋁

D

exists and that

b

≤

⋁

D

, then there is a

d

∈

D

such that

a

≤

d

.

First note that if

a

≪

b

, then

a

≤

b

since we can set

D

=

{

b

}

, and if

P

is finite, we have the

converse

(since

⋁

D

∈

D

). So, given any

element

b

∈

P

, what exactly are the elements that are way below

b

? Below are some examples that will throw some light:

Examples

1.

Let

P

be the poset given by the

Hasse diagram

below:

\xymatrix

⁢

&

⁢

&

⁢

b

⁢

\ar

⁢

@

.

[

d

]

⁢

\ar

⁢

@

.

[

d

⁢

l

⁢

l

]

⁢

\ar

⁢

@

.

[

d

⁢

r

]

⁢

&

⁢

p

⁢

\ar

⁢

@

-

[

d

]

⁢

\ar

⁢

@

-

[

d

⁢

r

]

⁢

&

⁢

&

⁢

q

⁢

\ar

⁢

@

-

[

d

⁢

l

]

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

r

⁢

\ar

⁢

@

-

[

d

⁢

l

]

⁢

s

⁢

&

⁢

t

⁢

\ar

⁢

@

-

[

d

]

⁢

&

⁢

u

⁢

\ar

⁢

@

-

[

d

⁢

l

]

⁢

&

⁢

v

⁢

&

⁢

&

where the dotted lines denote

infinite chains

between the

end points

. First, note that every element in

P

is below (

≤

)

b

. However, only

v

is way below

b

.

u

, for example, is not way below

b

, because

D

=

{

x

∣

p

≤

x

<

b

}

is a directed set such that

⋁

D

=

b

and non of the elements in

D

are above

u

. This illustrates the fact that if

P

has a bottom, it is way below everything else.

2.

Suppose

P

is a

lattice

. Then

a

≪

b

iff for any set

D

such that

⋁

D

exists and

b

≤

⋁

D

, there is a finite subset

F

⊆

D

such that

a

≤

⋁

F

.

Proof.

(

⇒

)

. Suppose

a

≪

b

. Let

D

be the set in the

assumption

. Let

E

be the set of all finite joins of elements of

D

. Then

D

⊆

E

. Also, every element of

E

is bounded above by

⋁

D

. If

t

is an

upper bound

of elements of

E

, then it is certainly an upper bound of elements of

D

, and hence

⋁

D

≤

t

. So

⋁

D

is the least upper bound of elements of

E

, or

⋁

E

=

⋁

D

. Furthermore,

E

is directed. So there is an element

e

∈

E

such that

a

≤

e

. But

e

=

⋁

F

for some finite subset of

D

, and this

completes

one side of the proof.

(

⇐

)

. Let

D

be a directed set such that

⋁

D

exists and

b

≤

⋁

D

. There is a finite subset

F

of

D

such that

a

≤

⋁

F

. Since

D

is directed, there is an element

d

∈

D

such that

d

is the upper bound of elements of

F

. So

a

≤

d

, completing the other side of the proof.
∎

3.

With the above assertion, we see that, for example, in the

lattice of subgroups

L

⁢

(

G

)

of a group

G

,

H

≪

K

iff

H

is

finitely generated

. Other

similar

examples can be found in the lattice of

two-sided ideals

of a ring, and the lattice of

subspaces

(projective geometry) of a

vector space

.

4.

In particular, if

P

is a chain, then

a

≤

b

implies that

a

≪

b

. If

D

is a set such that

⋁

D

exists and

b

≤

⋁

D

, then there is a

d

∈

D

such that

b

≤

d

(otherwise

b

is an upper bound of elements of

D

and

⋁

D

≤

b

), so

a

≤

d

.

5.

Here’s an example where

a

≪

b

in

P

but

a

is not the bottom of

P

. Take two complete infinite chains

C

1

and

C

2

with bottom

0

and

1

, and let

P

be their

product

(

http://planetmath.org/ProductOfPosets

)

P

=

C

1

×

C

2

. What elements are way below

(

1

,

1

)

? First, take

D

=

{

(

a

,

1

)

∣

0

≤

a

<

1

}

. Since

P

is complete,

⋁

D

=

(

1

,

1

)

, but every element of

D

is stricly less than

(

1

,

1

)

, so

(

1

,

1

)

is not way below itself. What about elements of the form

(

a

,

1

)

,

a

≠

1

? If we take

D

=

{

(

1

,

b

)

∣

0

≤

b

<

1

}

, then

⋁

D

=

(

1

,

1

)

once again. But no elements of

D

are above

(

a

,

1

)

. So

(

a

,

1

)

can not be way below

(

1

,

1

)

. Similarly, neither can

(

1

,

b

)

be way below

(

1

,

1

)

. Finally, what about

(

a

,

b

)

for

a

<

1

and

b

<

1

? If

D

is a set with

⋁

D

=

(

1

,

1

)

, then

⋁

D

1

=

1

and

⋁

D

2

=

1

, where

D

1

=

{

x

∣

(

x

,

1

)

∈

D

}

and

D

2

=

{

y

∣

(

1

,

y

)

∈

D

}

. Since

C

1

and

C

2

are chains,

a

≤

1

implies that there is an

s

∈

D

1

such that

a

≤

s

. Similarly, there is a

t

∈

D

2

such that

b

≤

t

. Together,

(

a

,

b

)

≤

(

s

,

t

)

∈

D

. So

(

a

,

b

)

≪

(

1

,

1

)

.

6.

Let

X

be a

topological space

and

L

⁢

(

X

)

be the lattice of

open sets

in

X

. Suppose

U

,

V

∈

L

⁢

(

X

)

and

U

≤

V

. If there is a

compact subset

C

such that

U

⊆

C

⊆

V

, then

U

≪

V

.

Remarks

.

•

In a lattice

L

,

a

≪

a

iff

a

is a

compact element

. This follows directly from the assertion above. In fact, a compact element can be defined in a general poset as an element that is way below itself.

•

If we remove the condition that

D

be directed in the definition above, then

a

is said to be

way way below

b

.

References

1

G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. W. Mislove, D. S. Scott,

Continuous

Lattices and Domains

, Cambridge University Press, Cambridge (2003).

Title

way below

Canonical name

WayBelow

Date of creation

2013-03-22 16:38:28

Last modified on

2013-03-22 16:38:28

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06B35

Classification

msc 06A99

Synonym

way way below

Synonym

way-below

Synonym

way-way-below

Defines

way below

relation

Defines

way way below relation

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