topological vector lattice

topological vector lattice

A

topological vector lattice

V

over

ℝ

is

•

a

Hausdorff

topological vector space

over

ℝ

,

•

a

vector lattice

, and

•

locally solid

. This means that there is a

neighborhood base

of

0

consisting of

solid sets

.

Proposition 1

.

A topological vector lattice

V

is a

topological lattice

.

Before proving this, we show the following equivalence on the continuity of various

operations

on a vector lattice

V

that is also a topological vector space.

Lemma 1

.

Let

V

be a vector lattice and a topological vector space. The following are

equivalent

:

1.

∨

:

V

2

→

V

is

continuous

(simultaneously in both

arguments

)

2.

∧

:

V

2

→

V

is continuous (simultaneously in both arguments)

3.

:

+

V

→

V

given by

x

+

:=

x

∨

0

is continuous

4.

:

-

V

→

V

given by

x

-

:=

-

x

∨

0

is continuous

5.

|

⋅

|

:

V

→

V

given by

|

x

|

:=

-

x

∨

x

is continuous

Proof.

(

1

⇔

2

)

. If

∨

is continuous, then

x

∧

y

=

x

+

y

-

x

∨

y

is continuous too, as

+

and

-

are both continuous under a topological vector space. This proof works in reverse too.

(

1

⇒

3

)

,

(

1

⇒

4

)

, and

(

3

⇔

4

)

are obvious. To see

(

4

⇒

5

)

, we see that

|

x

|

=

x

+

+

x

-

, since

-

is continuous,

+

is continuous also, so that

|

⋅

|

is continuous. To see

(

5

⇒

4

)

, we use the

identity

x

=

x

+

-

x

-

, so that

|

x

|

=

(

x

+

x

-

)

+

x

-

, which implies

x

-

=

1

2

⁢

(

|

x

|

-

x

)

is continuous. Finally,

(

3

⇒

1

)

is given by

x

∨

y

=

(

x

-

y

+

y

)

∨

(

0

+

y

)

=

(

x

-

y

)

∨

0

+

y

=

(

x

-

y

)

+

+

y

, which is continuous.
∎

In

addition

, we show an important

inequality

that is true on any vector lattice:

Lemma 2

.

Let

V

be a vector lattice. Then

|

a

+

-

b

+

|

≤

|

a

-

b

|

for any

a

,

b

∈

V

.

Proof.

|

a

+

-

b

+

|

=

(

b

+

-

a

+

)

∨

(

a

+

-

b

+

)

=

(

b

∨

0

-

a

∨

0

)

∨

(

a

∨

0

-

b

∨

0

)

. Next,

a

∨

0

-

b

∨

0

=

(

b

+

(

-

a

∧

0

)

∨

(

-

a

∧

0

)

=

(

(

b

-

a

)

∧

b

)

∨

(

-

a

∧

0

)

so that

|

a

+

-

b

+

|

=

(

(

b

-

a

)

∧

b

)

∨

(

-

a

∧

0

)

∨

(

(

a

-

b

)

∧

a

)

∨

(

-

b

∧

0

)

≤

(

b

-

a

)

∨

(

-

a

∧

0

)

∨

(

a

-

b

)

∨

(

-

b

∧

0

)

. Since

(

b

-

a

)

∨

(

a

-

b

)

=

|

a

-

b

|

and

a

∨

0

are both in the

positive cone

of

V

, so is their sum, so that

0

≤

(

b

-

a

)

∨

(

a

-

b

)

+

(

a

∨

0

)

=

(

b

-

a

)

∨

(

a

-

b

)

-

(

-

a

∧

0

)

, which means that

(

-

a

∧

0

)

≤

(

b

-

a

)

∨

(

a

-

b

)

. Similarly,

(

-

b

∧

0

)

≤

(

b

-

a

)

∨

(

a

-

b

)

. Combining these two inequalities, we see that

|

a

+

-

b

+

|

≤

(

b

-

a

)

∨

(

-

a

∧

0

)

∨

(

a

-

b

)

∨

(

-

b

∧

0

)

≤

(

b

-

a

)

∨

(

a

-

b

)

=

|

a

-

b

|

.
∎

We are now ready to prove the main assertion.

Proof.

To show that

V

is a topological lattice, we need to show that the

lattice

operations meet

∧

and join

∨

are continuous, which, by Lemma 1, is equivalent in showing, say, that

+

is continuous. Suppose

N

is a neighborhood base of 0 consisting of solid sets. We prove that

+

is continuous. This amounts to showing that if

x

is close to

x

0

, then

x

+

is close to

x

0

+

, which is the same as saying that if

x

-

x

0

is in a solid

neighborhood

U

of

0

(

U

∈

N

), then so is

x

+

-

x

0

+

in

U

. Since

x

-

x

0

∈

U

,

|

x

-

x

0

|

∈

U

. But

|

x

+

-

x

0

+

|

≤

|

x

-

x

0

|

by Lemma 2, and

U

is solid,

x

+

-

x

0

+

∈

U

as well, and therefore

+

is continuous.
∎

As a corollary, we have

Proposition 2

.

A topological vector lattice is an

ordered topological vector space

.

Proof.

All we need to show is that the positive cone is a

closed set

. But the positive cone is defined as

{

x

∣

0

≤

x

}

=

{

x

∣

x

-

=

0

}

, which is closed since

-

is continuous, and the positive cone is the

inverse image

of a singleton, a closed set in

ℝ

.
∎

Title

topological vector lattice

Canonical name

TopologicalVectorLattice

Date of creation

2013-03-22 17:03:51

Last modified on

2013-03-22 17:03:51

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Definition

Classification

msc 06F20

Classification

msc 46A40

Defines

locally solid

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