uniform neighborhood

uniform neighborhood

Let

X

be a

uniform space

with uniformity

𝒰

. For each

x

∈

X

and

U

∈

𝒰

, define the following items

•

U

⁢

[

x

]

:=

{

y

∣

(

x

,

y

)

∈

U

}

, and

•

𝔑

x

:=

{

(

x

,

U

⁢

[

x

]

)

∣

U

∈

𝒰

}

•

𝔑

=

⋃

x

∈

X

𝔑

x

.

Proposition

.

𝔑

x

is the

abstract neighborhood system

around

x

, hence

𝔑

is the abstract neighborhood system of

X

.

Proof.

We show that all five defining conditions of a

neighborhood system

on a set are met:

1.

For each

(

x

,

U

⁢

[

x

]

)

∈

𝔑

,

x

∈

U

⁢

[

x

]

, since every entourage contains the

diagonal relation

.

2.

Every

x

∈

X

and every entourage

U

∈

𝒰

,

U

⁢

[

x

]

⊆

X

with

(

x

,

U

⁢

[

x

]

)

∈

𝔑

3.

Suppose

(

x

,

U

⁢

[

x

]

)

∈

𝔑

and

U

⁢

[

x

]

⊆

Y

⊆

X

. Showing that

(

x

,

Y

)

∈

𝔑

amounts to showing

Y

=

V

⁢

[

x

]

for some

V

∈

𝒰

. First, note that each entourage

U

can be decomposed into

disjoint union

of sets “slices” of the form

{

a

}

×

U

⁢

[

a

]

. We replace the “slice”

{

x

}

×

U

⁢

[

x

]

by

{

x

}

×

Y

. The resulting disjoint union is a set

V

, which is a

superset

of

U

. Since

𝒰

is a filter,

V

∈

𝒰

. Furthermore,

V

⁢

[

x

]

=

Y

.

4.

a

∈

U

⁢

[

x

]

∩

V

⁢

[

x

]

iff

(

x

,

a

)

∈

U

∩

V

iff

a

∈

(

U

∩

V

)

⁢

[

x

]

. This implies that if

(

x

,

U

⁢

[

x

]

)

,

(

x

,

V

⁢

[

x

]

)

∈

𝔑

, then

(

x

,

U

⁢

[

x

]

∩

V

⁢

[

x

]

)

=

(

x

,

(

U

∩

V

)

⁢

[

x

]

)

∈

𝔑

.

5.

Suppose

(

x

,

U

⁢

[

x

]

)

∈

𝔑

. There is

V

∈

𝒰

such that

(

V

∘

V

)

⁢

[

x

]

⊆

U

⁢

[

x

]

. We show that

V

⁢

[

x

]

⊆

X

is what we want. Clearly,

x

∈

V

⁢

[

x

]

. For any

y

∈

V

⁢

[

x

]

, and any

a

∈

V

⁢

[

y

]

, we have

(

x

,

a

)

=

(

x

,

y

)

∘

(

y

,

a

)

∈

V

∘

V

, or

a

∈

(

V

∘

V

)

⁢

[

x

]

⊆

U

⁢

[

x

]

. So

V

⁢

[

y

]

⊆

U

⁢

[

x

]

for any

y

∈

V

⁢

[

x

]

. In order to show that

(

y

,

U

⁢

[

x

]

)

∈

𝔑

, we must find

W

∈

𝒰

such that

U

⁢

[

x

]

=

W

⁢

[

y

]

. By the third step above, since

V

⁢

[

y

]

⊆

U

⁢

[

x

]

, there is

W

∈

𝒰

with

W

⁢

[

y

]

=

U

⁢

[

x

]

. Thus

(

y

,

U

⁢

[

x

]

)

=

(

y

,

W

⁢

[

y

]

)

∈

𝔑

.

∎

Definition

. For each

x

in a uniform space

X

with uniformity

𝒰

, a

uniform neighborhood

of

x

is a set

U

⁢

[

x

]

for some entourage

U

∈

𝒰

. In general, for any

A

⊆

X

, the set

U

⁢

[

A

]

:=

{

y

∈

X

∣

(

x

,

y

)

∈

U

⁢

for some

⁢

x

∈

A

}

is called a

uniform neighborhood

of

A

.

Two immediate properties that we have already seen in the proof above are: (1). for each

U

∈

𝒰

,

x

∈

U

⁢

[

x

]

; and (2).

U

⁢

[

x

]

∩

V

⁢

[

x

]

=

(

U

∩

V

)

⁢

[

x

]

. More generally,

⋂

U

i

⁢

[

x

]

=

(

⋂

U

i

)

⁢

[

x

]

.

Remark

. If we define

T

𝒰

:=

{

A

⊆

X

∣

∀

x

∈

A

,

∃

U

∈

𝒰

⁢

such that

⁢

U

⁢

[

x

]

⊆

A

}

, then

T

𝒰

is a

topology

induced

by the uniform structure (

http://planetmath.org/TopologyInducedByAUniformStructure

)

𝒰

. Under this topology, uniform neighborhoods are synonymous with

neighborhoods

.

Title

uniform neighborhood

Canonical name

UniformNeighborhood

Date of creation

2013-03-22 16:42:31

Last modified on

2013-03-22 16:42:31

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Definition

Classification

msc 54E15

Related topic

TopologyInducedByAUniformStructure

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