neighborhood system on a set

neighborhood system on a set

In point-set topology, a

neighborhood system

is defined as the set of

neighborhoods

of some point in the

topological space

.

However, one can start out with the

definition

of a “

abstract neighborhood system

”

𝔑

on an arbitrary set

X

and define a

topology

T

on

X

based on this system

𝔑

so that

𝔑

is the neighborhood system of

T

. This is done as follows:

Let

X

be a set and

𝔑

be a subset of

X

×

P

⁢

(

X

)

, where

P

⁢

(

X

)

is the

power set

of

X

. Then

𝔑

is said to be a

abstract neighborhood system

of

X

if the following conditions are satisfied:

1.

if

(

x

,

U

)

∈

𝔑

, then

x

∈

U

,

2.

for every

x

∈

X

, there is a

U

⊆

X

such that

(

x

,

U

)

∈

𝔑

,

3.

if

(

x

,

U

)

∈

𝔑

and

U

⊆

V

⊆

X

, then

(

x

,

V

)

∈

𝔑

,

4.

if

(

x

,

U

)

,

(

x

,

V

)

∈

𝔑

, then

(

x

,

U

∩

V

)

∈

𝔑

,

5.

if

(

x

,

U

)

∈

𝔑

, then there is a

V

⊆

X

such that

–

(

x

,

V

)

∈

𝔑

, and

–

(

y

,

U

)

∈

𝔑

for all

y

∈

V

.

In

addition

, given this

𝔑

, define the

abstract neighborhood system around

x

∈

X

to be the subset

𝔑

x

of

𝔑

consisting of all those

elements

whose first coordinate is

x

. Evidently,

𝔑

is the

disjoint union

of

𝔑

x

for all

x

∈

X

. Finally, let

T

=

{

U

⊆

X

∣

for every

x

∈

U

,

(

x

,

U

)

∈

𝔑

}

=

{

U

⊆

X

∣

for every

x

∈

U

, there is a

V

⊆

U

, such that

(

x

,

V

)

∈

𝔑

}

.

The two definitions are the same by condition 3. We assert that

T

defined above is a topology on

X

. Furthermore,

T

x

:=

{

U

∣

(

x

,

U

)

∈

𝔑

x

}

is the set of neighborhoods of

x

under

T

.

Proof.

We first show that

T

is a topology. For every

x

∈

X

, some

U

⊆

X

, we have

(

x

,

U

)

∈

𝔑

by condition 2. Hence

(

x

,

X

)

∈

𝔑

by condition 3. So

X

∈

T

. Also,

∅

∈

T

is

vacuously

satisfied, for no

x

∈

∅

. If

U

,

V

∈

T

, then

U

∩

V

∈

T

by condition 4. Let

{

U

i

}

be a subset of

T

whose elements are

indexed by

I

(

i

∈

I

). Let

U

=

⋃

U

i

. Pick any

x

∈

U

, then

x

∈

U

i

for some

i

∈

I

. Since

U

i

∈

T

,

(

x

,

U

i

)

∈

𝔑

. Since

U

i

⊆

U

,

(

x

,

U

)

∈

𝔑

by condition 3, so

U

∈

T

.

Next, suppose

𝒩

is the set of neighborhoods of

x

under

T

. We need to show

𝒩

=

T

x

:

1.

(

𝒩

⊆

T

x

). If

N

∈

𝒩

, then there is

U

∈

T

with

x

∈

U

⊆

N

. But

(

x

,

U

)

∈

𝔑

, so by condition 3,

(

x

,

N

)

∈

𝔑

, or

(

x

,

N

)

∈

𝔑

x

, or

N

∈

T

x

.

2.

(

T

x

⊆

𝒩

). Pick any

U

∈

T

x

and set

W

=

{

z

∣

U

∈

T

z

}

. Then

x

∈

W

⊆

U

by condition 1. We show

W

is open. This means we need to find, for each

z

∈

W

, a

V

⊆

W

such that

(

z

,

V

)

∈

𝔑

. If

z

∈

W

, then

(

z

,

U

)

∈

𝔑

. By condition 5, there is

V

∈

𝔑

such that

(

z

,

V

)

∈

𝔑

, and for any

y

∈

V

,

(

y

,

U

)

∈

𝔑

, or

y

∈

U

by condition 1. So

y

∈

W

by the definition of

W

, or

V

⊆

W

. Thus

W

is open and

U

∈

𝒩

.

This

completes

the proof. By the way,

W

defined above is none other than the

interior

of

U

:

W

=

U

∘

.
∎

Remark

.

Conversely

, if

T

is a topology on

X

, we can define

𝔑

x

to be the set consisting of

(

x

,

U

)

such that

U

is a neighborhood of

x

. The the union

𝔑

of

𝔑

x

for each

x

∈

X

satisfies

conditions

1

through

5

above:

1.

(condition 1): clear

2.

(condition 2): because

(

x

,

X

)

∈

𝔑

for each

x

∈

X

3.

(condition 3): if

U

is a neighborhood of

x

and

V

a supserset of

U

, then

V

is also a neighborhood of

x

4.

(condition 4): if

U

and

V

are neighborhoods of

x

, there are open

A

,

B

with

x

∈

A

⊆

U

and

x

∈

B

⊆

V

, so

x

∈

A

∩

B

⊆

U

∩

V

, which means

U

∩

V

is a neighborhood of

x

5.

(condition 5): if

U

is a neighborhood of

x

, there is open

A

with

x

∈

A

⊆

U

; clearly

A

is a neighborhood of

x

and any

y

∈

A

has

U

as neighborhood.

So the definition of a neighborhood system on an arbitrary set gives an alternative way of defining a topology on the set. There is a

one-to-one correspondence

between the set of topologies on a set and the set of abstract neighborhood systems on the set.

Title

neighborhood system on a set

Canonical

name

NeighborhoodSystemOnASet

Date of creation

2013-03-22 16:41:34

Last modified on

2013-03-22 16:41:34

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

13

Author

CWoo (3771)

Entry type

Definition

Classification

msc 54-00

Defines

abstract neighborhood system

Generated on Sat Feb 10 11:11:43 2018 by

LaTeXML