Hausdorff space not completely Hausdorff

Hausdorff space not completely Hausdorff

On the set

ℤ

+

of

strictly

positive integers

, let

a

and

b

be two different

integers

b

≠

0

and consider the set

S

⁢

(

a

,

b

)

=

{

a

+

k

⁢

b

∈

ℤ

+

:

k

∈

ℤ

}

such set is the

infinite

arithmetic progression

of positive integers with

difference

b

and containing

a

.
The

collection

of all

S

⁢

(

a

,

b

)

sets is a basis for a

topology

on

ℤ

+

. We will use a

coarser

topology

induced

by the following basis:

𝔹

=

{

S

⁢

(

a

,

b

)

:

gcd

⁡

(

a

,

b

)

=

1

}

The collection

𝔹

is basis for a topology on

ℤ

+

We first prove such collection is a basis.
Suppose

x

∈

S

⁢

(

a

,

b

)

∩

S

⁢

(

c

,

d

)

. By Euclid’s algorithm we have

S

⁢

(

a

,

b

)

=

S

⁢

(

x

,

b

)

and

S

⁢

(

c

,

d

)

=

S

⁢

(

x

,

d

)

and

x

∈

S

⁢

(

x

,

b

⁢

d

)

⊂

S

⁢

(

x

,

d

)

∩

S

⁢

(

c

,

d

)

besides, since

gcd

⁡

(

x

,

b

)

=

1

and

gcd

⁡

(

x

,

d

)

=

1

then

gcd

⁡

(

x

,

b

⁢

d

)

=

1

so

x

and

b

⁢

d

are

coprimes

and

S

⁢

(

x

,

b

⁢

d

)

∈

𝔹

. This concludes the proof that

𝔹

is indeed a basis for a topology on

ℤ

+

.

The topology on

ℤ

+

induced by

𝔹

is Hausdorff

Let

m

,

n

integers two different integers.
We need to show that there are open

disjoint

neighborhoods

U

m

and

U

n

such that

m

∈

U

m

and

n

∈

U

n

, but it suffices to show the

existence

of disjoint basic

open sets

containing

m

and

n

.

Taking

d

=

|

m

-

n

|

, we can find an integer

t

such that

t

>

d

and such that

gcd

⁡

(

m

,

t

)

=

gcd

⁡

(

n

,

t

)

=

1

. A way to accomplish this is to take any

multiple

of

m

⁢

n

greater than

d

and add

1

.

The basic open sets

S

⁢

(

m

,

t

)

and

S

⁢

(

n

,

t

)

are disjoint, because they have common

elements

if and only if the

diophantine equation

m

+

t

⁢

x

=

n

+

t

⁢

y

has solutions. But it cannot have since

t

⁢

(

x

-

y

)

=

n

-

m

implies that

t

divides

n

-

m

but

t

>

|

n

-

m

|

makes it impossible.

We conclude that

S

⁢

(

m

,

t

)

∩

S

⁢

(

n

,

t

)

=

∅

and this means that

ℤ

+

becomes a

Hausdorff space

with the given topology.

Some properties of

S

⁢

(

a

,

b

)

¯

We need to determine first some facts about

S

⁢

(

a

,

b

)

¯

. in order to take an example, consider

S

⁢

(

3

,

5

)

first. Notice that if we had considered the former topology (where in

S

⁢

(

a

,

b

)

,

a

and

b

didn’t have to be coprime) the

complement

of

S

⁢

(

3

,

5

)

would have been

S

⁢

(

4

,

5

)

∪

S

⁢

(

5

,

5

)

∪

S

⁢

(

6

,

5

)

∪

S

⁢

(

7

,

5

)

which is open, and so

S

⁢

(

3

,

5

)

would have been closed. In general, in the finer topology, all basic sets were both open and closed. However, this is not true in our coarser topology (for

instance

S

⁢

(

5

,

5

)

is not open).

The key fact to prove

ℤ

+

is not a

completely Hausdorff space

is: given any

S

⁢

(

a

,

b

)

, then

b

⁢

ℤ

+

=

{

n

∈

ℤ

+

:

b

⁢

divides

⁢

n

}

is a subset of

S

⁢

(

a

,

b

)

¯

.

Indeed, any basic open set containing

b

⁢

k

is of the form

S

⁢

(

b

⁢

k

,

t

)

with

t

,

b

⁢

k

coprimes. This means

gcd

⁡

(

t

,

b

)

=

1

. Now

S

⁢

(

b

⁢

k

,

t

)

and

S

⁢

(

a

,

b

)

have common terms if an only if

b

⁢

k

+

t

⁢

x

=

a

+

b

⁢

y

for some integers

x

,

y

. But that diophantine equation can be rewritten as

t

⁢

x

-

b

⁢

y

=

a

-

b

⁢

k

and it always has solutions because

1

=

gcd

⁡

(

t

,

b

)

divides

a

-

b

⁢

k

.

This also proves

S

⁢

(

a

,

b

)

≠

S

⁢

(

a

,

b

)

¯

, because

b

is not in

S

⁢

(

a

,

b

)

but it is on the

closure

.

The topology on

ℤ

+

induced by

𝔹

is not completely Hausdorff

We will use the closed-neighborhood sense for

completely Hausdorff

, which will also imply the topology is not completely Hausdorff in the

functional

sense.

Let

m

,

n

different positive integers. Since

𝔹

is a basis, for any two disjoint neighborhoods

U

m

,

U

n

we can find basic sets

S

⁢

(

m

,

a

)

and

S

⁢

(

n

,

b

)

such that

m

∈

S

⁢

(

m

,

a

)

⊆

U

m

,

n

∈

S

⁢

(

n

,

b

)

⊆

U

n

and thus

S

⁢

(

m

,

a

)

∩

S

⁢

(

n

,

b

)

=

∅

.

But then

g

=

a

⁢

b

is both a multiple of

a

and

b

so it must be in

S

⁢

(

m

,

a

)

¯

and

S

⁢

(

n

,

b

)

¯

. This means

S

⁢

(

m

,

a

)

¯

∩

S

⁢

(

n

,

b

)

¯

≠

∅

and thus

U

m

¯

∩

U

n

¯

≠

∅

.

This proves the topology under consideration is not completely
Hausdorff (under both usual meanings).

Title

Hausdorff space not completely Hausdorff

Canonical

name

HausdorffSpaceNotCompletelyHausdorff

Date of creation

2013-03-22 14:16:05

Last modified on

2013-03-22 14:16:05

Owner

drini (3)

Last modified by

drini (3)

Numerical id

21

Author

drini (3)

Entry type

Example

Classification

msc 54D10

Synonym

T

2

space not

T

2

⁤

1

2

Synonym

example of a Hausdorff space that is not completely Hausdorff

Related topic

CompletelyHausdorff

Related topic

SeparationAxioms

Related topic

FrechetSpace

Related topic

RegularSpace

Related topic

FurstenbergsProofOfTheInfinitudeOfPrimes

Related topic

SeparationAxioms

Related topic

T2Space

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