aarjaneiro/planetmath huggingface dataset:
proof that products of connected spaces are connected

proof that products of connected spaces are connected

Let

{

X

α

⁢

for

⁢

α

∈

A

}

be

topological spaces

, and let

X

=

∏

X

α

be the

product

, with

projection maps

π

α

.

Using the

Axiom of Choice

, one can straightforwardly show that each

π

α

is

surjective

; they are

continuous

by

definition

, and the continuous image of a

connected space

is connected, so if

X

is connected, then all

X

α

are.

Let

{

X

α

⁢

for

⁢

α

∈

A

}

be connected topological spaces, and let

X

=

∏

X

α

be the product, with projection maps

π

α

.

First note that each

π

α

is an open map: If

U

is open, then it is the union of

open sets

of the form

⋂

β

∈

F

π

β

-

1

⁢

U

β

where

F

is a finite subset of

A

and

U

β

is an open set in

X

β

. But

π

α

⁢

(

U

β

)

is always open, and the image of a union is the union of the images.

Suppose the product is the

disjoint union

of open sets

U

and

V

, and suppose

U

and

V

are nonempty. Then there is an

α

∈

A

and an

element

u

∈

U

and an element

v

∈

V

that differ only in the

α

place. To see this, observe that for all but finitely many places

γ

, both

π

β

⁢

(

U

)

and

π

β

⁢

(

V

)

must be

X

γ

, so there are elements

u

and

v

that differ in finitely many places. But then since

U

and

V

are supposed to cover

X

, if

π

β

⁢

(

u

)

≠

π

β

⁢

(

v

)

, changing

u

in the

β

place lands us in either

U

or

V

. If it lands us in

V

, we have elements that differ in only one place. Otherwise, we can make a

u

′

∈

U

such that

π

β

⁢

(

u

′

)

=

π

β

⁢

(

v

)

and which otherwise agrees with

u

. Then by

induction

we can obtain elements

u

∈

U

and

v

∈

V

that differ in only one place. Call that place

α

.

We then have a map

ρ

:

X

α

→

X

such that

π

α

∘

ρ

is the

identity map

on

X

α

, and

(

ρ

∘

π

)

⁢

(

u

)

=

u

. Observe that since

π

α

is open,

ρ

is continuous. But

ρ

-

1

⁢

(

U

)

and

ρ

-

1

⁢

(

V

)

are

disjoint

nonempty open sets that cover

X

α

, which is impossible.

Note that if we do not assume the Axiom of Choice, the product may be empty, and hence connected, whether or not the

X

α

are connected; by taking the

discrete topology

on some

X

α

we get a

counterexample

to one direction of the

theorem

: we have a connected (empty!) space that is the product of non-connected spaces.
For the other direction, if the product is empty, it is connected; if it is not empty, then the

argument

below works unchanged.
So without the Axiom of Choice, this theorem becomes “If all

X

α

are connected, then

X

is.”

Title

proof that products of connected spaces are connected

Canonical

name

ProofThatProductsOfConnectedSpacesAreConnected

Date of creation

2013-03-22 14:10:05

Last modified on

2013-03-22 14:10:05

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

13

Author

yark (2760)

Entry type

Proof

Classification

msc 54D05

Generated on Sat Feb 10 11:21:00 2018 by

LaTeXML
quasicomponent

quasicomponent

Let

X

be a

topological space

. Define a

relation

∼

on

X

as follows:

x

∼

y

if there is no

partition

of

X

into

disjoint

open sets

U

and

V

such that

U

∪

V

=

X

,

x

∈

U

and

y

∈

V

.

This is an

equivalence relation

on

X

. The

equivalence classes

are called the

quasicomponents

of

X

.

Title

quasicomponent

Canonical name

Quasicomponent

Date of creation

2013-03-22 12:23:49

Last modified on

2013-03-22 12:23:49

Owner

Evandar (27)

Last modified by

Evandar (27)

Numerical id

4

Author

Evandar (27)

Entry type

Definition

Classification

msc 54D05

Synonym

quasi-component

Related topic

ConnectedSpace

Related topic

PathConnected

Related topic

ConnectedComponent

Related topic

PathComponent

Generated on Sat Feb 10 11:21:02 2018 by

LaTeXML
semilocally simply connected

semilocally simply connected

A

topological space

X

is

semilocally simply connected

if, for every point

x

∈

X

, there exists a

neighborhood

U

of

x

such that the map of

fundamental groups

π

1

⁢

(

U

,

x

)

⟶

π

1

⁢

(

X

,

x

)

induced by the

inclusion map

U

↪

X

is the

trivial

homomorphism

.

A topological space

X

is

connected

,

locally path connected

, and semilocally simply connected if and only if it has a

universal cover

.

Title

semilocally simply connected

Canonical

name

SemilocallySimplyConnected

Date of creation

2013-03-22 12:38:46

Last modified on

2013-03-22 12:38:46

Owner

djao (24)

Last modified by

djao (24)

Numerical id

6

Author

djao (24)

Entry type

Definition

Classification

msc 54D05

Classification

msc 57M10

Synonym

semilocally 1-connected

Synonym

locally relatively simply connected

Related topic

Connected2

Related topic

SimplyConnected

Related topic

ConnectedSpace

Related topic

LocallyConnected

Generated on Sat Feb 10 11:21:03 2018 by

LaTeXML
ultraconnected space

ultraconnected space

A

topological space

X

is said to be

ultraconnected

if no pair of nonempty

closed sets

of

X

is disjoint.

All ultraconnected spaces are

path-connected

, normal,

limit point compact

, and

pseudocompact

.

Title

ultraconnected space

Canonical name

UltraconnectedSpace

Date of creation

2013-03-22 14:20:33

Last modified on

2013-03-22 14:20:33

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

6

Author

yark (2760)

Entry type

Definition

Classification

msc 54D05

Synonym

ultra-connected space

Related topic

Hyperconnected

Defines

ultraconnected

Defines

ultra-connected

Generated on Sat Feb 10 11:21:05 2018 by

LaTeXML
union of non-disjoint connected sets is connected

union of non-disjoint connected sets is connected

Theorem 1

.

Suppose

A

,

B

are connected sets in a

topological
space

X

. If

A

,

B

are not

disjoint

, then

A

∪

B

is

connected

.

Proof.

By

assumption

, we have two

implications

.
First, if

U

,

V

are open in

A

and

U

∪

V

=

A

, then

U

∩

V

≠

∅

.
Second, if

U

,

V

are open in

B

and

U

∪

V

=

B

, then

U

∩

V

≠

∅

.
To prove that

A

∪

B

is connected, suppose

U

,

V

are open in

A

∪

B

and

U

∪

V

=

A

∪

B

.
Then

U

∪

V

=

(

(

U

∪

V

)

∩

A

)

∪

(

(

U

∪

V

)

∩

B

)

=

(

U

∩

A

)

∪

(

V

∩

A

)

∪

(

U

∩

B

)

∪

(

V

∩

B

)

Let us show that

U

∩

A

and

V

∩

A

are open in

A

.
To do this, we use this result (

http://planetmath.org/SubspaceOfASubspace

)
and notation from that entry too.
For example, as

U

∈

τ

A

∪

B

,

X

,

U

∩

A

∈

τ

A

,

A

∪

B

,

X

=

τ

A

,

X

,
and so

U

∩

A

,

V

∩

A

are open in

A

.
Since

(

U

∩

A

)

∪

(

V

∩

A

)

=

A

, it follows that

∅

≠

(

U

∩

A

)

∩

(

V

∩

A

)

=

(

U

∩

V

)

∩

A

.

If

U

∩

V

=

∅

, then this is a contradition, so

A

∪

B

must be connected.
∎

Title

union of non-disjoint connected sets is connected

Canonical name

UnionOfNondisjointConnectedSetsIsConnected

Date of creation

2013-03-22 15:17:50

Last modified on

2013-03-22 15:17:50

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

8

Author

matte (1858)

Entry type

Theorem

Classification

msc 54D05

Generated on Sat Feb 10 11:21:07 2018 by

LaTeXML
ℝ^2∖C is path connected if C is countable

ℝ

2

∖

C

is path connected if

C

is countable

Theorem 1

.

Let

C

be a

countable

subset of

R

2

. Then

R

2

∖

C

is

path connected

.

We use

ℝ

2

simply as an example; an analogous proof will work for any

ℝ

n

,

n

>

1

.

Proof.

Fix a point

P

not in

C

. The strategy of the proof is to construct a path

p

x

from any

x

∈

ℝ

2

∖

C

to

P

. If we can do this then for any

d

,

d

′

∈

ℝ

2

∖

C

we may construct a path from

d

to

d

′

by first following

p

d

and then following

p

d

′

in reverse.

Fix

x

∈

ℝ

2

∖

C

, and consider the set of all (straight) lines through

x

. There are uncountably many of these and they meet in the single point

x

, so not all of them contain a point of

C

. Choose one that doesn’t and move along it: your

distance

from

P

takes on uncountably many values, and hence at some point this distance

r

from

P

is not shared by any point of

C

. The whole of the circle with radius

r

, centre

P

, lies in

ℝ

2

∖

C

so we may move around it freely.

Consider all lines through

P

: these all

intersect

this circle, and there are uncountably many of them so we may choose one, say

L

, that contains no point of

C

. Moving around the circle until we meet

L

and then following it inwards

completes

our path form

x

to

P

.
∎

Corollary 1

.

Let

f

:

R

2

→

R

be

continuous

and onto. Then

f

-

1

⁢

(

0

)

is

uncountable

.

Proof.

Suppose that

f

-

1

⁢

(

0

)

is countable.

ℝ

2

can be written as the

disjoint union

f

-

1

⁢

(

0

)

∪

f

-

1

⁢

(

(

-

∞

,

0

)

)

∪

f

-

1

⁢

(

(

0

,

∞

)

)

where the last two sets are open (as

f

is continuous), non-empty (as

f

is onto) and

disjoint

. Since pathwise connected is the same as

connected

for

Hausdorff spaces

, we have that

ℝ

2

∖

f

-

1

⁢

(

0

)

is not path connected, contradicting the

theorem

.
∎

Title

ℝ

2

∖

C

is path connected if

C

is countable

Canonical name

mathbbR2setminusCIsPathConnectedIfCIsCountable

Date of creation

2013-03-22 16:09:11

Last modified on

2013-03-22 16:09:11

Owner

silverfish (6603)

Last modified by

silverfish (6603)

Numerical id

5

Author

silverfish (6603)

Entry type

Theorem

Classification

msc 54D05

Generated on Sat Feb 10 11:21:09 2018 by

LaTeXML
a compact set in a Hausdorff space is closed

a compact set in a Hausdorff space is closed

Theorem

.

A

compact set

in a

Hausdorff space

is closed.

Proof.

Let

A

be a compact set in a Hausdorff space

X

.
The case when

A

is empty is

trivial

, so let us
assume that

A

is non-empty.
Using this theorem (

http://planetmath.org/APointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods

),
it follows that each point

y

in

A

∁

has a

neighborhood

U

y

, which
is

disjoint

to

A

. (Here, we denote the

complement

of

A

by

A

∁

.)
We can therefore write

A

∁

=

⋃

y

∈

A

∁

U

y

.

Since an arbitrary union of

open sets

is open, it follows that

A

is
closed.

□

Note.

ÃÂ

The above theorem can, for

instance

, be found in

[

1

]

(page 141),
or

[

2

]

(

Section

2.1, Theorem 2).

References

1

J.L. Kelley,

General Topology

,
D. van Nostrand Company, Inc., 1955.

2

I.M. Singer, J.A.Thorpe,

Lecture Notes on

Elementary

Topology and Geometry

,
Springer-Verlag, 1967.

Title

a compact set in a Hausdorff space is closed

Canonical name

ACompactSetInAHausdorffSpaceIsClosed

Date of creation

2013-03-22 13:34:31

Last modified on

2013-03-22 13:34:31

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 54D10

Classification

msc 54D30

Related topic

ClosedSubsetsOfACompactSetAreCompact

Generated on Sat Feb 10 11:21:11 2018 by

LaTeXML
a space is T_1 if and only if distinct points are separated

a space is

T

1

if and only if distinct points are separated

Theorem 1

.

Let

X

be a

topological space

. Then

X

is a

T

1

-space
if and only if sets

{

x

}

,

{

y

}

are

separated

for all distinct

x

,

y

∈

X

.

Proof.

Suppose

X

is a

T

1

-space. Then every singleton is
closed and if

x

,

y

∈

X

are distinct, then

{

x

}

∩

{

y

}

¯

=

{

x

}

∩

{

y

}

=

∅

,

{

x

}

¯

∩

{

y

}

=

{

x

}

∩

{

y

}

=

∅

,

and

{

x

}

,

{

y

}

are separated.
On the other hand, suppose that

{

x

}

∩

{

y

}

¯

=

∅

for all

x

≠

y

. It follows that

{

y

}

¯

=

{

y

}

, so

{

y

}

is closed and

X

is a

T

1

-space.
∎

Title

a space is

T

1

if and only if distinct points are separated

Canonical name

ASpaceIsT1IfAndOnlyIfDistinctPointsAreSeparated

Date of creation

2013-03-22 15:16:49

Last modified on

2013-03-22 15:16:49

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

6

Author

matte (1858)

Entry type

Theorem

Classification

msc 54D10

Generated on Sat Feb 10 11:21:13 2018 by

LaTeXML
a space is T1 if and only if every singleton is closed

a space is T1 if and only if every singleton is closed

Say

X

is a

http://planetmath.org/node/1852

T

1

topological space

. Let’s show that

{

x

}

is closed for every

x

∈

X

:

The

T

1

axiom (

http://planetmath.org/T1Space

) gives us, for every

y

distinct from

x

, an open

U

y

that contains

y

but not

x

. Since we’re in a topological space, we can take the union of all these

open sets

to get a new open set,

U

=

⋃

y

≠

x

U

y

.

{

x

}

is the

complement

of

U

, closed because

U

is open: None of the

U

y

contain

x

, so

U

doesn’t contain

x

. But any

y

≠

x

is in

U

, since

y

∈

U

y

⊂

U

. That takes care of that.

Now let’s say we have a topological space

X

in which

{

x

}

is closed for every

x

∈

X

. We’d like to show that

T

1

holds:

Given

x

≠

y

, we want to find an open set that contains

x

but not

y

.

{

y

}

is closed by

hypothesis

, so its complement is open, and our search is over.

Title

a

space is T1 if and only if every singleton is closed

Canonical name

ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed

Date of creation

2013-03-22 14:20:15

Last modified on

2013-03-22 14:20:15

Owner

waj (4416)

Last modified by

waj (4416)

Numerical id

7

Author

waj (4416)

Entry type

Proof

Classification

msc 54D10

Related topic

ASpaceIsT1IfAndOnlyIfEverySubsetAIsTheIntersectionOfAllOpenSetsContainingA

Generated on Sat Feb 10 11:21:15 2018 by

LaTeXML
a space is T1 if and only if every subset A is the intersection of all open sets containing A

a space is T1 if and only if every subset A is the intersection of all open sets containing A

Say we have

X

, a

T

1

-space, and

A

, a subset of

X

. We aim to show that the

intersection

of all

open sets

containing

A

equals

A

. By de Morgan’s laws, that would be true if the

complement

of

A

,

A

c

, equalled the union of all

closed sets

in

A

c

. Let’s call this union of closed sets

C

.

Each set that makes up

C

is

contained

by

A

c

, so

C

⊂

A

c

. If we could show

A

c

⊂

C

, we’d be done.

Since

X

is

T

1

, each

singleton

in

A

c

is closed (

http://planetmath.org/ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed

). Their union, a subset of

C

, contains

A

c

, so we’re through.

Now suppose we know that in some

topological space

X

, any subset

A

of

X

is the intersection of all open sets containing

A

. Given

x

≠

y

, we’re looking for an open set containing

x

but not

y

, to show that

X

is

T

1

.

{

x

}

=

⋂

U

⁢

open

U

∋

x

U

by

hypothesis

. If all open sets containing

x

contained

y

,

y

would be in the intersection; since

y

isn’t in the intersection,

X

must be

T

1

.

Title

a space is T1 if and only if every subset A is the intersection of all open sets containing A

Canonical

name

ASpaceIsT1IfAndOnlyIfEverySubsetAIsTheIntersectionOfAllOpenSetsContainingA

Date of creation

2013-03-22 14:20:18

Last modified on

2013-03-22 14:20:18

Owner

waj (4416)

Last modified by

waj (4416)

Numerical id

4

Author

waj (4416)

Entry type

Proof

Classification

msc 54D10

Related topic

T1Space

Related topic

ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed

Generated on Sat Feb 10 11:21:16 2018 by

LaTeXML
a space X is Hausdorff if and only if Δ⁢(X) is closed

a space

X

is Hausdorff if and only if

Δ

⁢

(

X

)

is closed

Theorem

.

A space

X

is

Hausdorff

if and only if

{

(

x

,

x

)

∈

X

×

X

∣

x

∈

X

}

is closed in

X

×

X

under the

product topology

.

Proof.

First, some preliminaries: Recall that the

diagonal map

Δ

:

X

→

X

×

X

is defined as

x

⟼

Δ

(

x

,

x

)

. Also recall that in a

topology

generated by a basis (like the product topology), a set

Y

is open if and only if, for every point

y

∈

Y

, there’s a basis element

B

with

y

∈

B

⊂

Y

. Basis elements for

X

×

X

have the form

U

×

V

where

U

,

V

are

open sets

in

X

.

Now, suppose that

X

is Hausdorff. We’d like to show its image under

Δ

is closed. We can do that by showing that its

complement

Δ

⁢

(

X

)

c

is open.

Δ

⁢

(

X

)

consists of points with equal coordinates, so

Δ

⁢

(

X

)

c

consists of points

(

x

,

y

)

with

x

and

y

distinct.

For any

(

x

,

y

)

∈

Δ

⁢

(

X

)

c

, the Hausdorff condition gives us

disjoint

open

U

,

V

⊂

X

with

x

∈

U

,

y

∈

V

. Then

U

×

V

is a basis element containing

(

x

,

y

)

.

U

and

V

have no points in common, so

U

×

V

contains nothing in the image of the diagonal map:

U

×

V

is

contained

in

Δ

⁢

(

X

)

c

. So

Δ

⁢

(

X

)

c

is open, making

Δ

⁢

(

X

)

closed.

Now let’s suppose

Δ

⁢

(

X

)

is closed. Then

Δ

⁢

(

X

)

c

is open. Given any

(

x

,

y

)

∈

Δ

⁢

(

X

)

c

, there’s a basis element

U

×

V

with

(

x

,

y

)

∈

U

×

V

⊂

Δ

⁢

(

X

)

c

.

U

×

V

lying in

Δ

⁢

(

X

)

c

implies that

U

and

V

are disjoint.

If we have

x

≠

y

in

X

, then

(

x

,

y

)

is in

Δ

⁢

(

X

)

c

. The basis element containing

(

x

,

y

)

gives us open, disjoint

U

,

V

with

x

∈

U

,

y

∈

V

.

X

is Hausdorff, just like we wanted.
∎

Title

a space

X

is Hausdorff if and only if

Δ

⁢

(

X

)

is closed

Canonical name

ASpacemathnormalXIsHausdorffIfAndOnlyIfDeltaXIsClosed

Date of creation

2013-03-22 14:20:47

Last modified on

2013-03-22 14:20:47

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

9

Author

mathcam (2727)

Entry type

Proof

Classification

msc 54D10

Related topic

DiagonalEmbedding

Related topic

T2Space

Related topic

ProductTopology

Related topic

SeparatedScheme

Generated on Sat Feb 10 11:21:19 2018 by

LaTeXML
characterization of T⁢2 spaces

characterization of

T

⁢

2

spaces

Proposition 1

.

[

1

,

2

]

Suppose

X

is a

topological space

. Then

X

is a

T

2

space (

http://planetmath.org/T2Space

) if and only if
for all

x

∈

X

, we have

{

x

}

=

⋂

{

A

∣

A

⊆

X

⁢

𝑐𝑙𝑜𝑠𝑒𝑑

,

∃

open set

⁢

U

⁢

such that

⁢

x

∈

U

⊆

A

}

.

(1)

Proof.

By manipulating the

definition

using de Morgan’s laws, the claim
can be rewritten as

{

x

}

∁

=

⋃

{

V

∣

V

⊆

X

⁢

open

,

∃

open set

⁢

U

⁢

such that

⁢

x

∈

U

⊆

V

∁

}

.

Suppose

y

∈

{

x

}

∁

. As

X

is a

T

2

space,
there are

open sets

U

,

V

such that

x

∈

U

,

y

∈

V

, and

U

∩

V

=

∅

.
Thus, the

inclusion

from left to right holds.
On the other hand, suppose

y

∈

V

for some open

V

such that

{

x

}

⊆

V

∁

. Then

y

∈

V

⊆

{

x

}

∁

and the claim follows.
∎

Notes

If we adopt the notation that a

neighborhood

of

x

is any set containing an open set containing

x

, then the equation

1

can be written as

{

x

}

=

⋂

{

A

∣

A

⊆

X

⁢

is a closed neighborhood of

x

}

.

References

1

L.A. Steen, J.A.Seebach, Jr.,

Counterexamples in topology

,
Holt, Rinehart and Winston, Inc., 1970.

2

N. Bourbaki,

General Topology, Part 1

,
Addison-Wesley Publishing Company, 1966.

Title

characterization

of

T

⁢

2

spaces

Canonical

name

CharacterizationOfT2Spaces

Date of creation

2013-03-22 14:41:47

Last modified on

2013-03-22 14:41:47

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

7

Author

matte (1858)

Entry type

Theorem

Classification

msc 54D10

Related topic

LocallyCompactHausdorffSpace

Generated on Sat Feb 10 11:21:22 2018 by

LaTeXML
closed Hausdorff neighbourhoods, a theorem on

closed Hausdorff neighbourhoods, a theorem on

Theorem

.

If

X

is a

topological space

in which
every point has a closed

Hausdorff

neighbourhood

,
then

X

is Hausdorff.

Note.

In this theorem (and the proof that follows)
neighbourhoods are not assumed to be open.
That is, a neighbourhood of a point

x

is a set

A

such that

x

lies in the

interior

of

A

.

Proof of theorem.

Let

X

be a topological space in which
every point has a closed Hausdorff neighbourhood.
Suppose

a

,

b

∈

X

are distinct.
It suffices to show that

a

and

b

have

disjoint

neighbourhoods.
By

assumption

, there is a closed Hausdorff neighbourhood

N

of

b

.
If

a

∉

N

, then

X

∖

N

and

N

are disjoint neighbourhoods of

a

and

b

(as

N

is closed).

So suppose

a

∈

N

.
As

N

is Hausdorff,
there are

disjoint sets

U

0

,

V

0

⊆

N

that are open in

N

, such that

a

∈

U

0

and

b

∈

V

0

.
There are

open sets

U

and

V

of

X

such that

U

0

=

U

∩

N

and

V

0

=

V

∩

N

.
Note that

U

is a neighbourhood of

a

, and

V

is a neighbourhood of

b

.
As

N

is a neighbourhood of

b

,
it follows that

V

∩

N

(that is,

V

0

) is a neighbourhood of

b

.
We have

U

∩

V

0

=

U

0

∩

V

0

=

∅

.
So

U

and

V

0

are disjoint neighbourhoods of

a

and

b

.
QED.

Title

closed Hausdorff neighbourhoods, a theorem on

Canonical name

ClosedHausdorffNeighbourhoodsATheoremOn

Date of creation

2013-03-22 18:30:46

Last modified on

2013-03-22 18:30:46

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

7

Author

yark (2760)

Entry type

Theorem

Classification

msc 54D10

Generated on Sat Feb 10 11:21:24 2018 by

LaTeXML
completely Hausdorff

completely Hausdorff

Definition 1

.

[

1

]

Let

(

X

,

τ

)

be a

topological space

.
Suppose that for any two different points

x

,

y

∈

X

,

x

≠

y

,
we can find two

disjoint

neighborhoods

U

x

,

V

y

∈

τ

,

x

∈

U

x

,

y

∈

Y

y

such that their

closures

are also disjoint:

U

x

¯

∩

V

y

¯

=

∅

.

Then we say that

(

X

,

τ

)

is a

completely Hausdorff space

or a

T

2

⁤

1

2

space

.

Notes

A synonym for functionally Hausdorff space is

Urysohn space

[

1

]

.
Unfortunately, the definition of

completely Hausdorff

and

T

2

⁤

1

2

are not as standard as one would like since. For example, the
term completely Hausdorff space is also used to mean
a functionally Hausdorff space (e.g.

[

2

]

).
Nevertheless, in the present convention, we have the

implication

:

functionally Hausdorff

⇒

completely Hausdorff

⇒

T

2

=

Hausdorff

,

which suggests why the

T

2

⁤

1

2

name have been used to
denote both completely Hausdorff spaces and functionally Hausdorff spaces.

References

1

L.A. Steen, J.A.Seebach, Jr.,

Counterexamples in topology

,
Holt, Rinehart and Winston, Inc., 1970.

2

S. Willard,

General Topology

,
Addison-Wesley Publishing Company, 1970.

Title

completely Hausdorff

Canonical name

CompletelyHausdorff

Date of creation

2013-03-22 14:16:03

Last modified on

2013-03-22 14:16:03

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

15

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 54D10

Synonym

completely Hausdorff space

Synonym

T

2

⁤

1

2

Synonym

Urysohn space

Related topic

HausdorffSpaceNotCompletelyHausdorff

Generated on Sat Feb 10 11:21:26 2018 by

LaTeXML
Hausdorff property is hereditary

Hausdorff property is hereditary

Theorem 1

.

A

subspace

of a

Hausdorff space

is

Hausdorff

.

Proof.

Let

X

be a Hausdorff space , and let

Y

be a subspace of

X

. Let

y

1

,

y

2

∈

Y

where

y

1

≠

y

2

.
Since

X

is Hausdorff, there are disjoint

neighborhoods

U

1

of

y

1

and

U

2

of

y

2

∈

X

. Then

U

1

∩

Y

is a neighborhood of

y

1

in

Y

and

U

2

∩

Y

is a neighborhood of

y

2

in

Y

, and

U

1

∩

Y

and

U

2

∩

Y

are disjoint.
Therefore,

Y

is Hausdorff.
∎

Title

Hausdorff property is hereditary

Canonical name

HausdorffPropertyIsHereditary

Date of creation

2013-03-22 15:22:27

Last modified on

2013-03-22 15:22:27

Owner

georgiosl (7242)

Last modified by

georgiosl (7242)

Numerical id

8

Author

georgiosl (7242)

Entry type

Theorem

Classification

msc 54D10

Generated on Sat Feb 10 11:21:27 2018 by

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