aarjaneiro/planetmath huggingface dataset:
locally finite collection

locally finite collection

Let

𝒞

be a

collection

of subsets of a

topological space

X

.

𝒞

is said to be

locally finite

if for all

x

∈

X

there is a

neighbourhood

U

of

x

such that

U

∩

A

=

∅

for all but finitely many

A

∈

𝒞

.

Similarly,

𝒞

is said to be

locally countable

if for all

x

∈

X

there is a neighbourhood

U

of

x

such that

U

∩

A

=

∅

for all but countably many

A

∈

𝒞

.

Title

locally finite collection

Canonical name

LocallyFiniteCollection

Date of creation

2013-03-22 12:12:51

Last modified on

2013-03-22 12:12:51

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

11

Author

yark (2760)

Entry type

Definition

Classification

msc 54D20

Related topic

PointFinite

Defines

locally finite

Defines

locally countable collection

Defines

locally countable

Generated on Sat Feb 10 11:22:15 2018 by

LaTeXML
partition of unity

partition of unity

Let

X

be a

topological space

.
A

partition of unity

is a

collection

of

continuous functions

{

ε

i

:

X

→

[

0

,

1

]

}

such that

∑

i

ε

i

⁢

(

x

)

=

1

for all

x

∈

X

.

(1)

A partition of unity is

locally finite

if each

x

in

X

is

contained

in an

open set

on which only a finite number of

ε

i

are non-zero.
That is, if the cover

{

ε

i

-

1

⁢

(

(

0

,

1

]

)

}

is locally finite.

A partition of unity is

subordinate to an

open cover

{

U

i

}

of

X

if each

ε

i

is zero on the

complement

of

U

i

.

Example 1

(Circle)

A partition of unity for

S

1

is given by

{

sin

2

⁡

(

θ

/

2

)

,

cos

2

⁡

(

θ

/

2

)

}

subordinate to the covering

{

(

0

,

2

⁢

π

)

,

(

-

π

,

π

)

}

.

Application to integration

Let

M

be an

orientable manifold

with volume form

ω

and a partition of unity

{

ε

i

⁢

(

x

)

}

.
Then, the

integral

of a

function

f

⁢

(

x

)

over

M

is given by

∫

M

f

⁢

(

x

)

⁢

ω

=

∑

i

∫

U

i

ε

i

⁢

(

x

)

⁢

f

⁢

(

x

)

⁢

ω

.

It is of the choice of partition of unity.

Title

partition of unity

Canonical name

PartitionOfUnity

Date of creation

2013-03-22 13:29:23

Last modified on

2013-03-22 13:29:23

Owner

mhale (572)

Last modified by

mhale (572)

Numerical id

10

Author

mhale (572)

Entry type

Definition

Classification

msc 54D20

Classification

msc 58A05

Defines

locally finite partition of unity

Defines

subordinate to an open cover

Generated on Sat Feb 10 11:22:17 2018 by

LaTeXML
strongly paracompact space

strongly paracompact space

A

collection

𝒜

of sets is said to be

star-finite

if each member of

𝒜

meets only finitely many members of

𝒜

.

A

topological space

X

is said to be

strongly paracompact

if every

open cover

of

X

has a star-finite open refinement.

A star-finite open cover of a topological space
is clearly

locally finite

.
Therefore, every strongly paracompact space is

paracompact

(as the name suggests).

Title

strongly paracompact space

Canonical name

StronglyParacompactSpace

Date of creation

2013-03-22 17:09:01

Last modified on

2013-03-22 17:09:01

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

5

Author

yark (2760)

Entry type

Definition

Classification

msc 54D20

Synonym

strongly paracompact topological space

Defines

strongly paracompact

Defines

star-finite

Defines

star finite

Generated on Sat Feb 10 11:22:18 2018 by

LaTeXML
a space is compact iff any family of closed sets having fip has non-empty intersection

a space is compact iff any family of closed sets having fip has non-empty intersection

Theorem

.

A

topological space

is

compact

if and only if any

collection

of its

closed sets

having the

finite intersection property

has non-empty

intersection

.

The above theorem is essentially
the

definition

of a

compact space

rewritten using de Morgan’s laws.
The usual definition of a compact space is based on

open sets

and
unions. The above

characterization

, on the other hand, is written
using closed sets and intersections.

Proof.

Suppose

X

is compact, i.e., any collection of open subsets
that cover

X

has a finite collection that also cover

X

. Further, suppose

{

F

i

}

i

∈

I

is an arbitrary collection of closed subsets
with the finite intersection property. We claim that

∩

i

∈

I

F

i

is non-empty.
Suppose otherwise, i.e., suppose

∩

i

∈

I

F

i

=

∅

. Then,

X

=

(

⋂

i

∈

I

F

i

)

c

=

⋃

i

∈

I

F

i

c

.

(Here, the

complement

of a set

A

in

X

is written as

A

c

.)
Since each

F

i

is closed, the collection

{

F

i

c

}

i

∈

I

is an

open cover

for

X

. By

compactness

, there is a
finite subset

J

⊂

I

such
that

X

=

∪

i

∈

J

F

i

c

. But then

X

=

(

∩

i

∈

J

F

i

)

c

, so

∩

i

∈

J

F

i

=

∅

, which
contradicts the finite intersection property of

{

F

i

}

i

∈

I

.

The proof in the other direction is analogous.
Suppose

X

has the finite intersection property.
To prove that

X

is compact, let

{

F

i

}

i

∈

I

be a collection of open sets
in

X

that cover

X

. We claim that this collection contains a finite subcollection
of sets that also cover

X

.
The proof is by

contradiction

.
Suppose
that

X

≠

∪

i

∈

J

F

i

holds for all finite

J

⊂

I

.
Let us first show that the collection of closed subsets

{

F

i

c

}

i

∈

I

has the finite intersection property.
If

J

is a finite subset of

I

, then

⋂

i

∈

J

F

i

c

=

(

⋃

i

∈

J

F

i

)

c

≠

∅

,

where the last assertion follows since

J

was finite.
Then, since

X

has the finite intersection property,

∅

≠

⋂

i

∈

I

F

i

c

=

(

⋃

i

∈

I

F

i

)

c

.

This contradicts the

assumption

that

{

F

i

}

i

∈

I

is a cover for

X

.

□

References

1

R.E. Edwards,

Functional Analysis

: Theory and Applications

, Dover Publications, 1995.

Title

a

space is compact iff any family of closed sets having fip has non-empty intersection

Canonical

name

ASpaceIsCompactIffAnyFamilyOfClosedSetsHavingFipHasNonemptyIntersection

Date of creation

2013-03-22 13:34:10

Last modified on

2013-03-22 13:34:10

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

20

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 54D30

Generated on Sat Feb 10 11:22:20 2018 by

LaTeXML
alternative characterization of Stone-Čech compactification

alternative characterization of Stone-Čech compactification

The Stone-Čech compactificaton of a space is the set of all

ultrafilters

1

1

Here, the term “utrafilter” is used to denote ultrafilters with
respects to the

topology

of the space.

with the topology generated by the Stone-basis.

Title

alternative

characterization

of Stone-Čech

compactification

Canonical

name

AlternativeCharacterizationOfStonevCechCompactification

Date of creation

2013-03-22 15:51:42

Last modified on

2013-03-22 15:51:42

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 54D30

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

LaTeXML
closed set in a compact space is compact

closed set in a compact space is compact

Proof.

Let

A

be a

closed set

in a

compact space

X

.
To show that

A

is

compact

, we show that an arbitrary

open cover

has
a finite subcover. For this purpose, suppose

{

U

i

}

i

∈

I

be an arbitrary open cover for

A

.
Since

A

is closed, the

complement

of

A

,
which we denote by

A

c

, is open.
Hence

A

c

and

{

U

i

}

i

∈

I

together form an open cover for

X

.
Since

X

is compact, this cover has a finite subcover that
covers

X

. Let

D

be this subcover.
Either

A

c

is part of

D

or

A

c

is not.
In any case,

D

\

{

A

c

}

is a finite open cover
for

A

, and

D

\

{

A

c

}

is a subcover of

{

U

i

}

i

∈

I

. The claim follows.

□

Title

closed set in a compact space is compact

Canonical name

ClosedSetInACompactSpaceIsCompact

Date of creation

2013-03-22 13:34:02

Last modified on

2013-03-22 13:34:02

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

9

Author

mathcam (2727)

Entry type

Proof

Classification

msc 54D30

Related topic

ClosedSubsetsOfACompactSetAreCompact

Generated on Sat Feb 10 11:22:23 2018 by

LaTeXML
closed subsets of a compact set are compact

closed subsets of a compact set are compact

Theorem 1

.

Suppose

X

is a

topological space

. If

K

is a

compact subset

of

X

,

C

is a

closed set

in

X

, and

C

⊆

K

, then

C

is a compact set in

X

.

The below proof follows e.g. (

http://planetmath.org/Eg

)

[

3

]

. A proof based on the

finite intersection property

is given in

[

4

]

.

Proof.

Let

I

be an

indexing set

and

F

=

{

V

α

∣

α

∈

I

}

be an arbitrary

open cover

for

C

. Since

X

∖

C

is open, it follows that

F

together with

X

∖

C

is an open cover for

K

. Thus,

K

can be covered by a finite number of sets, say,

V

1

,

…

,

V

N

from

F

together with possibly

X

∖

C

. Since

C

⊂

K

,

V

1

,

…

,

V

N

cover

C

, and it follows that

C

is

compact

.
∎

The following proof uses the finite intersection property (

http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty

).

Proof.

Let

I

be an indexing set and

{

A

α

}

α

∈

I

be a

collection

of

X

-closed sets

contained

in

C

such that, for any finite

J

⊆

I

,

⋂

α

∈

J

A

α

is not empty. Recall that, for every

α

∈

I

,

A

α

⊆

C

⊆

K

. Thus, for every

α

∈

I

,

A

α

=

K

∩

A

α

. Therefore,

{

A

α

}

α

∈

I

are

K

-closed subsets of

K

(see this page (

http://planetmath.org/ClosedSetInASubspace

)) such that, for any finite

J

⊆

I

,

⋂

α

∈

J

A

α

is not empty. As

K

is compact,

⋂

α

∈

I

A

α

is not empty (again, by this result (

http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty

)).
This proves the claim.
∎

References

1

J.L. Kelley,

General Topology

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

2

S. Lang,

Analysis

II

,
Addison-Wesley Publishing Company Inc., 1969.

3

G.J. Jameson,

Topology

and

Normed Spaces

,
Chapman and Hall, 1974.

4

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

Lecture Notes on

Elementary

Topology and Geometry

,
Springer-Verlag, 1967.

Title

closed subsets of a compact set are compact

Canonical name

ClosedSubsetsOfACompactSetAreCompact

Date of creation

2013-03-22 13:55:56

Last modified on

2013-03-22 13:55:56

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

16

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 54D30

Related topic

AClosedSetInACompactSpaceIsCompact

Related topic

ACompactSetInAHausdorffSpaceIsClosed

Generated on Sat Feb 10 11:22:25 2018 by

LaTeXML
compact

compact

A

topological space

X

is

compact

if, for every

collection

{

U

i

}

i

∈

I

of

open sets

in

X

whose union is

X

, there exists a finite subcollection

{

U

i

j

}

j

=

1

n

whose union is also

X

.

A subset

Y

of a topological space

X

is said to be compact if

Y

with its

subspace topology

is a compact topological space.

Note:

Some authors require that a compact topological space be

Hausdorff

as well, and use the term

quasi-compact

to refer to a non-Hausdorff

compact space

. The modern convention seems to be to use compact in the sense given here, but the old definition is still occasionally encountered (particularly in the French school).

Title

compact

Canonical name

Compact

Date of creation

2013-03-22 11:53:35

Last modified on

2013-03-22 11:53:35

Owner

djao (24)

Last modified by

djao (24)

Numerical id

11

Author

djao (24)

Entry type

Definition

Classification

msc 54D30

Classification

msc 81-00

Classification

msc 83-00

Classification

msc 82-00

Classification

msc 46L05

Classification

msc 22A22

Related topic

QuasiCompact

Related topic

LocallyCompact

Related topic

HeineBorelTheorem

Related topic

TychonoffsTheorem

Related topic

Compactification

Related topic

SequentiallyCompact

Related topic

Lindelof

Related topic

NoetherianTopologicalSpace

Defines

compact set

Defines

compact subset

Generated on Sat Feb 10 11:22:27 2018 by

LaTeXML
compact subspace of a Hausdorff space is closed

compact subspace of a Hausdorff space is closed

Let

X

be a

Hausdorff space

, and

Y

be a

compact

subspace

of

X

. We prove that

X

∖

Y

is open, by finding for every point

x

∈

X

∖

Y

a

neighborhood

U

x

disjoint

from

Y

.

Let

y

∈

Y

.

x

≠

y

, so by the

definition

of a Hausdorff space, there exist open neighborhoods

U

x

(

y

)

of

x

and

V

x

(

y

)

of

y

such that

U

x

(

y

)

∩

V

x

(

y

)

=

∅

. Clearly

Y

⊆

⋃

y

∈

Y

V

x

(

y

)

but since

Y

is compact, we can select from these a finite

subcover

of

Y

Y

⊆

V

x

(

y

1

)

∪

⋯

∪

V

x

(

y

n

)

Now for every

y

∈

Y

there exists

k

∈

1

⁢

…

⁢

n

such that

y

∈

V

x

(

y

k

)

. Since

U

x

(

y

k

)

and

V

x

(

y

k

)

are disjoint,

y

∉

U

x

(

y

k

)

, therefore neither is it in the

intersection

U

x

=

⋂

j

=

1

n

U

x

(

y

j

)

A finite intersection of

open sets

is open, hence

U

x

is a neighborhood of

x

disjoint from

Y

.

Title

compact subspace of a Hausdorff space is closed

Canonical

name

CompactSubspaceOfAHausdorffSpaceIsClosed

Date of creation

2013-03-22 16:31:23

Last modified on

2013-03-22 16:31:23

Owner

ehremo (15714)

Last modified by

ehremo (15714)

Numerical id

7

Author

ehremo (15714)

Entry type

Proof

Classification

msc 54D30

Generated on Sat Feb 10 11:22:29 2018 by

LaTeXML
continuous image of a compact set is compact

continuous image of a compact set is compact

Theorem 1

.

The

continuous

image of a

compact set

is also

compact

.

Proof.

Let

X

and

Y

be

topological spaces

,

f

:

X

→

Y

be continuous,

A

be a compact subset of

X

,

I

be an

indexing set

, and

{

V

α

}

α

∈

I

be an

open cover

of

f

⁢

(

A

)

. Thus,

f

⁢

(

A

)

⊆

⋃

α

∈

I

V

α

. Therefore,

A

⊆

f

-

1

⁢

(

f

⁢

(

A

)

)

⊆

f

-

1

⁢

(

⋃

α

∈

I

V

α

)

=

⋃

α

∈

I

f

-

1

⁢

(

V

α

)

.

Since

f

is continuous, each

f

-

1

⁢

(

V

α

)

is an

open subset

of

X

. Since

A

⊆

⋃

α

∈

I

f

-

1

⁢

(

V

α

)

and

A

is compact, there exists

n

∈

ℕ

with

A

⊆

⋃

j

=

1

n

f

-

1

⁢

(

V

α

j

)

for some

α

1

,

…

,

α

n

∈

I

. Hence,

f

⁢

(

A

)

⊆

f

⁢

(

⋃

j

=

1

n

f

-

1

⁢

(

V

α

j

)

)

=

f

⁢

(

f

-

1

⁢

(

⋃

j

=

1

n

V

α

j

)

)

⊆

⋃

j

=

1

n

V

α

j

. It follows that

f

⁢

(

A

)

is compact.
∎

Title

continuous image of a compact set is compact

Canonical name

ContinuousImageOfACompactSetIsCompact

Date of creation

2013-03-22 15:53:14

Last modified on

2013-03-22 15:53:14

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

16

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 54D30

Related topic

CompactnessIsPreservedUnderAContinuousMap

Generated on Sat Feb 10 11:22:34 2018 by

LaTeXML
equivalent formulation of the tube lemma

equivalent formulation of the tube lemma

Let us recall the thesis of the

tube lemma

. Assume, that

X

and

Y

are

topological spaces

.

(TL)

If

U

⊆

X

×

Y

is open (in

product topology

) and if

x

∈

X

is such that

x

×

Y

⊆

U

, then there exists an open

neighbourhood

V

⊆

X

of

x

such that

V

×

Y

⊆

U

.

We wish to give a

relation

between (TL) and the the following thesis, concering closed projections:

(CP)

The projection

π

:

X

×

Y

→

X

given by

π

⁢

(

x

,

y

)

=

x

is a closed map.

The following

theorem

relates these two statements:

Theorem.

(TL) is

equivalent

to (CP).

Proof.

,,

⇒

” Let

F

⊆

X

×

Y

be a

closed set

and let

U

=

(

X

×

Y

)

\

F

be its open

complement

. We will show, that

π

⁢

(

F

)

is closed, by showing that

V

=

X

\

π

⁢

(

F

)

is open. So assume, that

x

∈

V

. Obviously

(

π

-

1

(

x

)

=

x

×

Y

)

∩

F

=

∅

.

Therefore

x

×

Y

⊆

U

and by (TL) there exists open neighbourhood

V

′

⊆

X

of

x

such that

V

′

×

Y

⊆

U

. It easily follows, that

V

′

⊆

V

and it is open, so (since

x

was chosen arbitrary)

V

is open.

,,

⇐

” Let

U

⊆

X

×

Y

be an

open subset

such that

x

×

Y

⊆

U

for some

x

∈

X

. Let

F

=

(

X

×

Y

)

\

U

. Then

F

is closed and by (CP) we have that

π

⁢

(

F

)

⊆

X

is closed. Also

x

∉

π

⁢

(

F

)

and thus

V

=

X

\

π

⁢

(

F

)

is an open neighbourhood of

x

. It can be easily checked, that

V

×

Y

⊆

U

, which

completes

the proof.

□

Remark.

The theorem doesn’t state that any of statements is true. It is well known (see tha parent object), that if both

X

and

Y

are

Hausdorff

with

Y

compact

, then both are true. On the other hand, for example for

X

=

Y

=

ℝ

, where

ℝ

denotes reals with

standard topology

, they are both false. For example consider

F

=

{

(

x

,

y

)

∈

ℝ

2

|

x

⁢

y

=

1

}

.

Of course

F

is closed, but

π

⁢

(

F

)

=

ℝ

\

{

0

}

is not closed, so the (CP) is false.

Title

equivalent formulation of the tube lemma

Canonical name

EquivalentFormulationOfTheTubeLemma

Date of creation

2013-03-22 19:15:18

Last modified on

2013-03-22 19:15:18

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Theorem

Classification

msc 54D30

Generated on Sat Feb 10 11:22:36 2018 by

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finite intersection property

finite intersection property

A

collection

𝒜

=

{

A

α

}

α

∈

I

of subsets of a set

X

is said to have the

finite intersection property

, abbreviated f.i.p., if every finite subcollection

{

A

1

,

A

2

,

…

,

A

n

}

of

𝒜

satisifes

⋂

i

=

1

n

A

i

≠

∅

.

The finite intersection property is most often used to give the following

http://planetmath.org/node/3769

equivalent

condition for the

http://planetmath.org/node/503

compactness

of a

topological space

(a proof of which may be found

http://planetmath.org/node/4181

here):

Proposition

.

A topological space

X

is

compact

if and only if for every collection

C

=

{

C

α

}

α

∈

J

of

closed subsets

of

X

having the finite intersection property,

⋂

α

∈

J

C

α

≠

∅

.

An important special case of the preceding is that in which

𝒞

is a

countable

collection of non-empty nested sets, i.e., when we have

C

1

⊃

C

2

⊃

C

3

⊃

⋯

⁢

.

In this case,

𝒞

automatically has the finite intersection property, and if each

C

i

is a closed subset of a compact topological space, then, by the

proposition

,

⋂

i

=

1

∞

C

i

≠

∅

.

The f.i.p.

characterization

of may be used to prove a general result on the uncountability of certain compact

Hausdorff spaces

, and is also used in a proof of

Tychonoff

’s

Theorem

.

References

1

J. Munkres,

Topology

, 2nd ed. Prentice Hall, 1975.

Title

finite intersection property

Canonical

name

FiniteIntersectionProperty

Date of creation

2013-03-22 13:34:05

Last modified on

2013-03-22 13:34:05

Owner

azdbacks4234 (14155)

Last modified by

azdbacks4234 (14155)

Numerical id

17

Author

azdbacks4234 (14155)

Entry type

Definition

Classification

msc 54D30

Synonym

finite intersection condition

Synonym

f.i.c.

Synonym

f.i.p.

Related topic

Compact

Related topic

Intersection

Related topic

Finite

Defines

finite intersection property

Generated on Sat Feb 10 11:22:39 2018 by

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Heine-Borel theorem

Heine-Borel theorem

A subset

A

of

ℝ

n

is

compact

if and only if

A

is closed (

http://planetmath.org/TopologicalSpace

) and

bounded

.

Title

Heine-Borel theorem

Canonical name

HeineBorelTheorem

Date of creation

2013-03-22 12:05:09

Last modified on

2013-03-22 12:05:09

Owner

Evandar (27)

Last modified by

Evandar (27)

Numerical id

8

Author

Evandar (27)

Entry type

Theorem

Classification

msc 54D30

Related topic

Compact

Related topic

ProofOfLeastAndReatestValueOfFunction

Generated on Sat Feb 10 11:22:41 2018 by

LaTeXML
point and a compact set in a Hausdorff space have disjoint open neighborhoods.

point and a compact set in a Hausdorff space have disjoint open neighborhoods.

Theorem

.

Let

X

be a

Hausdorff space

, let

A

be a

compact

non-empty set in

X

, and let

y

a point in the

complement

of

A

.
Then there exist

disjoint

open sets

U

and

V

in

X

such that

A

⊂

U

and

y

∈

V

.

Proof.

First we use the fact that

X

is a Hausdorff space.
Thus, for all

x

∈

A

there exist
disjoint open sets

U

x

and

V

x

such that

x

∈

U

x

and

y

∈

V

x

.
Then

{

U

x

}

x

∈

A

is an

open cover

for

A

.
Using this

characterization

of

compactness

(

http://planetmath.org/YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover

),
it follows that there
exist a

finite set

A

0

⊂

A

such that

{

U

x

}

x

∈

A

0

is a finite open cover for

A

.
Let us define

U

=

⋃

x

∈

A

0

U

x

,

⁢

V

=

⋂

x

∈

A

0

V

x

.

Next we show that these sets

satisfy

the given conditions
for

U

and

V

. First, it is clear
that

U

and

V

are open. We also have that

A

⊂

U

and

y

∈

V

.
To see that

U

and

V

are disjoint, suppose

z

∈

U

. Then

z

∈

U

x

for some

x

∈

A

0

.
Since

U

x

and

V

x

are disjoint,

z

can not be in

V

x

, and
consequently

z

can not be in

V

.
∎

The above result and proof follows

[

1

]

(Chapter 5,

Theorem

7) or

[

2

]

(page 27).

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

point and a compact set in a Hausdorff space have disjoint open neighborhoods.

Canonical

name

PointAndACompactSetInAHausdorffSpaceHaveDisjointOpenNeighborhoods

Date of creation

2013-03-22 13:34:27

Last modified on

2013-03-22 13:34:27

Owner

drini (3)

Last modified by

drini (3)

Numerical id

13

Author

drini (3)

Entry type

Theorem

Classification

msc 54D30

Classification

msc 54D10

Generated on Sat Feb 10 11:22:43 2018 by

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proof of Tychonoff’s theorem

proof of Tychonoff’s theorem

This is a proof in of nets. Recall the following facts:

1 -

A net

(

x

α

)

α

∈

𝒜

in

∏

i

∈

I

X

i

converges

to

x

∈

∏

i

∈

I

X

i

if and only if each coordinate

(

x

α

i

)

α

∈

𝒜

converges to

x

i

∈

X

i

2 -

A

topological space

X

is

compact

if and only if every net in

X

has a convergent subnet.

3 -

Every net has a

universal

subnet.

4 -

A

universal net

(

http://planetmath.org/Ultranet

)

(

x

α

)

α

∈

𝒜

in a

compact space

X

is convergent. (see this entry (

http://planetmath.org/UniversalNetsInCompactSpacesAreConvergent

))

We now prove

Tychonoff

’s

theorem

.

Proof (Tychonoff’s theorem) :

Let

(

x

α

)

α

∈

𝒜

be a net in

∏

i

∈

I

X

i

.

Using Lemma 3 we can find a subnet

(

y

β

)

β

∈

ℬ

of

(

x

α

)

α

∈

𝒜

.

It is easily seen that each coordinate net

(

y

β

i

)

β

∈

ℬ

is a net in

X

i

.

Using Lemma 4 we see that each coordinate net converges, because

X

i

is compact.

Using Lemma 1 we see that the whole net

(

y

β

)

β

∈

ℬ

converges in

∏

i

∈

I

X

i

.

We conclude that every net in

∏

i

∈

I

X

i

has a convergent subnet, so, by Lemma 2,

∏

i

∈

I

X

i

must be compact.

□

Title

proof of Tychonoff’s theorem

Canonical name

ProofOfTychonoffsTheorem

Date of creation

2013-03-22 17:25:24

Last modified on

2013-03-22 17:25:24

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

8

Author

asteroid (17536)

Entry type

Proof

Classification

msc 54D30

Generated on Sat Feb 10 11:22:47 2018 by

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