aarjaneiro/planetmath huggingface dataset:
uncountable Polish spaces contain Cantor space

uncountable Polish spaces contain Cantor space

Cantor space

is an example of a

compact

and

uncountable

Polish space

. In fact, every

uncountable Polish space contains Cantor space

, as stated by the following

theorem

.

Theorem

.

Let

X

be an uncountable Polish space. Then, it contains a subset

S

which is

homeomorphic

to Cantor space.

For example, the set

ℝ

of

real numbers

contains the Cantor middle thirds set (

http://planetmath.org/CantorSet

). Note that, being homeomorphic to Cantor space,

S

must be a compact and hence

closed subset

of

X

.
The result is

trivial

in the case of

Baire space

𝒩

, in which case we may take

S

to be the set of all

s

∈

𝒩

satisfying

s

n

∈

{

1

,

2

}

for all

n

.
Then, for any uncountable Polish space

X

there exists a

continuous

and one-to-one function

f

:

𝒩

→

X

(see here (

http://planetmath.org/InjectiveImagesOfBaireSpace

)). Then

f

gives a continuous

bijection

from

S

to

f

⁢

(

S

)

. The

inverse function theorem

(

http://planetmath.org/InverseFunctionTheoremTopologicalSpaces

) implies that

f

is a

homeomorphism

between

S

and

f

⁢

(

S

)

and, therefore,

f

⁢

(

S

)

is homeomorphic to Cantor space.

Title

uncountable Polish spaces contain Cantor space

Canonical name

UncountablePolishSpacesContainCantorSpace

Date of creation

2013-03-22 18:48:33

Last modified on

2013-03-22 18:48:33

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

6

Author

gel (22282)

Entry type

Theorem

Classification

msc 54E50

Related topic

PolishSpace

Generated on Sat Feb 10 11:29:00 2018 by

LaTeXML
uniformly equicontinuous

uniformly equicontinuous

A family

ℱ

of functions from a

metric space

(

X

,

d

)

to a metric space

(

X

′

,

d

′

)

is

uniformly equicontinuous

if, for each

ε

>

0

there exists

δ

>

0

such that,

∀

f

∈

ℱ

,

∀

x

,

y

∈

X

,

d

⁢

(

x

,

y

)

<

δ

⇒

d

′

⁢

(

f

⁢

(

x

)

,

f

⁢

(

y

)

)

<

ε

.

Title

uniformly equicontinuous

Canonical

name

UniformlyEquicontinuous

Date of creation

2013-03-22 13:14:33

Last modified on

2013-03-22 13:14:33

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

6

Author

Koro (127)

Entry type

Definition

Classification

msc 54E50

Related topic

Equicontinuous

Generated on Sat Feb 10 11:29:02 2018 by

LaTeXML
another proof of the non-existence of a continuous function that switches the rational and the irrational numbers

another proof of the non-existence of a continuous function that switches the rational and the irrational numbers

Let

𝕁

=

ℝ

∖

ℚ

denote the

irrationals

.
There is no

continuous function

f

:

ℝ

→

ℝ

such that

f

⁢

(

ℚ

)

⊆

𝕁

and

f

⁢

(

𝕁

)

⊆

ℚ

.

Proof

Suppose

f

is such a

function

. Since

ℚ

is

countable

,

f

⁢

(

ℚ

)

and

f

⁢

(

𝕁

)

are also countable. Therefore the image of

f

is countable. If

f

is not a

constant function

, then by the

intermediate value theorem

the image of

f

contains a nonempty

interval

, so the image of

f

is

uncountable

. We have just shown that this isn’t the case, so there must be some

c

such that

f

⁢

(

x

)

=

c

for all

x

∈

ℝ

. Therefore

f

⁢

(

ℚ

)

=

{

c

}

⊂

𝕁

and

f

⁢

(

𝕁

)

=

{

c

}

⊂

ℚ

. Obviously no number is both

rational and irrational

, so no such

f

exists.

Title

another proof of the non-existence of a continuous function that switches the

rational

and the irrational numbers

Canonical name

AnotherProofOfTheNonexistenceOfAContinuousFunctionThatSwitchesTheRationalAndTheIrrationalNumbers

Date of creation

2013-03-22 16:23:54

Last modified on

2013-03-22 16:23:54

Owner

neapol1s (9480)

Last modified by

neapol1s (9480)

Numerical id

7

Author

neapol1s (9480)

Entry type

Proof

Classification

msc 54E52

Generated on Sat Feb 10 11:29:04 2018 by

LaTeXML
Baire category theorem

Baire category theorem

In a non-empty

complete metric space

, any

countable

intersection

of dense,

open subsets

is non-empty.

In fact, such countable intersections of dense, open subsets are dense. So the

theorem

holds also for any non-empty open subset of a complete metric space.

Alternative formulations:

Call a set

first category

, or a

meagre

set, if it is a countable union of nowhere

dense sets

, otherwise

second category

. The

Baire category theorem

is often stated as “no non-empty complete metric space is of first category”, or, trivially, as “a non-empty, complete metric space is of second category”. In short, this theorem says that every nonempty complete metric space is a

Baire space

.

In

functional analysis

, this important property of complete metric spaces forms the basis for the proofs of the important principles of

Banach spaces

: the

open mapping theorem

and the

closed graph theorem

.

It may also be taken as giving a

concept

of “small sets”,

similar

to sets of measure zero: a countable union of these sets remains “small”. However, the real line

ℝ

may be partitioned into a set of measure zero and a set of first category; the two concepts are distinct.

Note that, apart from the requirement that the set be a complete metric space, all conditions and

conclusions

of the theorem are phrased topologically. This “metric requirement” is thus something of a disappointment. As it turns out, there are two ways to reduce this requirement.

First, if a

topological space

𝒯

is

homeomorphic

to a non-empty open subset of a complete metric space, then we can transfer the Baire property through the homeomorphism, so in

𝒯

too any countable intersection of open dense sets is non-empty (and, in fact, dense). The other formulations also hold in this case.

Second, the Baire category theorem holds for a

locally compact

,

Hausdorff

1

1

Some authors only define a locally compact space to be a Hausdorff space; that is the sense required for this theorem.

topological space

𝒯

.

Title

Baire category theorem

Canonical name

BaireCategoryTheorem

Date of creation

2013-03-22 12:43:32

Last modified on

2013-03-22 12:43:32

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

13

Author

Koro (127)

Entry type

Theorem

Classification

msc 54E52

Related topic

SardsTheorem

Related topic

Meager

Related topic

Residual

Generated on Sat Feb 10 11:29:05 2018 by

LaTeXML
Baire space

Baire space

A

Baire space

is a

topological space

such that the

intersection

of any

countable

family of open and

dense sets

is dense.

Title

Baire space

Canonical

name

BaireSpace

Date of creation

2013-03-22 13:40:33

Last modified on

2013-03-22 13:40:33

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

5

Author

Koro (127)

Entry type

Definition

Classification

msc 54E52

Generated on Sat Feb 10 11:29:07 2018 by

LaTeXML
equivalent statement of Baire category theorem

equivalent statement of Baire category theorem

Let

(

X

,

d

)

be a

complete metric space

. Then every subset

B

⊂

X

of

first category

has empty

interior

.

Corollary

: Every non empty complete metric space is of second category.

Title

equivalent statement of Baire category theorem

Canonical name

EquivalentStatementOfBaireCategoryTheorem

Date of creation

2013-03-22 14:04:49

Last modified on

2013-03-22 14:04:49

Owner

gumau (3545)

Last modified by

gumau (3545)

Numerical id

8

Author

gumau (3545)

Entry type

Theorem

Classification

msc 54E52

Generated on Sat Feb 10 11:29:12 2018 by

LaTeXML
every subspace of a normed space of finite dimension is closed

every subspace of a normed space of finite dimension is closed

Let

(

V

,

∥

⋅

∥

)

be a

normed vector space

, and

S

⊂

V

a finite dimensional

subspace

. Then

S

is closed.

Proof

Let

a

∈

S

¯

and choose a

sequence

{

a

n

}

with

a

n

∈

S

such that

a

n

converges

to

a

. Then

{

a

n

}

is a

Cauchy sequence

in

V

and
is also a Cauchy sequence in

S

. Since a finite dimensional

normed
space

is a

Banach space

,

S

is

complete

, so

{

a

n

}

converges to an
element of

S

. Since limits in a normed space are unique, that limit
must be

a

, so

a

∈

S

.

Example

The result depends on the field being the real or

complex numbers

.
Suppose the

V

=

Q

×

R

, viewed as a vector space over

Q

and

S

=

Q

×

Q

is the finite dimensional subspace. Then clearly

(

1

,

2

)

is in

V

and is a

limit point

of

S

which is not in

S

. So

S

is not closed.

Example

On the other hand, there is an example where

Q

is the underlying
field and we can still show a finite dimensional subspace is closed. Suppose
that

V

=

Q

n

, the set of

n

-tuples of

rational numbers

, viewed
as vector space over

Q

. Then if

S

is a finite dimensional subspace
it must be that

S

=

{

x

|

A

⁢

x

=

0

}

for some matrix

A

.
That is,

S

is the

inverse image

of the

closed set

{

0

}

.
Since the map

x

→

A

⁢

x

is

continuous

, it follows that

S

is a closed set.

Title

every subspace of a normed space of finite

dimension

is closed

Canonical name

EverySubspaceOfANormedSpaceOfFiniteDimensionIsClosed

Date of creation

2013-03-22 14:56:28

Last modified on

2013-03-22 14:56:28

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

12

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 54E52

Classification

msc 15A03

Classification

msc 46B99

Generated on Sat Feb 10 11:29:14 2018 by

LaTeXML
example of a meager set

example of a meager set

Note that

ℚ

is meager in

ℝ

under the

usual topology

. Let

{

r

n

}

n

∈

ℕ

be an enumeration of

ℚ

. Then

ℚ

=

⋃

n

∈

ℕ

{

r

n

}

and, for every

n

∈

ℕ

,

int

⁡

{

r

n

}

¯

=

int

⁡

{

r

n

}

=

∅

.

Title

example of a meager set

Canonical name

ExampleOfAMeagerSet

Date of creation

2013-03-22 17:07:08

Last modified on

2013-03-22 17:07:08

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

4

Author

Wkbj79 (1863)

Entry type

Example

Classification

msc 54E52

Related topic

ExamplesOfNowhereDenseSets

Generated on Sat Feb 10 11:29:16 2018 by

LaTeXML
generic

generic

A property that holds for all

x

in some

residual

subset of a

Baire space

X

is said to be

generic

in

X

, or to

hold generically

in

X

. In the study of generic properties, it is common to state “generically,

P

⁢

(

x

)

”, where

P

⁢

(

x

)

is some

proposition

about

x

∈

X

. The useful fact about generic properties is that, given countably many generic properties

P

n

, all of them hold simultaneously in a residual set, i.e. we have that, generically,

P

n

⁢

(

x

)

holds for each

n

.

Title

generic

Canonical name

Generic

Date of creation

2013-03-22 13:40:30

Last modified on

2013-03-22 13:40:30

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

10

Author

Koro (127)

Entry type

Definition

Classification

msc 54E52

Defines

generically

Generated on Sat Feb 10 11:29:17 2018 by

LaTeXML
meager

meager

A meager or Baire

first category

set in a

topological space

is one which is a countable union of nowhere

dense sets

. A Baire

second category

set is one which contains a countable union of open and dense sets.

Title

meager

Canonical name

Meager

Date of creation

2013-03-22 13:12:57

Last modified on

2013-03-22 13:12:57

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

8

Author

Koro (127)

Entry type

Definition

Classification

msc 54E52

Synonym

meagre

Synonym

first category

Synonym

second category

Related topic

BaireCategoryTheorem

Generated on Sat Feb 10 11:29:19 2018 by

LaTeXML
no continuous function switches the rational and the irrational numbers

no continuous function switches the rational and the irrational numbers

Let

𝕁

=

ℝ

∖

ℚ

denote the irrationals.
There is no

continuous function

f

:

ℝ

→

ℝ

such that

f

⁢

(

ℚ

)

⊆

𝕁

and

f

⁢

(

𝕁

)

⊆

ℚ

.

Proof

Suppose there is such a

function

f

.

First,

ℚ

=

⋃

i

∈

ℕ

{

q

i

}

implies

f

⁢

(

ℚ

)

=

⋃

i

∈

ℕ

{

f

⁢

(

q

i

)

}

.

This is because functions

preserve

unions (see

properties of functions

).

And then,

f

⁢

(

ℚ

)

is

first category

, because every

singleton

in

ℝ

is nowhere dense (because

ℝ

with the

Euclidean metric

has no

isolated points

, so the

interior

of a singleton is empty).

But

f

⁢

(

𝕁

)

⊆

ℚ

, so

f

⁢

(

𝕁

)

is first category too.
Therefore

f

⁢

(

ℝ

)

is first category, as

f

⁢

(

ℝ

)

=

f

⁢

(

ℚ

)

∪

f

⁢

(

𝕁

)

.
Consequently, we have

f

⁢

(

ℝ

)

=

⋃

i

∈

ℕ

{

z

i

}

.

But functions preserve unions in both ways, so

ℝ

=

f

-

1

⁢

(

⋃

i

∈

ℕ

{

z

i

}

)

=

⋃

i

∈

ℕ

f

-

1

⁢

(

{

z

i

}

)

.

(1)

Now,

f

is

continuous

, and as

{

z

i

}

is closed for every

i

∈

ℕ

, so is

f

-

1

⁢

(

{

z

i

}

)

. This means that

f

-

1

⁢

(

{

z

i

}

)

¯

=

f

-

1

⁢

(

{

z

i

}

)

. If

int

⁡

(

f

-

1

⁢

(

{

z

i

}

)

)

≠

∅

, we have that there is an

open interval

(

a

i

,

b

i

)

⊆

f

-

1

⁢

(

{

z

i

}

)

, and this implies that there is an irrational number

x

i

and a

rational number

y

i

such that both lie in

f

-

1

⁢

(

{

z

i

}

)

, which is not possible because this would imply that

f

⁢

(

x

i

)

=

f

⁢

(

y

i

)

=

z

i

, and then

f

would map an irrational and a rational number to the same

element

, but by

hypothesis

f

⁢

(

ℚ

)

⊆

𝕁

and

f

⁢

(

𝕁

)

⊆

ℚ

.

Then, it must be

int

⁡

(

f

-

1

⁢

(

{

z

i

}

)

)

=

∅

for every

i

∈

ℕ

, and this implies that

ℝ

is first category (by (1)).
This is absurd, by the

Baire Category Theorem

.

Title

no continuous function switches the rational and the irrational numbers

Canonical

name

NoContinuousFunctionSwitchesTheRationalAndTheIrrationalNumbers

Date of creation

2013-03-22 14:59:15

Last modified on

2013-03-22 14:59:15

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

13

Author

yark (2760)

Entry type

Result

Classification

msc 54E52

Generated on Sat Feb 10 11:29:21 2018 by

LaTeXML
no countable dense subset of a complete metric space is a G_δ

no countable dense subset of a complete metric space is a

G

δ

Let

(

X

,

d

)

be a complete metric space with no

isolated points

, and let

D

⊂

X

be a

countable

dense set

. Then

D

is not a

G

δ

set (

http://planetmath.org/G_deltaSet

).

Proof

First, we will prove that

D

is

first category

. Then, supposing that

D

is a

G

δ

, we will conclude that

X

-

D

must be first category. But then so must be

X

, which is absurd because

X

is

complete

.

1)

D

is a first category set:

By

hypothesis

D

=

{

x

i

}

i

∈

N

. Let’s see that each

singleton

is nowhere dense if

X

has no isolated points:

{

x

i

}

¯

=

{

x

i

}

(trivially). Suppose that

{

x

i

}

o

=

{

x

i

}

. Then there is a ball aisolating the point. Absurd (

X

has no isolated points). Then

{

x

i

}

o

=

∅

and we have that

(

{

x

i

}

¯

)

o

=

∅

so every singleton is nowhere dense and

D

is of first category because it is a countable union of nowhere dense sets.

2) Suppose

D

is a

G

δ

, that is,

D

=

⋂

i

=

1

∞

U

i

such that every

U

is open. As

D

is dense, then each

U

is dense, because

D

⊂

U

i

⇒

D

¯

=

X

⊂

U

i

¯

∀

i

. But then

X

-

D

=

X

-

⋂

i

=

1

∞

U

i

=

⋃

i

=

1

∞

(

X

-

U

i

)

and

(

X

-

U

i

¯

)

o

=

(

X

-

U

i

o

)

o

=

(

X

-

U

i

)

o

=

X

-

U

i

¯

=

∅

which implies that

X

-

D

is of first category. Then

D

⁢

⋃

(

X

-

D

)

=

X

is of first category. Absurd, because

X

is complete.

Title

no countable dense subset of a complete metric space is a

G

δ

Canonical

name

NoCountableDenseSubsetOfACompleteMetricSpaceIsAGdelta

Date of creation

2013-03-22 14:59:06

Last modified on

2013-03-22 14:59:06

Owner

gumau (3545)

Last modified by

gumau (3545)

Numerical id

8

Author

gumau (3545)

Entry type

Result

Classification

msc 54E52

Generated on Sat Feb 10 11:29:23 2018 by

LaTeXML
proof for one equivalent statement of Baire category theorem

proof for one equivalent statement of Baire category theorem

First, let’s assume Baire’s category

theorem

and prove the alternative statement.

We have

B

=

⋃

n

=

1

∞

B

n

, with

int

⁢

(

B

k

¯

)

=

∅

⁢

∀

k

∈

𝐍

.

Then

X

=

X

-

int

⁢

(

B

k

¯

)

=

X

-

B

k

¯

¯

⁢

∀

k

∈

𝐍

Then

X

-

B

k

¯

is dense in

X

for every

k

. Besides,

X

-

B

k

¯

is open because

X

is open and

B

k

¯

closed. So, by Baire’s Category Theorem, we have that

⋂

n

=

1

∞

(

X

-

B

n

¯

)

=

X

-

⋃

n

=

1

∞

B

n

¯

is dense in

X

.
But

B

⊂

⋃

n

=

1

∞

B

n

¯

⟹

X

-

⋃

n

=

1

∞

B

n

¯

⊂

X

-

B

, and then

X

=

X

-

⋃

n

=

1

∞

B

n

¯

¯

⊂

X

-

B

¯

=

X

-

int

⁢

(

B

)

⟹

int

⁢

(

B

)

=

∅

.

Now, let’s assume our alternative statement as the

hypothesis

, and let

(

B

k

)

k

∈

N

be a

collection

of open

dense sets

in
a

complete metric space

X

.
Then

int

⁢

(

X

-

B

k

¯

)

=

int

⁢

(

X

-

int

⁢

(

B

k

)

)

=

int

⁢

(

X

-

B

k

)

=

X

-

B

k

¯

=

∅

and so

X

-

B

k

is nowhere dense for every

k

.

Then

X

-

⋂

n

=

1

∞

B

n

¯

=

int

⁢

(

X

-

⋂

n

=

1

∞

B

n

)

=

int

⁢

(

⋃

n

=

1

∞

X

-

B

n

)

=

∅

due to our hypothesis. Hence Baire’s category theorem holds.

QED

Title

proof for one equivalent statement of Baire category theorem

Canonical

name

ProofForOneEquivalentStatementOfBaireCategoryTheorem

Date of creation

2013-03-22 14:04:52

Last modified on

2013-03-22 14:04:52

Owner

gumau (3545)

Last modified by

gumau (3545)

Numerical id

8

Author

gumau (3545)

Entry type

Proof

Classification

msc 54E52

Generated on Sat Feb 10 11:29:25 2018 by

LaTeXML
proof of Baire category theorem

proof of Baire category theorem

Let

(

X

,

d

)

be a

complete metric space

, and

U

k

a

countable

collection

of dense,

open subsets

. Let

x

0

∈

X

and

ϵ

0

>

0

be
given. We must show that there exists a

x

∈

⋂

k

U

k

such that

d

⁢

(

x

0

,

x

)

<

ϵ

0

.

Since

U

1

is dense and open, we may choose an

ϵ

1

>

0

and an

x

1

∈

U

1

such that

d

⁢

(

x

0

,

x

1

)

<

ϵ

0

2

,

ϵ

1

<

ϵ

0

2

,

and such that the

open ball

of
radius

ϵ

1

about

x

1

lies entirely
in

U

1

. We then choose an

ϵ

2

>

0

and a

x

2

∈

U

2

such that

d

⁢

(

x

1

,

x

2

)

<

ϵ

1

2

,

ϵ

2

<

ϵ

1

2

,

and such that the open ball
of radius

ϵ

2

about

x

2

lies
entirely in

U

2

. We continue by

induction

, and construct a

sequence

of points

x

k

∈

U

k

and

positive

ϵ

k

such that

d

⁢

(

x

k

-

1

,

x

k

)

<

ϵ

k

-

1

2

,

ϵ

k

<

ϵ

k

-

1

2

,

and such that the open
ball of radius

ϵ

k

lies entirely in

U

k

.

By construction, for

0

≤

j

<

k

we have

d

⁢

(

x

j

,

x

k

)

<

ϵ

j

⁢

(

1

2

+

⋯

+

1

2

k

-

j

)

<

ϵ

j

≤

ϵ

0

2

j

.

Hence the
sequence

x

k

,

k

=

1

,

2

,

…

is Cauchy, and

converges

by

hypothesis

to some

x

∈

X

. It is clear that for every

k

we have

d

⁢

(

x

,

x

k

)

≤

ϵ

k

.

Moreover it follows that

d

⁢

(

x

,

x

k

)

≤

d

⁢

(

x

,

x

k

+

1

)

+

d

⁢

(

x

k

,

x

k

+

1

)

<

ϵ

k

+

1

+

ϵ

k

2

,

and hence a fortiori

d

⁢

(

x

,

x

k

)

<

ϵ

k

for every

k

. By construction then,

x

∈

U

k

for all

k

=

1

,

2

,

…

, as well. QED

Title

proof of Baire category theorem

Canonical name

ProofOfBaireCategoryTheorem

Date of creation

2013-03-22 13:06:55

Last modified on

2013-03-22 13:06:55

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

10

Author

rmilson (146)

Entry type

Proof

Classification

msc 54E52

Generated on Sat Feb 10 11:29:28 2018 by

LaTeXML
residual

residual

A

subspace

A

of a

topological space

X

is called

residual

(or

comeager

) if and only if it is

second category

and its

complement

X

∖

A

is first category. Equivalently, a set is residual if and only if it contains a

countable

intersection

of open (

http://planetmath.org/OpenSet

)

dense sets

.

Title

residual

Canonical

name

Residual

Date of creation

2013-03-22 13:04:05

Last modified on

2013-03-22 13:04:05

Owner

mps (409)

Last modified by

mps (409)

Numerical id

7

Author

mps (409)

Entry type

Definition

Classification

msc 54E52

Related topic

BaireCategoryTheorem

Related topic

SardsTheorem

Defines

comeager

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

LaTeXML
ℝ^n is not a countable union of proper vector subspaces

ℝ

n

is not a countable union of proper vector subspaces

ℝ

n

is not a

countable

union of

proper vector subspaces

.

Proof

We know that every finite dimensional proper

subspace

of a

normed space

is nowhere dense. Besides,

ℝ

n

is a

Banach space

, so the results follows directly.

Title

ℝ

n

is not a countable union of proper vector subspaces

Canonical name

mathbbRnIsNotACountableUnionOfProperVectorSubspaces

Date of creation

2013-03-22 14:59:03

Last modified on

2013-03-22 14:59:03

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

6

Author

rspuzio (6075)

Entry type

Result

Classification

msc 54E52

Generated on Sat Feb 10 11:29:31 2018 by

LaTeXML
characterization of compactly generated space

characterization of compactly generated space

We give

equivalent

conditions when a

Hausdorff topological space

is

compactly generated

. First we need a definition.

Definition.

Let

X

be a

T

1

space and

Y

a

topological space

. We define

C

k

⁢

(

X

,

Y

)

to be the set of

functions

f

:

X

→

Y

such that for every

compact

,

closed set

K

⊂

X

the

restriction

f

|

K

is

continuous

.

Clearly,

C

⁢

(

X

,

Y

)

⊂

C

k

⁢

(

X

,

Y

)

for such spaces

X

,

Y

. With this we have the following

theorem

.

Theorem.

Let

X

be a Hausdorff space. Then the following conditions are equivalent.

i)

X

is compactly generated

ii)

X

carries the

final topology

generated by the family of

inclusion mappings

(

i

K

:

K

→

X

)

K

⊂

X

⁢

compact

.

iii)

For every topological space

Y

we have

C

k

⁢

(

X

,

Y

)

=

C

⁢

(

X

,

Y

)

.

iv)

X

is an image of a locally compact space under a quotient mapping.

Remark.

It follows easily from this that if there is a quotient mapping

f

:

X

→

Y

which maps a compactly generated space

X

onto a Hausdorff space

Y

then

Y

is compactly generated.

Title

characterization of compactly generated space

Canonical name

CharacterizationOfCompactlyGeneratedSpace

Date of creation

2013-03-22 19:10:04

Last modified on

2013-03-22 19:10:04

Owner

karstenb (16623)

Last modified by

karstenb (16623)

Numerical id

4

Author

karstenb (16623)

Entry type

Theorem

Classification

msc 54E99

Defines

C_k(X

Defines

Y)

Generated on Sat Feb 10 11:29:42 2018 by

LaTeXML
compactly generated

compactly generated

A

Hausdorff topological space

X

is said to be

compactly generated

if for each

A

⊂

X

,

A

is closed if and only if

A

∩

C

is closed for every

compact

C

⊂

X

.

Title

compactly generated

Canonical name

CompactlyGenerated

Date of creation

2013-03-22 12:16:00

Last modified on

2013-03-22 12:16:00

Owner

RevBobo (4)

Last modified by

RevBobo (4)

Numerical id

5

Author

RevBobo (4)

Entry type

Definition

Classification

msc 54E99

Generated on Sat Feb 10 11:29:43 2018 by

LaTeXML