proof of equivalent definitions of analytic sets for Polish spaces

proof of equivalent definitions of analytic sets for Polish spaces

Let

A

be a nonempty subset of a

Polish space

X

. Then, letting

𝒩

denote

Baire space

and

Y

be any

uncountable

Polish space, we show that the following are

equivalent

.

1.

A

is

ℱ

-

analytic

(

http://planetmath.org/AnalyticSet2

).

2.

A

is the projection (

http://planetmath.org/GeneralizedCartesianProduct

) of a

closed subset

of

X

×

𝒩

onto

X

.

3.

A

is the image (

http://planetmath.org/DirectImage

) of a

continuous function

f

:

Z

→

X

for some Polish space

Z

.

4.

A

is the image of a continuous function

f

:

𝒩

→

X

.

5.

A

is the image of a

Borel measurable function

f

:

Y

→

X

.

6.

A

is the projection of a

Borel subset

of

X

×

Y

onto

X

.

(

1

) implies (

2

)

:
Let

ℱ

be the paving consisting of closed subsets of

X

. The

collection

of

𝒦

-

analytic sets

contains the Borel

σ

-

algebra

of

X

(see

countable unions and intersections of analytic sets are analytic

) and, as the analytic sets are given by a

closure operator

(

http://planetmath.org/AnalyticSetsDefineAClosureOperator

) it follows that it contains all analytic subsets of

X

. So, any analytic subset

A

of

X

is

ℱ

-analytic.
Then, there is a closed

S

⊆

𝒩

and a

function

θ

:

ℕ

→

ℱ

such that

A

=

⋃

s

∈

S

⋂

n

=

1

∞

θ

⁢

(

s

n

)

(see

proof of equivalent definitions of analytic sets for paved spaces

).
For each

m

,

n

∈

ℕ

let

K

m

,

n

denote the closed subset of

s

∈

𝒩

with

s

n

=

m

. Then, we can rearrange the above

expression

to get

A

=

π

X

⁢

(

B

)

where

π

X

:

X

×

𝒩

→

X

is the

projection map

and

B

=

(

X

×

S

)

∩

⋂

n

=

1

∞

⋃

m

=

1

∞

θ

⁢

(

m

)

×

K

m

,

n

.

It is easily seen that

⋃

m

θ

⁢

(

m

)

×

K

m

,

n

is a closed subset of

X

×

𝒩

for each

n

, and therefore

B

is closed, as required.

(

2

) implies (

3

)

:
Suppose that

A

=

π

X

⁢

(

S

)

for a closed subset

S

of

X

×

𝒩

, where

π

X

:

X

×

𝒩

→

X

is the projection map. As the

product

of Polish spaces is Polish, and every closed subset of a Polish space is Polish, then

S

will be a Polish space under the

subspace topology

. So, we can take

Z

=

S

and let

f

:

Z

→

X

be the

restriction

of

π

X

to

Z

.

(

3

) implies (

4

)

:
Suppose that

A

is the image of a continuous function

g

:

Z

→

X

, for a Polish space

Z

. As

Baire space is universal for Polish spaces

, there exists a

continuous

and onto (

http://planetmath.org/

Surjective

) function

h

:

𝒩

→

Z

. The result follows by taking

f

=

g

∘

h

.

(

4

) implies (

5

)

:
Suppose that

A

is the image of a continuous function

g

:

𝒩

→

X

. Since uncountable Polish spaces are all Borel

isomorphic

(see

Polish spaces up to Borel isomorphism

), there is a

Borel isomorphism

h

:

Y

→

𝒩

. The result follows by taking

f

=

g

∘

h

.

(

5

) implies (

6

)

:
Suppose that

A

is the image of a Borel measurable function

f

:

Y

→

X

, and let

Γ

be its graph (

http://planetmath.org/Graph2

)

Γ

≡

{

(

f

⁢

(

y

)

,

y

)

:

y

∈

Y

}

⊆

X

×

Y

.

The projection of

Γ

onto

X

is equal to

f

⁢

(

Y

)

=

A

, so the result will follow once it is shown that

Γ

is a Borel set.

Choose a

countable

and

dense subset

{

x

1

,

x

2

,

…

}

of

X

, and let

d

be a metric generating the

topology

on

X

. Then, for integers

m

,

n

≥

1

, denote the

open ball

about

x

m

of radius

1

/

n

by

B

m

,

n

. Since the

x

m

form a dense set,

⋃

m

B

m

,

n

=

X

for each

n

. Let us define

Γ

n

≡

⋃

m

=

1

∞

B

m

,

n

×

f

-

1

⁢

(

B

m

,

n

)

⊆

X

×

Y

,

which contains

Γ

. Furthermore, since

f

-

1

⁢

(

B

m

,

n

)

are Borel,

Γ

n

are Borel sets.
Suppose that

(

x

,

y

)

∈

⋂

n

Γ

n

. Then, for each

n

, there is an

m

such that

x

∈

B

m

,

n

and

y

∈

f

-

1

⁢

(

B

m

,

n

)

. So,

d

⁢

(

x

,

f

⁢

(

y

)

)

≤

d

⁢

(

x

,

x

m

)

+

d

⁢

(

x

m

,

f

⁢

(

y

)

)

≤

2

/

n

.

This holds for all

n

, showing that

y

=

f

⁢

(

x

)

and so

(

x

,

y

)

∈

Γ

. We have shown that

Γ

=

⋂

n

Γ

n

is Borel, as required.

(

6

) implies (

1

)

:
This is an immediate

consequence

of the result that

projections of analytic sets are analytic

.

Title

proof of equivalent definitions of analytic sets for Polish spaces

Canonical

name

ProofOfEquivalentDefinitionsOfAnalyticSetsForPolishSpaces

Date of creation

2013-03-22 18:48:48

Last modified on

2013-03-22 18:48:48

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 28A05

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