proof of equivalent definitions of analytic sets for paved spaces

proof of equivalent definitions of analytic sets for paved spaces

Let

(

X

,

ℱ

)

be a

paved space

with

∅

∈

ℱ

, let

𝒩

be

Baire space

, and let

Y

be any

uncountable

Polish space

. For a subset

A

of

X

, we show that the following statements are

equivalent

.

1.

A

is

ℱ

-

analytic

(

http://planetmath.org/AnalyticSet2

).

2.

There is a

closed subset

S

of

𝒩

and

θ

:

ℕ

2

→

ℱ

such that

A

=

⋃

s

∈

S

⋂

n

=

1

∞

θ

⁢

(

n

,

s

n

)

.

3.

There is a closed subset

S

of

𝒩

and

θ

:

ℕ

→

ℱ

such that

A

=

⋃

s

∈

S

⋂

n

=

1

∞

θ

⁢

(

s

n

)

.

4.

A

is the result of a

Souslin scheme

on

ℱ

.

5.

A

is the projection of a set in

(

ℱ

×

𝒦

)

σ

⁢

δ

onto

X

, where

𝒦

is the

collection

of

compact subsets

of

Y

.

6.

A

is the projection of a set in

(

ℱ

×

𝒢

)

σ

⁢

δ

onto

X

, where

𝒢

is the collection of closed subsets of

Y

.

(

1

) implies (

2

)

:
As

A

is analytic, there exists a

compact

paved space

(

K

,

𝒦

)

and a set

B

∈

(

ℱ

×

𝒦

)

σ

⁢

δ

such that

A

=

π

X

⁢

(

B

)

, where

π

X

:

X

×

K

→

X

is the

projection map

.
Write

B

=

⋂

n

=

1

∞

⋃

m

=

1

∞

A

n

,

m

×

K

n

,

m

for

A

n

,

m

∈

ℱ

and

K

n

,

m

∈

𝒦

.
Rearranging this expression,

B

=

⋃

s

∈

𝒩

⋂

n

=

1

∞

A

n

,

s

n

×

K

n

,

s

n

.

So, defining

S

⊆

𝒩

by

S

=

{

s

∈

𝒩

:

⋂

n

=

1

∞

K

n

,

s

n

≠

∅

}

.

gives

A

=

π

X

⁢

(

B

)

=

⋃

s

∈

S

⋂

n

=

1

∞

A

n

,

s

n

.

Setting

θ

⁢

(

n

,

m

)

=

A

n

,

m

gives the required expression, and it only remains to show that

S

is closed.
So, let

s

1

,

s

2

,

…

be a

sequence

in

S

converging to a limit

s

∈

𝒩

. For any

k

≥

0

then

s

n

r

=

s

n

for all

n

≤

k

and large enough

r

. Hence,

⋂

n

≤

k

K

n

,

s

n

=

⋂

n

≤

k

K

n

,

s

n

r

⊇

⋂

n

=

1

∞

K

n

,

s

n

≠

∅

.

So, the collection of sets

K

n

,

s

n

for

n

=

1

,

2

,

…

satisfies

the

finite intersection property

, and

compactness

(

http://planetmath.org/PavedSpace

) of the paving

𝒦

gives

⋂

n

=

1

∞

K

n

,

s

n

≠

∅

,

showing that

s

∈

S

and that

S

is indeed closed.

(

2

) implies (

3

)

:
Supposing that

A

satisfies the required expression, choose any

bijection

ϕ

:

ℕ

→

ℕ

2

. Then define

θ

~

≡

θ

∘

ϕ

and

f

:

𝒩

→

𝒩

by

f

⁢

(

s

)

=

t

where

t

n

=

ϕ

-

1

⁢

(

n

,

s

n

)

. As

S

is closed, it follows that

S

~

=

f

⁢

(

S

)

will also be closed and,

A

=

⋃

s

∈

S

⋂

n

θ

⁢

(

n

,

s

n

)

=

⋃

s

∈

S

⋂

n

θ

~

⁢

(

ϕ

-

1

⁢

(

n

,

s

n

)

)

=

⋃

s

∈

S

~

⋂

n

θ

~

⁢

(

s

n

)

as required.

(

3

) implies (

4

)

:
Suppose that

A

satisfies the required expression and define a Souslin scheme

(

A

s

)

as follows. For any

n

≥

1

and

s

∈

ℕ

n

let us set

A

s

=

{

θ

⁢

(

s

n

)

,

if

s

=

t

|

n

for some

t

∈

𝒩

,

∅

,

otherwise

.

Then, for

s

∈

𝒩

,

⋂

n

=

1

∞

A

s

|

n

=

{

⋂

n

=

1

∞

θ

⁢

(

s

n

)

,

if

s

∈

S

,

∅

,

otherwise

.

Here, if

s

∉

S

, we have used the fact that

S

is closed to deduce that for large

n

, there is no

t

∈

S

with

t

|

n

=

s

|

n

and, therefore,

A

s

|

n

=

∅

.
The result of the Souslin scheme

(

A

s

)

is then

⋃

s

∈

𝒩

⋂

n

=

1

∞

A

s

|

n

=

⋃

s

∈

S

⋂

n

=

1

∞

θ

⁢

(

s

n

)

=

A

as required.

(

4

) implies (

5

)

:
Suppose that

A

is the result of a Souslin scheme

(

A

s

)

. Let us first consider the case where

Y

is

Cantor space

,

𝒞

=

{

0

,

1

}

ℕ

, which is a compact Polish space.
Then, for any

s

∈

ℕ

n

, let

K

s

be the set of

t

∈

𝒞

such that

t

k

=

1

if

k

=

s

1

+

⋯

+

s

m

for some

m

≤

n

and

t

k

=

0

for all other

k

<

s

1

+

⋯

+

s

n

.
These are closed and, therefore, compact sets.

Given any sequence

s

1

∈

ℕ

1

,

s

2

∈

ℕ

2

,

…

it is easily seen that

⋂

n

K

s

n

is nonempty if and only if there is an

s

∈

𝒩

such that

s

|

n

=

s

n

for all

n

.
Define the set

B

in

(

ℱ

×

𝒦

)

σ

⁢

δ

by

B

=

⋂

n

=

1

∞

⋃

s

∈

ℕ

n

A

s

×

K

s

=

⋃

s

1

∈

ℕ

1

,

s

2

∈

ℕ

2

,

…

⋂

n

=

1

∞

A

s

n

×

K

s

n

=

⋃

s

∈

𝒩

⋂

n

=

1

∞

A

s

|

n

×

K

s

|

n

.

The projection of

B

onto

X

is then

π

X

⁢

(

S

)

=

⋃

s

∈

𝒩

⋂

n

=

1

∞

A

s

|

n

,

which is the result

A

of the scheme

(

A

s

)

as required.
The result then generalizes to any uncountable Polish space

Y

, as all such spaces contain Cantor space (

http://planetmath.org/UncountablePolishSpacesContainCantorSpace

).

(

5

) implies (

6

)

:
This is

trivial

, since all compact sets are closed.

(

6

) implies (

1

)

:
This is a

consequence

of the result that

projections of analytic sets are analytic

.

Title

proof of equivalent definitions of analytic sets for paved spaces

Canonical name

ProofOfEquivalentDefinitionsOfAnalyticSetsForPavedSpaces

Date of creation

2013-03-22 18:48:36

Last modified on

2013-03-22 18:48:36

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 28A05

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