proof that forcing notions are equivalent to their composition

proof that forcing notions are equivalent to their composition

This is a long and complicated proof, the more so because the meaning of

Q

shifts depending on what

generic

subset of

P

is being used. It is therefore broken into a number of steps. The core of the proof is to prove that, given any generic subset

G

of

P

and a generic subset

H

of

Q

^

⁢

[

G

]

there is a corresponding generic subset

G

*

H

of

P

*

Q

such that

𝔐

⁢

[

G

]

⁢

[

H

]

=

𝔐

⁢

[

G

*

H

]

, and

conversely

, given any generic subset

G

of

P

*

Q

we can find some generic

G

P

of

P

and a generic

G

Q

of

Q

^

⁢

[

G

P

]

such that

𝔐

⁢

[

G

P

]

⁢

[

G

Q

]

=

𝔐

⁢

[

G

]

.

We do this by constructing

functions

using

operations

which can be performed within the forced

universes

so that, for example, since

𝔐

⁢

[

G

]

⁢

[

H

]

has both

G

and

H

,

G

*

H

can be calculated, proving that it contains

𝔐

⁢

[

G

*

H

]

. To ensure equality, we will also have to ensure that our operations are

inverses

; that is, given

G

,

G

P

*

G

H

=

G

and given

G

and

H

,

(

G

*

H

)

P

=

P

and

(

G

*

H

)

Q

=

H

.

The

remainder

of the proof merely defines the precise operations, proves that they give generic sets, and proves that they are inverses.

Before beginning, we prove a lemma which comes up several times:

Lemma: If

G

is generic in

P

and

D

is dense above some

p

∈

G

then

G

∩

D

≠

∅

Let

D

′

=

{

p

′

∈

P

∣

p

′

∈

D

∨

p

′

⁢

is incompatible with

⁢

p

}

. This is dense, since if

p

0

∈

P

then either

p

0

is incompatible with

p

, in which case

p

0

∈

D

′

, or there is some

p

1

such that

p

1

≤

p

,

p

0

, and therefore there is some

p

2

≤

p

1

such that

p

2

∈

D

, and therefore

p

2

≤

p

0

. So

G

intersects

D

′

. But since a generic set is directed, no two

elements

are incompatible, so

G

must contain an element of

D

′

which is not incompatible with

p

, so it must contain an element of

D

.

G

*

H

is a generic filter

First, given generic subsets

G

and

H

of

P

and

Q

^

⁢

[

G

]

, we can define:

G

*

H

=

{

⟨

p

,

q

^

⟩

∣

p

∈

G

∧

q

^

⁢

[

G

]

∈

H

}

G

*

H

is closed

Let

⟨

p

1

,

q

^

1

⟩

∈

G

*

H

and let

⟨

p

1

,

q

^

1

⟩

≤

⟨

p

2

,

q

^

2

⟩

. Then we can conclude

p

1

∈

G

,

p

1

≤

p

2

,

q

^

1

⁢

[

G

]

∈

H

, and

p

1

⊩

q

^

1

≤

q

^

2

, so

p

2

∈

G

(since

G

is closed) and

q

^

2

⁢

[

G

]

∈

H

since

p

1

∈

G

and

p

1

forces both

q

^

1

≤

q

^

2

and that

H

is downward closed. So

⟨

p

2

,

q

^

2

⟩

∈

G

*

H

.

G

*

H

is directed

Suppose

⟨

p

1

,

q

^

1

⟩

,

⟨

p

1

,

q

^

1

⟩

∈

G

*

H

. So

p

1

,

p

2

∈

G

, and since

G

is directed, there is some

p

3

≤

p

1

,

p

2

. Since

q

^

1

⁢

[

G

]

,

q

^

2

⁢

[

G

]

∈

H

and

H

is directed, there is some

q

^

3

⁢

[

G

]

≤

q

^

1

⁢

[

G

]

,

q

^

2

⁢

[

G

]

. Therefore there is some

p

4

≤

p

3

,

p

4

∈

G

, such that

p

4

⊩

q

^

3

≤

q

^

1

,

q

^

2

, so

⟨

p

4

,

q

^

3

⟩

≤

⟨

p

1

,

q

^

1

⟩

,

⟨

p

1

,

q

^

1

⟩

and

⟨

p

4

,

q

^

3

⟩

∈

G

*

H

.

G

*

H

is generic

Suppose

D

is a dense subset of

P

*

Q

^

. We can project it into a dense subset of

Q

using

G

:

D

Q

=

{

q

^

⁢

[

G

]

∣

⟨

p

,

q

^

⟩

∈

D

}

⁢

for some

⁢

p

∈

G

Lemma:

D

Q

is dense in

Q

^

⁢

[

G

]

Given any

q

^

0

∈

Q

^

, take any

p

0

∈

G

. Then we can define yet another dense subset, this one in

G

:

D

q

^

0

=

{

p

∣

p

≤

p

0

∧

p

⊩

q

^

≤

q

^

0

∧

⟨

p

,

q

^

⟩

∈

D

}

⁢

for some

⁢

q

^

∈

Q

^

Lemma:

D

q

^

0

is dense above

p

0

in

P

Take any

p

∈

P

such that

p

≤

p

0

. Then, since

D

is dense in

P

*

Q

^

, we have some

⟨

p

1

,

q

^

1

⟩

≤

⟨

p

,

q

^

0

⟩

such that

⟨

p

1

,

q

^

1

⟩

∈

D

. Then by

definition

p

1

≤

p

and

p

1

∈

D

q

^

0

.

From this lemma, we can conclude that there is some

p

1

≤

p

0

such that

p

1

∈

G

∩

D

q

^

0

, and therefore some

q

^

1

such that

p

1

⊩

q

^

1

≤

q

^

0

where

⟨

p

1

,

q

^

1

⟩

∈

D

. So

D

Q

is indeed dense in

Q

^

⁢

[

G

]

.

Since

D

Q

is dense in

Q

^

⁢

[

G

]

, there is some

q

^

such that

q

^

⁢

[

G

]

∈

D

Q

∩

H

, and so some

p

∈

G

such that

⟨

p

,

q

^

⟩

∈

D

. But since

p

∈

G

and

q

^

∈

H

,

⟨

p

,

q

^

⟩

∈

G

*

H

, so

G

*

H

is indeed generic.

G

P

is a generic filter

Given some generic subset

G

of

P

*

Q

^

, let:

G

P

=

{

p

∈

P

∣

p

′

≤

p

∧

⟨

p

′

,

q

^

⟩

∈

G

}

⁢

for some

⁢

p

′

∈

P

⁢

and some

⁢

q

^

∈

Q

G

P

is closed

Take any

p

1

∈

G

P

and any

p

2

such that

p

1

≤

p

2

. Then there is some

p

′

≤

p

1

satisfying the definition of

G

P

, and also

p

′

≤

p

2

, so

p

2

∈

G

P

.

G

P

is directed

Consider

p

1

,

p

2

∈

G

P

. Then there is some

p

1

′

and some

q

^

1

such that

⟨

p

1

′

,

q

^

1

⟩

∈

G

and some

p

2

′

and some

q

^

2

such that

⟨

p

2

′

,

q

^

2

⟩

∈

G

. Since

G

is directed, there is some

⟨

p

3

,

q

^

3

⟩

∈

G

such that

⟨

p

3

,

q

^

3

⟩

≤

⟨

p

1

′

,

q

^

1

⟩

,

⟨

p

2

′

,

q

^

2

⟩

, and therefore

p

3

∈

G

P

,

p

3

≤

p

1

,

p

2

.

G

P

is generic

Let

D

be a dense subset of

P

. Then

D

′

=

{

⟨

p

,

q

^

⟩

∣

p

∈

D

}

. Clearly this is dense, since if

⟨

p

,

q

^

⟩

∈

P

*

Q

^

then there is some

p

′

≤

p

such that

p

′

∈

D

, so

⟨

p

′

,

q

^

⟩

∈

D

′

and

⟨

p

′

,

q

^

⟩

≤

⟨

p

,

q

^

⟩

. So there is some

⟨

p

,

q

^

⟩

∈

D

′

∩

G

, and therefore

p

∈

D

∩

G

P

. So

G

P

is generic.

G

Q

is a generic filter

Given a generic subset

G

⊆

P

*

Q

^

, define:

G

Q

=

{

q

^

⁢

[

G

P

]

∣

⟨

p

,

q

^

⟩

∈

G

}

⁢

for some

⁢

p

∈

P

(Notice that

G

Q

is dependant on

G

P

, and is a subset of

Q

^

⁢

[

G

P

]

, that is, the

forcing

notion inside

𝔐

⁢

[

G

P

]

, as opposed to the set of names

Q

which we’ve been primarily working with.)

G

Q

is closed

Suppose

q

^

1

⁢

[

G

P

]

∈

G

Q

and

q

^

1

⁢

[

G

P

]

≤

q

^

2

⁢

[

G

P

]

. Then there is some

p

1

∈

G

P

such that

p

1

⊩

q

^

1

≤

q

^

2

. Since

p

1

∈

G

P

, there is some

p

2

≤

p

1

such that for some

q

^

3

,

⟨

p

2

,

q

^

3

⟩

∈

G

. By the definition of

G

Q

, there is some

p

3

such that

⟨

p

3

,

q

^

1

⟩

∈

G

, and since

G

is directed, there is some

⟨

p

4

,

q

^

4

⟩

∈

G

and

⟨

p

4

,

q

^

4

⟩

≤

⟨

p

3

,

q

^

1

⟩

,

⟨

p

2

,

q

^

3

⟩

. Since

G

is closed and

⟨

p

4

,

q

^

4

⟩

≤

⟨

p

4

,

q

^

2

⟩

, we have

q

^

2

⁢

[

G

P

]

∈

G

Q

.

G

Q

is directed

Suppose

q

^

1

⁢

[

G

P

]

,

q

^

2

⁢

[

G

P

]

∈

G

Q

. Then for some

p

1

,

p

2

,

⟨

p

1

,

q

^

1

⟩

,

⟨

p

2

,

q

^

2

⟩

∈

G

, and since

G

is directed, there is some

⟨

p

3

,

q

^

3

⟩

∈

G

such that

⟨

p

3

,

q

^

3

⟩

≤

⟨

p

1

,

q

^

1

⟩

,

⟨

p

2

,

q

^

2

⟩

. Then

q

^

3

⁢

[

G

P

]

∈

G

Q

and since

p

3

∈

G

and

p

3

⊩

q

^

3

≤

q

^

1

,

q

^

2

, we have

q

^

3

⁢

[

G

P

]

≤

q

^

1

⁢

[

G

P

]

,

q

^

2

⁢

[

G

P

]

.

G

Q

is generic

Let

D

be a dense subset of

Q

⁢

[

G

P

]

(in

𝔐

⁢

[

G

P

]

). Let

D

^

be a

P

-name for

D

, and let

p

1

∈

G

P

be a such that

p

1

⊩

D

^

is dense

. By the definition of

G

P

, there is some

p

2

≤

p

1

such that

⟨

p

2

,

q

^

2

⟩

∈

G

for some

q

2

. Then

D

′

=

{

⟨

p

,

q

^

⟩

∣

p

⊩

q

^

∈

D

∧

p

≤

p

2

}

.

Lemma:

D

′

is dense (in

G

) above

⟨

p

2

,

q

^

2

⟩

Take any

⟨

p

,

q

^

⟩

∈

P

*

Q

such that

⟨

p

,

q

^

⟩

≤

⟨

p

2

,

q

^

2

⟩

. Then

p

⊩

D

^

is dense

, and therefore there is some

q

^

3

such that

p

⊩

q

^

3

∈

D

^

and

p

⊩

q

^

3

≤

q

^

. So

⟨

p

,

q

^

3

⟩

≤

⟨

p

,

q

^

⟩

and

⟨

p

,

q

^

3

⟩

∈

D

′

.

Take any

⟨

p

3

,

q

^

3

⟩

∈

D

′

∩

G

. Then

p

3

∈

G

P

, so

q

^

3

∈

D

, and by the definition of

G

Q

,

q

^

3

∈

G

Q

.

G

P

*

G

Q

=

G

If

G

is a generic subset of

P

*

Q

, observe that:

G

P

*

G

Q

=

{

⟨

p

,

q

^

⟩

∣

p

′

≤

p

∧

⟨

p

′

,

q

^

′

⟩

∈

G

∧

⟨

p

0

,

q

^

⟩

∈

G

}

⁢

for some

⁢

p

′

,

q

^

′

,

p

0

If

⟨

p

,

q

^

⟩

∈

G

then obviously this holds, so

G

⊆

G

P

*

G

Q

. Conversely, if

⟨

p

,

q

^

⟩

∈

G

P

*

G

Q

then there exist

p

′

,

q

^

′

and

p

0

such that

⟨

p

′

,

q

^

′

⟩

,

⟨

p

0

,

q

^

⟩

∈

G

, and since

G

is directed, some

⟨

p

1

,

q

1

^

⟩

∈

G

such that

⟨

p

1

,

q

^

1

⟩

≤

⟨

p

′

,

q

^

′

⟩

,

⟨

p

0

,

q

^

⟩

. But then

p

1

≤

p

and

p

1

⊩

q

^

1

≤

q

^

, and since

G

is closed,

⟨

p

,

q

^

⟩

∈

G

.

(

G

*

H

)

P

=

G

Assume that

G

is generic in

P

and

H

is generic in

Q

⁢

[

G

]

.

Suppose

p

∈

(

G

*

H

)

P

. Then there is some

p

′

∈

P

and some

q

^

∈

Q

such that

p

′

≤

p

and

⟨

p

′

,

q

^

⟩

∈

G

*

H

. By the definition of

G

*

H

,

p

′

∈

G

, and then since

G

is closed

p

∈

G

.

Conversely, suppose

p

∈

G

. Then (since

H

is non-trivial),

⟨

p

,

q

^

⟩

∈

G

*

H

for some

q

^

, and therefore

p

∈

(

G

*

H

)

P

.

(

G

*

H

)

Q

=

H

Assume that

G

is generic in

P

and

H

is generic in

Q

⁢

[

G

]

.

Given any

q

∈

H

, there is some

q

^

∈

Q

such that

q

^

⁢

[

G

]

=

q

, and so there is some

p

such that

⟨

p

,

q

^

⟩

∈

G

*

H

, and therefore

q

^

⁢

[

G

]

∈

H

.

On the other hand, if

q

∈

(

G

*

H

)

Q

then there is some

⟨

p

,

q

^

⟩

∈

G

*

H

, and therefore some

q

^

⁢

[

G

]

∈

H

.

Title

proof that forcing notions are equivalent to their composition

Canonical

name

ProofThatForcingNotionsAreEquivalentToTheirComposition

Date of creation

2013-03-22 12:54:26

Last modified on

2013-03-22 12:54:26

Owner

Henry (455)

Last modified by

Henry (455)

Numerical id

5

Author

Henry (455)

Entry type

Proof

Classification

msc 03E40

Classification

msc 03E35

Related topic

EquivalenceOfForcingNotions

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