forcings are equivalent if one is dense in the other

forcings are equivalent if one is dense in the other

Suppose

P

and

Q

are

forcing

notions and that

f

:

P

→

Q

is a

function

such that:

•

p

1

≤

P

p

2

implies

f

⁢

(

p

1

)

≤

Q

f

⁢

(

p

2

)

•

If

p

1

,

p

2

∈

P

are

incomparable

then

f

⁢

(

p

1

)

,

f

⁢

(

p

2

)

are incomparable

•

f

⁢

[

P

]

is dense (

http://planetmath.org/DenseInAPoset

) in

Q

then

P

and

Q

are

equivalent

.

Proof

We seek to provide two

operations

(

computable

in the appropriate

universes

) which convert between

generic

subsets of

P

and

Q

, and to prove that they are

inverses

.

F

⁢

(

G

)

=

H

where

H

is generic

Given a generic

G

⊆

P

, consider

H

=

{

q

∣

f

⁢

(

p

)

≤

q

}

for some

p

∈

G

.

If

q

1

∈

H

and

q

1

≤

q

2

then

q

2

∈

H

by the definition of

H

. If

q

1

,

q

2

∈

H

then let

p

1

,

p

2

∈

P

be such that

f

⁢

(

p

1

)

≤

q

1

and

f

⁢

(

p

2

)

≤

q

2

. Then there is some

p

3

≤

p

1

,

p

2

such that

p

3

∈

G

, and since

f

is order preseving

f

⁢

(

p

3

)

≤

f

⁢

(

p

1

)

≤

q

1

and

f

⁢

(

p

3

)

≤

f

⁢

(

p

2

)

≤

q

2

.

Suppose

D

is a

dense subset

of

Q

. Since

f

⁢

[

P

]

is

dense in

Q

, for any

d

∈

D

there is some

p

∈

P

such that

f

⁢

(

p

)

≤

d

. For each

d

∈

D

, assign (using the

axiom of choice

) some

d

p

∈

P

such that

f

⁢

(

d

p

)

≤

d

, and call the set of these

D

P

. This is dense in

P

, since for any

p

∈

P

there is some

d

∈

D

such that

d

≤

f

⁢

(

p

)

, and so some

d

p

∈

D

P

such that

f

⁢

(

d

p

)

≤

d

. If

d

p

≤

p

then

D

P

is dense, so suppose

d

p

≰

p

. If

d

p

≤

p

then this provides a member of

D

P

less than

p

; alternatively, since

f

⁢

(

d

p

)

and

f

⁢

(

p

)

are

compatible

,

d

p

and

p

are compatible, so

p

≤

d

p

, and therefore

f

⁢

(

p

)

=

f

⁢

(

d

p

)

=

d

, so

p

∈

D

P

. Since

D

P

is dense in

P

, there is some

element

p

∈

D

P

∩

G

. Since

p

∈

D

P

, there is some

d

∈

D

such that

f

⁢

(

p

)

≤

d

. But since

p

∈

G

,

d

∈

H

, so

H

intersects

D

.

G

can be recovered from

F

⁢

(

G

)

Given

H

constructed as above, we can recover

G

as the set of

p

∈

P

such that

f

⁢

(

p

)

∈

H

. Obviously every element from

G

is included in the new set, so consider some

p

such that

f

⁢

(

p

)

∈

H

. By definition, there is some

p

1

∈

G

such that

f

⁢

(

p

1

)

≤

f

⁢

(

p

)

. Take some dense

D

∈

Q

such that there is no

d

∈

D

such that

f

⁢

(

p

)

≤

d

(this can be done easily be taking any dense subset and removing all such elements; the resulting set is still dense since there is some

d

1

such that

d

1

≤

f

⁢

(

p

)

≤

d

). This set intersects

f

⁢

[

G

]

in some

q

, so there is some

p

2

∈

G

such that

f

⁢

(

p

2

)

≤

q

, and since

G

is directed, some

p

3

∈

G

such that

p

3

≤

p

2

,

p

1

. So

f

⁢

(

p

3

)

≤

f

⁢

(

p

1

)

≤

f

⁢

(

p

)

. If

p

3

≰

p

then we would have

p

≤

p

3

and then

f

⁢

(

p

)

≤

f

⁢

(

p

3

)

≤

q

, contradicting the definition of

D

, so

p

3

≤

p

and

p

∈

G

since

G

is directed.

F

-

1

⁢

(

H

)

=

G

where

G

is generic

Given any generic

H

in

Q

, we define a corresponding

G

as above:

G

=

{

p

∈

P

∣

f

⁢

(

p

)

∈

H

}

. If

p

1

∈

G

and

p

1

≤

p

2

then

f

⁢

(

p

1

)

∈

H

and

f

⁢

(

p

1

)

≤

f

⁢

(

p

2

)

, so

p

2

∈

G

since

H

is directed. If

p

1

,

p

2

∈

G

then

f

⁢

(

p

1

)

,

f

⁢

(

p

2

)

∈

H

and there is some

q

∈

H

such that

q

≤

f

⁢

(

p

1

)

,

f

⁢

(

p

2

)

.

Consider

D

, the set of elements of

Q

which are

f

⁢

(

p

)

for some

p

∈

P

and either

f

⁢

(

p

)

≤

q

or there is no element greater than both

f

⁢

(

p

)

and

q

. This is dense, since given any

q

1

∈

Q

, if

q

1

≤

q

then (since

f

⁢

[

P

]

is dense) there is some

p

such that

f

⁢

(

p

)

≤

q

1

≤

q

. If

q

≤

q

1

then there is some

p

such that

f

⁢

(

p

)

≤

q

≤

q

1

. If neither of these and

q

there is some

r

≤

q

1

,

q

then any

p

such that

f

⁢

(

p

)

≤

r

suffices, and if there is no such

r

then any

p

such that

f

⁢

(

p

)

≤

q

suffices.

There is some

f

⁢

(

p

)

∈

D

∩

H

, and so

p

∈

G

. Since

H

is directed, there is some

r

≤

f

⁢

(

p

)

,

q

, so

f

⁢

(

p

)

≤

q

≤

f

⁢

(

p

1

)

,

f

⁢

(

p

2

)

. If it is not the case that

f

⁢

(

p

)

≤

f

⁢

(

p

1

)

then

f

⁢

(

p

)

=

f

⁢

(

p

1

)

=

f

⁢

(

p

2

)

. In either case, we confirm that

H

is directed.

Finally, let

D

be a dense subset of

P

.

f

⁢

[

D

]

is dense in

Q

, since given any

q

∈

Q

, there is some

p

∈

P

such that

p

≤

q

, and some

d

∈

D

such that

d

≤

p

≤

q

. So there is some

f

⁢

(

p

)

∈

f

⁢

[

D

]

∩

H

, and so

p

∈

D

∩

G

.

H

can be recovered from

F

-

1

⁢

(

H

)

Finally, given

G

constructed by this method,

H

=

{

q

∣

f

⁢

(

p

)

≤

q

}

for some

p

∈

G

. To see this, if there is some

f

⁢

(

p

)

for

p

∈

G

such that

f

⁢

(

p

)

≤

q

then

f

⁢

(

p

)

∈

H

so

q

∈

H

. On the other hand, if

q

∈

H

then the set of

f

⁢

(

p

)

such that either

f

⁢

(

p

)

≤

q

or there is no

r

∈

Q

such that

r

≤

q

,

f

⁢

(

p

)

is dense (as shown above), and so intersects

H

. But since

H

is directed, it must be that there is some

f

⁢

(

p

)

∈

H

such that

f

⁢

(

p

)

≤

q

, and therefore

p

∈

G

.

Title

forcings are equivalent if one is dense in the other

Canonical name

ForcingsAreEquivalentIfOneIsDenseInTheOther

Date of creation

2013-03-22 12:54:43

Last modified on

2013-03-22 12:54:43

Owner

Henry (455)

Last modified by

Henry (455)

Numerical id

6

Author

Henry (455)

Entry type

Result

Classification

msc 03E35

Classification

msc 03E40

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