The Main Results

Our context is the study of -models of subsystems of second
order arithmetic [5, Chapter VIII]. As in [5, Chapter
VII], a *-model* is an -model *M* such
that, for all
sentences
with parameters from
*M*,
is true if and only if
.
Theorems
1.1 and 1.3 below are an interesting
supplement to the results on -models which have been presented
in Simpson [5, §§VII.2 and VIII.6].

Let
HYP denote the set of hyperarithmetical reals. It is well
known that, for any -model *M*,
HYP is properly included in
*M*, and each
is definable in *M*.

if *X* is definable, then
.

*Proof.*Fix a recursive enumeration *S*_{e},
,
of the
sets of reals. If *p* is a finite subset of
,
say that
*meets* *p* if
for all
.
Let
be the set of *p* such that
there exists
meeting *p*. Put
if and only if
.
Say that
is *dense* if for all
there exists
such that
.
Say that
is *definable* if it is definable over
the -model
HYP, i.e., arithmetical in the complete
subset of .
Say that
is *generic* if for every dense
definable
there exists
such that
meets *p*. We can show that for every
there exists a generic
meeting *p*. (This is a fusion argument, a la Gandy forcing.)
Clearly
is a -model. We can also show
that, if *C* is countable and
,
then there
exists a generic
such that
.
Let
be the language of
second order arithmetic with additional set constants *X*_{n},
.
Let
be a sentence of
.
We say that *p* *forces* ,
written
,
if for all generic
meeting *p*, the -model
satisfies
.
It can be shown that, for all
generic
,
the -model
satisfies
if and only if
meets some *p* such that
.
If
is a permutation of ,
define an action of
on
and
by
and
.
It is
straightforward to show that
if and only if
.
The *support* of
is
defined by
.
Clearly if
and
,
then
.
We claim that if
and
are generic, then the -models
and
satisfy the same
*L*_{2}-sentences. Suppose not. Then for some
we have
and
,
for some
*L*_{2}-sentence .
Let
be a permutation of
such that
.
Since
,
we have
,
hence
,
a contradiction. This proves our
claim.
Finally, let
where
is generic. Suppose
is definable
in *M*. Let
be generic such that
has
.
Let
be
an *L*_{2}-formula with *X* as its only free variable, such that
exactly one
,
and
.
Then
exactly one
.
Let
be such that
.
Then for each
,
we have that
if and only if
and ,
if and only if
and ,
if and only if .
Thus *A*=*A*'. Hence
.
This
completes the proof.

We now improve Theorem 1.1 as follows.

Let
denote hyperarithmetical reducibility, i.e.,
if and only if *X* is hyperarithmetical in *Y*.

ifXis definable fromY, then .

The -model which we shall use to prove Theorem
1.3 is the same as for Theorem 1.1,
namely
where
is generic. In order to see that *M* has the desired property, we
first relativize the proof of Theorem 1.1, as follows.
Given *Y*, let
be the set of
such that there exists
meeting *p* with *X*_{0}=*Y*.
(Obviously 0 plays no special role here.) Say that
is *generic over* *Y* if, for every
dense
definable from *Y* over
,
there exists
such that
meets *p*.

*Proof.*The proof of this lemma is a straightforward relativization to *Y* of the proof of Theorem 1.1.

Consequently, in order to prove Theorem 1.3, it suffices to prove the following lemma.

*Proof.*It suffices to show that, for all *p* forcing
is dense
in
,
there exists
forcing
and
meets *r*).
Assume
is dense in
.
Since
,
it follows that
and
.
Fix
and
such that
.
Put
meets .
Then *S*' is a
set, so let
be such that *S*'=*S*_{e}. Claim 1:
.
If not, let
be such that
.
Let
be a permutation such that
and
.
Then
and
,
a
contradiction. This proves Claim 1.
Claim 2:
.
To see this, let
be generic meeting
.
By Claim 1 we have
.
Hence
,
i.e., there exists
meeting *q*' with *X*_{0}=*G*'_{0}. Thus
meets
.
This proves Claim 2. Finally, put
.
Then
and
and
meets *q*'). This proves our lemma.

The proof of Theorem 1.3 is immediate from Lemmas 1.4 and 1.5.