16.1
Let be a partial order, a name for a partial order.
Let be the completion of and a name for the
completion of . Let . Any can be viewed as
an element of and is mapped to some by the embedding
of lemma 16.3. For any such that ,
we can find a name such that
and . Again, we can
associate the corresponding element .
Consequently, we can define a mapping
by
.
Let .
Suppose . Then
and .
But and so
.
This inequality is clearly forced by since
and so
that is .
Conversely, if
then
in particular
.
Since , we get
and so
.
Moreover,
. But
we have . By definition of the
embedding we have (in ),
and so
. This means that
. But
that is
and finally
. Hence in , and a fortiori in
, we have . Finally .
By the previous paragraph, if
then
and
. Hence
by the usual identification
(see for example Akihiro Kanamori’s book ”The Higher Infinite”, p 123)
and so the mapping is one-to-one.
It remains to show that the embedding is dense. If and
then we can find such that
. Let . We have
. Hence
.
Finally, since is a complete boolean algebra, we have
.
16.2
Consider the application from to defined
by . We have
that is
and so the application is well-defined.
If then and
so the application is one-to-one.
If then
and so
and
. Conversely, if this latter
equality holds, then
and . But
and so actually
and consequently .
So we get an embedding from to .
To show that the embedding is dense, consider
that is , and
. In particular, we have
.
Hence there is and such that
. Thus .
16.4
Let be a name for a subset of and
. Let is such that
. Then and so .
Let such that . Then for such a we have
and so that is
. Hence and . Hence
and is complete.
Let such that .
We have
.
If then and so
. Conversely, if this later equality is true,
. Otherwise,
and we could find with both
and . Hence
. As a consequence
and so
.
16.5
Let be defined by .
If then
and so . If
and then
so and so
. Hence is compatible
with and . So satisfies the
hypothesis of Lemma 15.45 and .
16.6
Suppose has property (K) and .
Let be an uncountable set of pairwise
distinct elements of that is
.
Assume there is at most countably many distinct . Then by the
pigeon principle, there is
uncountable on which all the are equal.
Otherwise, since has property (K), we can still find
uncountable on which the are pairwise compatible
(but pairwise distinct).
Suppose any subset of on which
is at most countable.
Then by the pigeon principle, there is
uncountable and such that .
In particular, we are in the case where the are pairwise
distinct or otherwise for any distinct , we get
and .
Then is uncountable set of pairwise
distinct and compatible elements of .
Otherwise, let uncountable on which
.
Since satisfies the ccc, is still uncountable in and since
, there is
uncountable such that ,
.
Then the set is uncountable set of
pairwise distinct (because ) of .
For any , we can find (because
) and such that
(because is full).
Then and so the
are pairwise compatible.
Finally, has property (K).
16.9
Let be a finite support iteration. The first part is a particular
case of exercise 16.17. So let’s show by induction on that
for all , .
The successor step is deduced from theorem 16.2 so suppose is limit.
Let . Suppose and
. Let such that .
Define by
if and
if . We have
and so . Finally .
Suppose instead that and that
are such that and .
Let such that . Then
and .
Finally, if is dense then
is dense in .
So and a fortiori .
Hence is a generic filter on .
16.10
Let be the notion of forcing producing a Cohen generic real. satisfies
the countable chain condition. For each
define .
Each is dense since any function can be extended to some
such that for some
. If
then there is such that .
Then so
. Hence
is a family of dense subset of size .
We adjoin to this family, the countable family of dense subsets
.
If is generic for the final family then
is a function and there is .
. A contradiction.
16.13
This is the same as Lemma 16.12. The set mentioned in the lemma can be
defined by where is defined by
induction:
and
where is a refinement of if there is one (and an arbitrary
element of , say , otherwise). Each is at most countable and
is of size at most so is of size at most
. We deduce that each is is of size at most and
finally . The rest of the proof is not changed.
16.14
Let be a Suslin tree, the associated forcing notion and let
be the notion of forcing that adjoins Cohen generic reals.
and satisfy ccc. In particular, preserves cardinal and
the definition of remains the same in any generic extension:
. We apply corollary 16.6 in
the two directions to deduce that satisfy ccc first and then
that .
Suppose that is actually a normal Suslin tree. Then since preserves
and since and other concepts involved in (9.9) are absolute,
remains a normal -tree in . By the previous paragraph,
satisfy the countable chain condition in
and by exercise 9.7, we deduce that is a Suslin tree in .
We have and so the existence of Suslin
tree is consistent with .
16.15
Let be a generic filter for the forcing notion that collapses
to and . In , is preserved
and is collapsed. is of size
and so is its transitive closure. Let be isomorphic
to and define .
We have and so . Conversely, if
is the transitive collapse of in it is also the
transitive collapse of in and so
. Finally . In particular
.
We can apply the almost disjoint forcing method
given p 277 to find such that
and . In particular
and so is collapsed in .
16.16
Assume and let be a
sequence of infinite subsets of such that
is finite for . We consider the forcing notion
and iff
, and
for all .
It is clear that the relation is reflexive and antisymmetric. The transitivity
is almost obvious, just note that if and
then
for all we have
.
The forcing notion satisfies ccc: since is countable,
for any uncountable subset there is such
that is uncountable. Then any
have a common refinement .
For all , define .
Let the elements of .
We show by induction on that
is infinite. This is true for by assumption. If it is true for
then
|
|
|
The left hand side is infinite by induction hypothesis. The second term
of the right hand side is included in and thus
is finite. Hence the first term is infinite and the result is true for .
Finally, for , we get that
is infinite. Pick distinct
elements from that set and define .
We have , , and
for all ,
. This shows that
is dense. For each ,
the set is also dense:
for any consider .
By there is a generic filter for the previous family
of dense sets. Let .
For all , there is and so
. Hence is infinite. Let
and .
For any , there is such that .
Hence there is a refinement of .
We have and .
Hence is finite.
16.19
Let be a -closed ideal on and and iteration
of with -support.
Suppose that . Let and
() a descending
sequence in . By definition for all and
, and the
form a descending sequence in .
We shall construct by induction on an -sequence
such that and
where
. In particular,
.
Since is -closed, we let be a lower bound
of and even if ).
Then .
Clearly and .
If is constructed then
and since is full there is such that
and
. We let
and can take if ).
Again, .
Then and
, .
Now suppose is limit and that
has been constructed for all .
We just let .
By induction hypothesis,
each and
. Hence
Moreover for all and we have
and so
.
Finally at stage , we obtain such that
.
Hence is -closed.
16.20
Let be a regular cardinal. Let be a countable
support iteration of length such that for all ,
has a dense subset of size .
If is an antichain, then for each we can
find such that . If then
since is an antichain. Hence . Hence
satisfies the -chain condition.
is a countable support iteration and
so is a direct limit.
The set is stationary in
and for each we have and
so is a direct limit.
Finally, by theorem 16.30, satisfies the -chain conditon.