New solutions to exercises from Thomas Jech’s book “Set Theory”:
Last November, I tried to provide
of the proof given in chapter 7,
regarding the fact that the continuum hypothesis implies the
existence of a Ramsey ultrafilter.
pointed out that the proof could
probably work assuming only Martin’s Axiom.
This was indeed proved by Booth in 1970 and the missing argument is actually
given in exercise 16.16. For completeness, I copy the details on this blog
Remember that the proof involves contructing a sequence
of infinite subsets of .
The induction hypothesis is that at step ,
for all we have is finite.
It is then easy to show the result for the successor step,
since the construction satisfies
. However at limit step, to ensure that
is finite for all , the proof
relies on the continunum hypothesis. This is the only place where it is used.
Assume instead Martin’s Axiom and consider a limit step
. Define the forcing notion
for all .
It is clear that the relation is reflexive and antisymmetric. The transitivity
is almost obvious, just note that if and
for all we have
The forcing notion satisfies ccc or even property (K):
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
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 Martin’s Axiom there is a generic filter for the family
For all , there is and so
. Hence is infinite. Let
For any , there is such that .
Hence there is a refinement of .
We have and .
Hence is finite and the induction
hypothesis is true at step .