(This is the sixteenth lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the fifteenth lecture here. Quotes by Stevo are in dark blue; some are deep, some are funny, some are paraphrased so use your judgement. As always I appreciate any type of feedback, including reporting typos, in the comments below.)
- This lecture introduces the idea of forcing things that force things to be true.
- Five different types of these theorems are stated.
- We prove one of them. Namely that assuming PFA, after forcing with Sacks, MA is true.
Preserving (Consequences of) Forcing Axioms
We have two branches of forcing axioms, those that give CH type results, and those that give PFA type results.
Example: Katetov Problem. Is the following an equivalence?
A compact space is metrizable iff satisfies the separation axiom .
(i.e. such that .)
“Forcing with (1) Sacks Forcing or (2) Souslin Tree Forcing tend to preserve some consequences of forcing axioms.”
For the Katetov Problem, all 6 of the forcing axioms in the above table will give counterexamples.
“You want to force somebody to force. It is second order.”
“We start with Sacks because it is relatively easy.“
Theorem 1 (Carlson-Laver, 1989). Assuming PFA, .
Theorem 2 (Todorcevic, 1989, published in dichotomies in Set Theory) OCA fails.
(“You show no new reals are added. It is oscillation.“)
“Forcing with a Souslin tree preserves a lot of things.”
Theorem 4. Assume PFA(S), then SH (and PID and , so MA fails.)
Definition. Fix a downward closed coherent Souslin Tree. PFA(S) is the forcing axiom for -dense sets for the class of all proper posets that preserve .
“Question: Why Coherent? What fails for arbitrary? It is a bit more complicated because you have to preserve a tree.”
“What about OCA?“
Theorem 5. Assume PFA(S). Then both OCA and OCA.
Theorem 3 (Carlson-Laver, 1989). Assuming , .
“Interesting. You might need CH. That is how they stated it.“
On to the Show.
Proof of Theorem 1. This uses Sacks Amoemba (Introduced by Solovay in the Random Real setting.)
Ordering. Say iff .
“I strongly suggest you go through this proof. There are many wrong proofs in the literature.”
Fact 1. is proper.
Fact 2. (Assuming PFA.) Every uncountable set of ordinals in contains an uncountable subset in , i.e. for every uncountable there is an uncountable such that . (“The Poset has caliber .“)
proof of Fact 2. Let be an uncountable colection of -names for ordinals. Let . Note that each is dense open (for ).
“The corresponding fact about measure algebras is assuming MA(), then if there is an uncountable set in some random extension, then there is an uncountable subset in the ground model.”
“Ah! We need more dense sets. (To ensure perfect.)“
So let .
Let be a filter that intersects all . Let .
Then is perfect.
“Now what does this force?“
Note. For every there is such that decides . (Look at , and decides . So decides .)
“Now what is uncountable set?“
Fix an so is uncountable.
“Figure out the rest.” [End of Fact 2]
Now, back to the proof… preserving MA().
Let be an -name for a ccc poset and for be -names for -many dense open subsets of . We need to find an -name for a filter of which is forced by a coordinate in to intersect all these dense sets.
“We use Amoeba to help us.“
Identify all -names for elements of .
We go to and order as follows: iff in the generic filter such that .
(Shocking) Lemma: (In ). The poset is ccc.
(“You are requiring to be in the generic; you have fewer options.“)
By Fact 2, we can go to and assume to have uncountable .
“There is a price for this:”
Need to find and such that “ and are compatible in ”
List as . Find an uncountable and , (where )
By using the maximal principle, we have “ and are compatible in ” (by using )
“Be careful. There is a little bit of funny going on here.“
Take so that “ and are compatible in ”
Now take . So for we have “ is a common extension of and ”.