(These are the
voyages are the starship enterprisenotes from Arnie Miller’s guest lecture in Alan Dow’s Forcing course, held on October 11, 2012. The lecture was a self-contained exploration of strange new worldsNamba forcing. (Alright, I will stop doing that. For those who don’t get the reference.) Arnie Miller’s quotes are in Grey. Yes, the forcing checks are ugly. I have yet to figure that one out.)
- Motivation with posets that collapse cardinals.
- Namba Trees are defined.
- Two 0-extension lemmas are proved.
- The Fat Namba Filter is introduced.
- An important theorem of Namba’s is proved.
It starts with a theorem in Prikry’s Thesis.
Theorem (Prikry). We can have such that
- ; and
- such that “” but .
Of course here, would have to be a limit cardinal.
Changing the Cofinality of a Successor
Example. (~1962) Levy collapse of to .
Let where . The ordering is iff . The generic function here is , which collapses the cofinality of .
This poset is countably closed, so is not collapsed. So “, because it can’t be .
Namba is Something In-Between
The majority of this lecture will be spent proving the second part of this theorem.
Theorem (Namba, 1971). If and is -generic, then
- and “
- (Assuming CH) or equivalently, .
It doesn’t add reals, but it isn’t countably closed. (Recall that for separative posets these are equivalent. See XX.XX.X in Newnen.)
“I really like it as a bizarre way to collapse .”
“I’m really into proofs, so I think I’m done with my introduction.”
Defining Namba Forcing
First we define what a Namba Tree is, and the Namba poset will be the collection of all Namba trees, together with a non-obvious ordering.
Notation. For a poset , we write .
Definition. is a Namba Tree if
- is a tree, (i.e. we have )
- such that (i.e. )
- After the root there is -splitting. I.e. We have
“It is very bushy and wide as soon as you pass the root.”
Ordering: iff and there is still the immediate splitting.
What does the generic do? If is -generic, then let .
- is cofinal;
- where .
“The root is the working part, the splitting is the side condition.”
“You need to fix the root. If you move it countably many times then you are in trouble.”
To control this we point out a stronger extension property.
Definition. iff and . (I.e. we only change the side condition.)
Lemma 1. If , then and such that .
“Of course this is totally false for ”
“Here is the combinatorics: is not .”
proof of Lemma 1.
First a useful notion. Say is good iff and such that . If is not good, then it is bad.
Claim. If is bad, then .
“This says there are at most many good successors. The reason is the tree technology.”
“All combinatorial arguments start with ‘suppose not’, so…“
Suppose , (where stands for good), and
“This is where the contrast comes in.”
Now and .
“Do the obvious thing.“
Let and note “because it is impossible to force to be anything else.” But this contradicts “ is bad.” [End of Claim]
“Now look at the things that are hereditary. Start with the root, take all the bad ones beneath it, throw everything else out, then keep taking bad.”
Define is bad, and every such that root. Now and we have is bad.
Also, , and with . This implies is good, a contradiction. [QED]
“Start with something not good, and make it so bad it is impossible.”
“We want to soup this up.”
(Souped Up) Lemma 2. If , then there is a , and such that
- , with , , ; and
- If , then implies that .
“So this is a continuous map.”
“We get this by a fusion argument. You can work locally and fuse everything together.”
“For ease of the pictures, the root will be assumed to be empty.“
Assume . Construct of stronger conditions such that and so .
Note that is still a Namba tree as previous levels are preserved.
Take with .
“A fusion type argument. It turns something not countably closed into something close.” [QED]
We will be concerned with , and its topology. The basic clopen sets are where .
Then is a closed set which is “a super fat, (stupid) Namba tree.”
For with and implies then
“Let’s take the CH example. Find a subcondition of that is forced to be equal to .”
Recall: For a filter on ,
- For the measure 1 sets on , is the ideal of measure 0 sets and is the positive measure sets.
- For the cofinite sets on , is the ideal of finite sets and is the infinite sets.
- For the club filter on , is the ideal of non-stationary sets and is the stationary sets.
- For is the ideal of meager real subsets, is the comeager sets, is the non-meager sets.
“What’s this got to do with Namba?”
Definition. is a Fat Namba Tree if
- is a Namba tree; and
- The splitting is not just -splitting, but misses only many nodes.
Proposition. If , for are Fat Namba Trees, with the same roots, then is a Fat Namba Tree.
This tells us that , the filter generated by (with Fat Namba having root ) is closed under intersections.
For a Namba Tree with root, then , because if , then is Namba (because is fat) and . This is also true for any .
“The Namba closed set meets every Fat Namba tree.”
“How do things in relate to each other?”
Lemma 3. Suppose with . THEN for we have .
“What does the filter technology tell me about extensions?“
proof. “This proof starts in a novel way…” Suppose not. Let so for all we have . So there is a Fat Namba tree with root , so .
“Note that a fat union of Namba trees is Fat.“
For , Fat Nambas with root , . This is a contradiction.
“This isn’t just an abstract union, it is a clopen union of separated…“
Lemma 4. Suppose that where and closed. “I’m going to combine two lemmas.” THEN .
proof. WLOG, .
Look at .
Now implies . Also, .
For all (it is large). So .
In the world of topology, because is closed. [QED]
Take a look again at the (souped up) Lemma 2.
- are closed sets;
- where ;
- such that
So by the lemma, such that .
Recall: . So you can’t force it to disagree.
For all if , then such that .
So if CH, then .
Exercise. “Without CH, is still not collapsed. There would be too many . You will get an sequence of reals.”
Theorem (Bukovsky) If and “, then .