(This is the second lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the first lecture here. Quotes by Stevo are in dark blue; some are deep, some are funny, some are paraphrased so use your judgement.)
(Also, there are bound to be typos. I would appreciate anyone who could point them out in the comments and I will promptly fix them.)
- Give the classic example of a -centred poset. (The one that adds a “small” subset of .)
- Define the various “Martin numbers”: and .
- Define some small cardinals and .
- Show that and .
Last time we saw that ccc is not always productive, but there are some versions that are:
Fact 1. If is countable, and is ccc, then is ccc.
proof. Apply the pigeonhole principle.
An important weakening of ccc is -centred. “Seeing a proof that a poset is ccc but not -centred is a mark of sophistication.” Many posets we see that are ccc are actually -centred.
Definition. A poset is -centred iff where each is centred. A poset is centred iff .
Example (“The most natural -centred poset“). Let be a family, of infinite sets, with the finite intersection property.
The ordering is: iff
- (Growth condition) .
Exercise 1. Show that is -centred.
Exercise 2. If is a (sufficiently generic) filter, then is an infinite subset of such that we have .
“Sufficiently generic” means that satisfies many conditions. More specifically, intersects at least many dense sets. Notably we will want to intersect:
- , for each ,
- , for each .
[END OF EXAMPLE]
The Martin Numbers
The previous example gives rise to a forcing axiom MA(-centred) which is ” the least cardinality such that the Baire Category theorem fails for compact separable spaces”.
Similarly, which is the “Baire category number for ccc compacta”.
Finally, compact, ccc, nonseparable .
“Everything in set-theory is Martin’s Axiom” -Stevo said while trying to figure out the letter to use for the previous cardinal.
“SH [Suslin’s Hypothesis] tells you that certain compact spaces are separable.”
“In the dual form, this is some sort of internal forcing statement.”
We will show that and .
Exercise. Show that .
Erercise. Show that
“So MA is nothing more that SH, nothing more than you can diagonalize against sets.”
Exercise. Show that if then ccc is productive.
Corollary. If and is eventually constant , then is still ccc.
Cardinal Invariants and a Lemma
Here are definitions of some small cardinal invariants:
Definition 1 (pseudo-intersection number). has no pseudo-intersection . (Having a pseudo intersection means there is something almost contained in every element of the family.)
Definition 2 (tower number). minimum length of an almost increasing chain of subsets of , with no pseudo-intersection.
Definition 3 (bounding number). unbounded .
Recently, Shelah proved that .
Exercise. Show that .
The main lemma here is “a sort of simultaneous diagonalization”.
Diagonalization Lemma. Suppose and suppose that for every the family has the finite intersection property. THEN there is a function such that for each there is an such that for all we have .
proof. We start by finding a way to use .
For each take such that for all .
“Next a change of index set, countable, but different.”
For , define by letting .
Since , take such that (“now one must be careful…“) for all .
“I put this strange notation. It means for all but finitely many coordinates .”
Assume also that for all we have .
“Clear definition by recursion. Properties, maybe not so clear.”
“What is going to be ‘m’? Before, I promised you an ‘m’.“
For let be such that for all .
Here , and “the ‘B’ stands for the bad set”. [QED]
Martin’s Axiom for Separable Spaces
Proof that .
We leave the direction as an exercise and only show . Without loss of generality we assume that has no atoms.
Let be a given -centred poset and let be a collection of open dense subsets of where .
We need to find a filter such that for all .
Without loss of generality (by a Löwenheim-Skolem argument), . Moreover, we do not need to construct a filter with this property, as a 2-linked subset will be enough, by the following fact:
FACT. is a generic filter iff it is the upwards closure of a generic 2-linked set.
Only the backwards direction needs showing, and we do this “by testing“. Consider for each the dense set . There are only many of these dense sets, and 2-linked kills all the possibilities except .
“This is actually quite natural.”
Build a sequence and families where , and .
- If then for all . (As Chris said: “The largest element of s knows about it.”)
- “is a very natural set, which is infinite (as has no atoms).“
- implies that and .
Check that for a fixed , has the finite intersection property. This will allow us to apply the diagonalization lemma to find an such that we get .
Put . Note that by (2) it is generic, so all we need is the following exercise:
Exercise. Show that is linked.
“Draw a picture. The -centred posets will help you.”
“Use classical things to help you in forcing.”
Exercise. If -linked posets have size at most continuum.
Question: Can a Boolean Algebra supporting a continuous submeasure algebra distinguish -finite-cc and -bounded-cc? (Compare with the chart at the end of class 1 notes.)
Martin’s Number for SH
We now attempt to show that which, by a previous exercise, will give us . We first show the existence of a nice space, which will show . Stevo claimed in class that a further corollary is that is impossible, but I can’t see how this follows.
Theorem. There is a -linked poset of size without a centred subset of size . Topologically, this corresponds to finding a compact, ccc, non-separable space of cardinality .
“The key in forcing is your ability to make partially ordered sets.”
(For ease, in this proof we will be assuming that , but you can make the obvious adjustments.)
Here are some things you should think about before the proof begins. Convince yourself that if you have infinitely many branches in the Cantor tree then there are infinitely many nodes where two branches first split. Next, convince yourself that if you have 7 branches in the Cantor tree then there are at most 7 choose 2 many different first splits of branches. Finally, we will think of each branch of the Cantor tree as corresponding to an infinite subset of . In fact, each branch will correspond to a member of a family with the finite intersection property. Truncating these branches at level 100 will give us more information about the intersection of the corresponding subsets of as opposed to truncating at level .
The proof will play with the tension that these intersections can code as large a portion of an infinite subset of as we want, even though the set of “first branch splits” will be fixed and small.
Let be a non-extendible tower. For ease of notation, for we denote .
“Most important is the metric structure. The distance set: ” where .
is the first natural number where and do not agree about its membership. In a picture of a tree, it is the first place where two branches split. So just records the (finite) list of first splits of the branches in .
“Here I don’t have the options to meet dense sets.“
“This doesn’t look promising. The order is key. It is reverse-inclusion.“
We will be taking finite subsets of , but these correspond to infinite subsets of , which we really want to think of as branches in the Cantor tree. We only add to the poset if it has a small number of distinct distances relative to the size of the intersection of the corresponding infinite sets. Remember that the size of the distance set of will stop growing at some point, but the corresponding set intersection will continue to grow.
Claim 1. is -k-linked for all .
Claim 2 (the promise that your generic object fills a tower). If is centred in then is an infinite subset of .
Thus if then will be almost contained in every element of our (supposedly) non-extendible tower.
“This is the non-obvious way to fill a tower. The obvious way is -centred.”
“Exercise. Show this for . Draw the picture. You need to come up with countably many invariants such that if two finite sets of branches have the same invariant they are compatible. Try looking at at the countably many different subtrees of bounded height. This won’t work, but it is close. You also need to ensure that there are a large number of intersections above that bounded height.“
proof of claim 2. Suppose is finite (where ). So there is a such that . Also, since form a tower, it must be that is infinite. Thus there is a such that .
“Find a place where you see more than that many distances. The analogy is that in the Cantor tree if you have infinitely many branches, then you have infinitely many distances.“
Find an such that:
- ; and
We can do this because we only need finitely many branches to do the witnessing of finitely much. Thus we can conclude that (by chasing the inequalities). But this violates the fact that is centred. [QED]