Stevo’s Forcing Class Fall 2012 – Class 1

(In the fall of 2012 I will be taking Stevo Todorcevic’s class in Forcing at the University of Toronto. I will try to publish my notes here, although that won’t always be possible.)

Summary of class 1.

  • Discuss examples of ccc posets whose product is not ccc.
  • Prove a theorem of Baumgartner’s that relates the branches in a tree to its antichain structure.
  • Display the differences in chain conditions.

This class is going to be held Wed 1-2 and Friday 3:30-5:30 in BA 6183.

The theme for today is “Things that are ccc, but whose products are not”. So first some examples:

Example 1. If T is a Suslin tree, then T^2 is not a Suslin tree. (See Lemma III.2.18 in Newnen.)
A Suslin Tree is a tree with uncountably many nodes, no uncountable chains and no uncountable antichains. The existence of these objects is independent of ZFC, but follows from the combinatorial principle ‘diamond’ or Godel’s axiom V=L . Easing the requirement from “no uncountable antichains” to “no uncountable levels” gives the definition of an Aronszajn tree, whose existence can be shown in ZFC.

Example 2. Let \mathcal{A}(T) = finite antichains of T , where T is a Suslin Tree. Then \mathcal{A}(T) is powerfully ccc (its finite products are ccc), but T \times \mathcal{A}(T) is not ccc.

“A Suslin tree is the skeleton of a Suslin continuum. If you take x-rays, that’s what you get.” – Stevo

This example was first studied by Baumgartner in his thesis:

Theorem (Baumgartner, 1969): TFAE for a tree T :

  • \mathcal{A}(T) is ccc;
  • T has no uncountable branches.

It is clear that NOT (2) imples NOT (1), as the singletons of a branch form an antichain.

“There are many proofs of this fact, but Baumgartner’s original proof is still the most useful.” – Stevo

Proof. Like most proofs that a poset is ccc, we start with an uncountable subset of the poset and refine, refine, refine.

Let \{p_\xi : \xi \in \omega_1\} \subseteq \mathcal{A}(T) , we want to find \xi \neq \eta such that p_\xi \cup p_\eta \in \mathcal{A}(T) . We refine 3 times to an uncountable set \Gamma \subseteq \omega_1 so that:

  1. (All conditions are the same length) \vert p_\xi \vert = n for all \xi \in \Gamma ,
  2. (\Gamma is a \Delta -sytem) There is an r \in \mathcal{A}(T) such that for \xi \neq \eta in \Gamma we have p_\xi \cap p_\eta = r .
  3. (Heights are increasing) The ht(p_\xi) < ht(p_\eta) for \xi < \eta .

We want to assume that r is actually empty, and we can do this by considering q_\xi := p_\xi \setminus r \neq \emptyset . Note that q_\xi \not\perp q_\eta implies p_\xi \not \perp p_\eta .

WLOG, we get r = \emptyset .

We suppose that p_\xi \perp p_\eta for all \xi \neq \eta in \Gamma (and look for a contradiction). This means that if \xi < \eta then there is an x \in p_\xi and a y \in p_\eta such that x <_T y .

Now take a uniform ultrafilter \mathcal{U} on \omega_1 such that \Gamma \in \mathcal{U} .

A uniform ultrafilter is one where all the elements of \mathcal{U} have the same cardinality. In this case they are all uncountable. Such ultrafilters exist in ZFC, and we can ensure that there is a uniform ultrafilter that contains any given uncountable subset of \omega_1 .

 

Fix \xi \in \Gamma . Write p_\xi = \{p_\xi (0), \dots , p_\xi (n-1) \} .

There is an i_\xi < n such that \Gamma_\xi := \{\eta \in \Gamma : \exists y \in p_\eta, p_\xi (i_\xi) <_T y \} \in \mathcal{U} .

Define X := \{p_\xi (i_\xi) : \xi \in \Gamma\} .

Question: Is there a large antichain in X?

No! There is no antichain of cardinality n+1 , because \Gamma_\xi = \bigcup_{j<n} \Gamma_\xi (j) which is in \mathcal{U} by choice. Where \Gamma_\xi (j) = \{\eta \in \Gamma_\xi : p_\xi (i_\xi) <_T p_\eta (j)\} .

 

Now for each \xi \in \Gamma there is a j_\xi < n so that \Gamma_\xi (j_\xi) \in \mathcal{U} . Thus \bigcap_{i \leq n+1} \Gamma_\xi (i_\xi) \in \mathcal{U} and we can choose an \eta in this intersection so that ht(p_\eta) > ht(p_{\xi_i}) for all i \leq n .

Now if j_\xi does not depend on \xi then we are done. This follows from exercise 2.

[QED]

Some Exercises:

Exercise 1: Suppose that \mathbb{P} is a ccc poset and let \{ p_\xi : \xi < \omega_1\} \subseteq \mathbb{P} . Show that there is an infinite set I \subseteq \omega_1 such that \{p_i : i \in I\} are pairwise compatible.

Exercise 2: Let T be a tree with an uncountable subset that has no antichains of size \geq N . Show that T has an uncountable chain.

Some chain conditions, listed from easiest to satisfy to hardest to satisfy:

  • ccc
  • powerfully ccc
  • productively ccc
  • \sigma -finite-cc
  • \sigma -bounded-cc
  • \sigma -2-linked
  • \sigma -n-linked
  • \sigma -n-linked (\forall n )
  • \sigma -centred
  • countable

Note that Borel posets can distinguish all of the blue and dark blue properties. By the following exercise, the dark blue posets are small.

Exercise 3. If \mathbb{P} is an atomless \sigma -(2)-linked poset then \vert \mathbb{P} \vert \leq \mathfrak{c} .

Hint. Use the Erdös-Rado theorem.

Question. Can a boolean algebra supporting a continuous submeasure algebra distinguish \sigma -finite-cc from \sigma -bounded-cc?


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6 thoughts on “Stevo’s Forcing Class Fall 2012 – Class 1”

  1. Thanks for sharing these notes Mike!

    Small typo: In “[…] such that \Gamma_\xi := \lbrace \eta \in \Gamma : \exists y \in P_\eta ,p_\xi(i_\xi ) <_T y \rbrace \in \mathcal{U} " the first P should be lower case.

    I also have a very hard time seeing the difference between "blue" and "dark blue" on my laptop. Could you use more contrasting colors?

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    1. Thanks for the feedback. These kind of comments are very helpful. I’ve made the light blue lighter, so hopefully there is more “pop”.

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  2. Do we need to add some additional assumptions in exercise 3? (like having the poset be separative) As it stands we could just add a huge blob of mutually compatible guys at the top of some centered forcing, say Cohen forcing, and violate the conclusion of cardinality bounded by the continuum

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    1. It should just be inclusion. I.e. One antichain is compatible with another, if their union is an antichain.

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