# Stevo’s Forcing Class Fall 2012 – Class 1

By | Published: September 14, 2012

(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.)

Show Summary
• 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.)

Show Definitions

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

Let $\{p_\xi : \xi \in \omega_1\} \sse \mathcal{A}(T)$, 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 $\Gamma \sse \omega_1$ 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 $\Gamma \in \mathcal{U}$.

Show Ultrafilter Facts

An 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?

Show Answer

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 $\{ p_\xi : \xi < \omega_1\} \sse \mathbb{P}$. Show that there is an infinite set $I \sse \omega_1$ 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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### 5 Comments

1. Posted October 18, 2012 at 8:58 am | Permalink

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?

• Micheal Pawliuk
Posted October 18, 2012 at 10:52 am | Permalink

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

2. Anonymous
Posted November 27, 2012 at 4:06 pm | Permalink

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

3. Set Theorist
Posted January 25, 2013 at 12:15 am | Permalink

What is the ordering of the A(T) posets?

• Micheal Pawliuk
Posted January 30, 2013 at 8:14 pm | Permalink

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

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• By Stevo’s Forcing Class Fall 2012 – Class 2 on September 18, 2012 at 11:00 pm

[...] 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 [...]
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