# Stevo’s Forcing Class Fall 2012 – Class 11

(This is the eleventh lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the tenth 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.)

## The Problem with Density

We test this on the usual place where people get stuck, which is Baumgartner’s Axiom.

Recall
: A set $D \subseteq \mathbb{R}$ is $\theta$-dense if every open interval contains $\theta$-many elements of $D$.

##### Theorem (Cantor, ???): BA($\aleph_0$) is true.

Try to find out for yourself when he discovered this. It is interesting to see what depends on what.

##### Theorem (Baumgartner, 1973): BA($\aleph_1$) is consistent, and is a consequence of PFA.

The proof of this is really a great advance. It is the beginning of OCA and many great things in set theory.

##### Question (Baumgartner, 1973): Is BA($\aleph_2$) consistent?

PFA is really Baumgartner’s axiom.

The ccc in Baumgartner’s original proof can be seen (“although this was probably very difficult to see“) as the higher dimensional analogue of the classical construction of $f : \mathbb{R} \rightarrow \mathbb{R}$ which is not continuous on any set of reals of size $\mathfrak{c}$.

##### Theorem (Galvin): The ccc is not productive under the CH.

(Personal Aside – Mike. When Matthew Foreman visited Toronto in November 2012 he gave 3 nice talks which focused on large cardinals in mathematics. He refered to the ccc and the CH. I think it has stuck with me.)

Of course this follows from the Sierpinski function.

So the observation is that the poset is barely ccc.

So in particular, cof$(2^{\aleph_0})$-cc is not productive.

## The Work

Definition. A finite partial function $f : \mathbb{R} \rightarrow \mathbb{R}$ is separated by an $\epsilon$-chain $\mathcal{N}$ of CESMs of $H_{\cc^+}$ if:

• $\forall x \in \textrm{dom}(f), \exists N \in \mathcal{N}$ such that $x \in N$ and $f(x) \notin N$;
• $\forall x \neq y \in \textrm{ran}(f), \exists N \in \mathcal{N}$ such that $x\in N$ iff $y \notin N$;
• $\forall N \in \mathcal{N}, \forall x \in \textrm{ran}(f) \setminus N, \forall g \in N$ partial function from some $\mathbb{R}^k$ to $\mathbb{R}$, $x \neq g(s)$ for all $s \in \mathbb{R}^k \cap N$. (This $\mathbb{R}^k$ is “the higher dimension business.”)

Forget about the density, focus on the combinatorics.”

##### Lemma. Suppose $f, \mathcal{N}$ are as above, with $f$ increasing, and $\mathfrak{dom}(f) \cap \textrm{ran}(f) \cap \min(\mathcal{N}) = \emptyset$. Also, let $\mathfrak{X} \in \min(\mathcal{N})$ be a family of finite, partial, increasing functions from $\mathbb{R}$ into $\mathbb{R}$ such that $f \in \mathfrak{X}$. THEN there is an $\overline{f} \in \mathfrak{X} \cap \min(\mathcal{N})$ such that $\overline{f} \cup f$ is increasing.

This is a ccc or properness lemma.”

$\overline{f}$ should be thought of as a copy.”

Note. “Increasing” is not needed; you just need to know its behaviour.

proof (by induction on $k$).

Let $k+1 = \vert \textrm{dom}(f) \vert$. May assume $\vert \textrm{dom}(x) \vert = k+1$ for all $x \in \mathfrak{X}$.

#### $k=0$, The Classical Case.

Suppose we can’t copy, then for every $(b, x(b)) \in \mathfrak{X} \cap N_0$ we have $b < a \Rightarrow x(b) > f(a)$ and $b > a \Rightarrow x(b) < f(a)$.

You can approach it arbitrarily close from the left or the right. I don’t care which; I just need one side.

Suppose this approaches arbitrarily close to $a$ and define $g(c) := \inf\{x(b) : (b, x(b)) \in \mathfrak{X}, b < c\}$. Now $g \in N_0$ and $g(c) = f(c)$, a contradiction.

This is the classical argument in today’s language. The key is that $f$ is outside of the model, so $\mathfrak{X}$ is immense.

#### Inductive step $k-1$ to $k$.

The reduction is the same. We are going to remove something and go to a smaller dimension. What do we remove?

There is an “a” such that $f(a)$ is outside the model, the furthest. The models linearly order the range.

Now it becomes tricky. What is it you’re going to say “approaches from the left”?

$f^\prime := f \setminus \{(a, f(a))\}$ so $\overline{f^\prime} \in \mathfrak{X} \cap N_0$ and $\overline{f^\prime} \rho f$. Here $\rho$ is just a relation that means “is close to”, in some way.

If for every such $\overline{f^\prime} \in \mathfrak{X} \cap N_0$, when you supplement $\overline{f^\prime}$ to $\overline{f}$, the union $f \cup \overline{f^\prime}$ is not increasing then we can define $g \in N_0$ a function from $\mathbb{R}^{2k +1} \rightarrow \mathbb{R}$ such that $g(\overline{s}) = f(a_0)$.

Here $\overline{s} = (a_0, a_1, ..., a_k, f(a_1), f(a_2), ..., f(a_k)) \in M$. So the “inf” definition ($b < c$) becomes $(\rho)$ close and increasing. [END OF PROOF]

So that’s that: The first ZFC example of a chain condition that is not productive. Actually … it is the second.

##### Corollary. $\textrm{cof}(2^{\aleph_0})\textrm{-cc}\times \textrm{cof}(2^{\aleph_0})\textrm{-cc} \neq \textrm{cof}(2^{\aleph_0})\textrm{-cc}$.

I produced a set of entangled reals for you, of size $\mathfrak{c}$.

This lemma immediately implies some obvious poset is proper.

Baumgartner was suffering to get density. Somehow we are cheating, ignoring the range. What this is really doing? We need to call it something.

## 3 thoughts on “Stevo’s Forcing Class Fall 2012 – Class 11”

1. Ari Brodsky says:

Wow. How did you draw those diagrams?

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