Stevo’s Forcing Class Fall 2012 – Class 7

(This is the seventh lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the sixth 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 Combinatorial CH Statements

Recall the combinatorial versions of CH:

  • CH^\omega : \cc_d = \aleph_1 , i.e. every ccc poset of size \leq \mathfrak{c} is \aleph_1 -centred.
  • CH^n : Every ccc poset of size \leq \mathfrak{c} is \aleph_1 -n-linked.
  • CH^2 : Every ccc poset of size \leq \mathfrak{c} is \aleph_1 -2-linked.

They are linearly ordered, but maybe they collapse.”

Axiom Consequences
CH Everything, right?
CH^\omega Density of Lebesgue measure algebra is \aleph_1
CH^n
CH^3
  • \textrm{cof}(2^{\aleph_0}) = \aleph_1
  • Every f: \mathbb{R} \rightarrow \mathbb{R} is the union of \aleph_1 many continuous subfunctions.
  • \theta_2 = \aleph_2
CH^2 \bb = \aleph_1 ;

Question 1: Which of these tells you that 2^{2^{\aleph_0}} = 2^{\aleph_1} ?
Question 2: Which one tells you that every subset of the reals is \aleph_1 borel?

One of These Consequences

Theorem: CH^3 implies that f: \mathbb{R} \rightarrow \mathbb{R} is the union of \aleph_1 many continuous subfunctions.

What about the union of monotone functions? “Oh, no, I do not think that is possible. Monotone functions are nice. They are always continuous, modulo something small. In general, it is a two dimensional object.

Corollary: CH^3 implies \textrm{cof}(2^{\aleph_0}) = \aleph_1 .

proof. “The proof is important, but easy by today’s standards. Look at Sierpinski’s function that is not continuous on any set of size \mathfrak{c} .

If \{r_\alpha : \alpha < 2^{\aleph_0}\} = \mathbb{R} , then there is a function f : \mathbb{R} \rightarrow \mathbb{R} such that for every partial Borel function g : \mathbb{R} \rightarrow \mathbb{R} there is an \alpha < 2^{\aleph_0} such that f(r_\beta) \neq g (r_\beta) for all \beta \geq \alpha . [QED]

This trick is important. When you do this for all finite dimensions, you enter the world of Baumgartner’s theorem. All \aleph_1 subsets of the reals are isomorphic.

proof of Theorem. We may assume that f: \mathbb{N}^\N \rightarrow \mathbb{N}^\N . Define p \in \mathbb{P} iff p \subseteq \mathbb{N}^\N is finite and for all distinct x,y,z \in \mathbb{N}^\N we have \Delta(x,z) \neq \Delta (y,z) implies \Delta(f(x),f(z)) \neq \Delta (f(y),f(z)) . Call this the metric property.

You really don’t want this property. You really want to say that f preserves the splitting structure of the tree. Unfortunately, this is probably not ccc. This is the short blanket phenomenon.

Note: If X \subseteq \mathbb{N}^\N satisfies the metric property, then f\upharpoonright X is continuous.

Now can you show ccc? You have to make it \sigma -linked… almost. It is highly not \sigma -3-linked. (Plug in the Sierpinski function). As always, the countable invariants are freezing the metric structure below some height.

Now comes the payoff. Checking the metric property for the union. Draw the picture.” [QED]

What is CH^3 telling you?” Suppose \{\{x\}: x \in X\} is 3-linked in \mathbb{P} for some X \subseteq \mathbb{N}^\N . By CH^3 , \mathbb{P} = \bigcup_{\xi<\omega_1} \PP_\xi , where each \PP_\xi is 3-linked. In fact, \mathbb{N}^\N = \bigcup_{\xi <\omega_1} X_\xi , with each X_\xi satisfying the metric property.

The MA Analogue for CH^n

Whereas the CH^n were “global” properties, we describe “local” properties

Axiom Definition Consequences
MA(\aleph_1) Every ccc poset has an \aleph_1 -generic filter. Everything else, right?
K^n Every ccc poset has an uncountable n -linked family.
K^3 Every ccc poset has an uncountable 3 -linked family. (2^{\omega_1}, <_{\textrm{lex}}) embeds into all ultrapowers (\N^\N, <_\mathcal{U})
K^2 Every ccc poset has an uncountable 2 -linked family. \bb > \aleph_1

The interesting thing is connecting these to things that aren’t related to the continuum.

Chang’s Conjecture and the Countable Chain Condition, i.e. cc&ccc

Let us do something different. Now for something completely different.”

Theorem: TFAE

  • Chang’s Conjecture;
  • Every ccc poset forces \theta_2 = \aleph_2 .

There are also strengthenings of Chang’s Conjecture denoted C^{**} \Rightarrow C^* \Rightarrow C^+ \Rightarrow 2^{\aleph_0} \leq \aleph_2 .

Recall: Chang’s conjecture states – For every structure of the form \mathfrak{A} = (\omega_2, \omega_1,) there is an elementary substructure \mathfrak{B} = (B, B\cap \omega_1,) of \mathfrak{A} such that \vert B \vert = \aleph_1 and \vert b \cap \omega_1 \vert \leq \aleph_0 .

Useful Note: Chang’s Conjecture is equivalent to the statement that for all sufficiently large regular cardinals \kappa and a \in H(\theta) there is M \prec (H(\theta), \in) , with a \in M , \vert M \cap \omega_1 \vert = \aleph_0 and \vert M \cap \omega_2 \vert = \aleph_1 . “You first find a code for the closure of \omega_2 in M …(?)” It seems you want to use a well-ordering of something.

Lemma: Chang’s Conjecture is preserved by any ccc forcing.

proof. “Nice exercise to see if you really understand forcing.

I’ll do it unless anyone complains. You should complain if you want. Complaining shows that someone is alive.

Let \mathbb{P} be ccc and f_n for n < \omega be \mathbb{P} -names for functions from \omega_2^{k_n} to \omega_2 .

Let \kappa be such a \mathbb{P} with f_n \in H(\kappa) for all n < \omega . By the note, choose M \prec H(\kappa) , with \vert M \cap \omega_1 \vert = \aleph_0 and \vert M \cap \omega_2 \vert = \aleph_1 .

Claim: Liberally applying checks, \Vdash_\mathbb{P} (M \cap \omega_2, M \cap \omega_1, \dot f_n \upharpoonright M \cap \omega_2)_{n < \omega} \prec (\omega_2, \omega_1, \dot f_n)_{n < k}

Closed under functions is easy as everything is countable. Elementary submodel takes work.

proof. For (\alpha_0, ... \alpha_{k-1}) \in M \cap \omega_2 , again with checks,
\displaystyle  \Vdash_\mathbb{P} \dot f_n (\omega_0, ..., \alpha_{k-1}) \in M \cap \omega_2

Now ccc has to show up. All the possibilities, by elementarity, have to be in M .” [QED]

Most posets destroy Chang’s Conjecture. For example, adding an \omega_1 -analogue of a Hechler real kills CC.

I did not do anything I planned. This is the definition of a good lecture. You just talk.”

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