(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.)

$\sigma$-centred Posets

Last time we saw that ccc is not always productive, but there are some versions that are:

Fact 1. If $\mathbb{P}$ is countable, and $\Q$ is ccc, then $\mathbb{P} \times \Q$ is ccc.

proof. Apply the pigeonhole principle.

An important weakening of ccc is $\sigma$-centred. “Seeing a proof that a poset is ccc but not $\sigma$-centred is a mark of sophistication.” Many posets we see that are ccc are actually $\sigma$-centred.

Definition. A poset $\mathbb{P}$ is $\sigma$-centred iff $\mathcal{P} = \bigcup_{n<\omega} \mathbb{P}_n$ where each $\mathbb{P}_n$ is centred. A poset $\mathbb{P}$ is centred iff $(\forall F \in [\mathbb{P}]^{<\omega})(\exists q \in \mathbb{P})[q \leq p, \forall p \in F]$.

Example (“The most natural $\sigma$-centred poset“). Let $\mathcal{F} \sse [\omega]^{\omega}$ be a family, of infinite sets, with the finite intersection property.

Define $\mathbb{P} := \mathbb{P}(\mathcal{F}) := \{(s,F) : s \in [\omega]^{<\omega}, F \in [\mathcal{F}]^{<\omega} \}$.

The ordering is: $(t, G) \leq (s,F)$ iff

• $t \supseteq s$,
• $G \supseteq F$,
• (Growth condition) $t \setminus s \subset \bigcap F$.

Exercise 1. Show that $\mathbb{P}$ is $\sigma$-centred.

Exercise 2. If $\mathcal{G} \sse \mathbb{P}$ is a (sufficiently generic) filter, then $s_\mathcal{G} := \{ s: \exists F, (s, F) \in \mathcal{G}\}$ is an infinite subset of $\omega$ such that $\forall M \in \mathcal{F}$ we have $s_\mathcal{G} \sse^* M$.

“Sufficiently generic” means that $\mathcal{G}$ satisfies many conditions. More specifically, $\mathcal{G}$ intersects at least $\vert \mathcal{F} \vert \cdot \aleph_0$ many dense sets. Notably we will want $\mathcal{G}$ to intersect:

• $\mathcal{D}_M := \{(s,F)\in \mathbb{P} : M \in F\}$, for each $M \in \mathcal{F}$,
• $\mathcal{E}_n := \{(s,F)\in \mathbb{P} : \vert s \vert \geq n\}$, for each $n \in \omega$.

[END OF EXAMPLE]

The Martin Numbers

The previous example gives rise to a forcing axiom MA$_\kappa$($\sigma$-centred) $\equiv \mathfrak{m}_{\sigma-centred} \equiv \mathfrak{m}_{sep}$ which is ” the least cardinality such that the Baire Category theorem fails for $T_2$ compact separable spaces”.

Similarly, $\mathfrak{m} \equiv \mathfrak{m}_{ccc}$ which is the “Baire category number for $T_2$ ccc compacta”.

Finally, $\mathfrak{m}_{SH} := \min\{w(K) : K$ 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 $\mathfrak{m} = \mathfrak{m}_{SH}$ and $\mathfrak{m}_{sep} = \mathfrak{p}$.

Exercise. Show that $\mathfrak{m} \leq \mathfrak{m}_{SH}$.

Erercise. Show that $\mathfrak{m} \leq \mathfrak{m}_{SEP} \leq \mathfrak{p}$

“So MA is nothing more that SH, nothing more than you can diagonalize against sets.”

Exercise. Show that if $\mathfrak{m} > \omega_1$ then ccc is productive.

Corollary. If $\vert \mathbb{P} \vert < \mathfrak{m}$ and $\mathbb{P}^{<\omega} := \{p \in \mathbb{P} : p$ is eventually constant $\}$, then $\mathbb{P}^{<\omega}$ is still ccc.

Cardinal Invariants and a Lemma

Here are definitions of some small cardinal invariants:

Definition 1 (pseudo-intersection number). $\mathfrak{p} := \min\{ \vert \mathcal{F} \vert : \mathcal{F} \sse \omega^\omega, \mathcal{F}$ has no pseudo-intersection $\}$. (Having a pseudo intersection means there is something almost contained in every element of the family.)

Definition 2 (tower number). $\mathfrak{t} := $ minimum length of an almost increasing chain of subsets of $\omega$, with no pseudo-intersection.

Definition 3 (bounding number). $\mathfrak{b} := \min\{ \vert \mathcal{F} \vert : \mathcal{F} \sse \omega^\omega, \mathcal{F} <^*$ unbounded $\}$.

Recently, Shelah proved that $\mathfrak{p} = \mathfrak{t}$.

Exercise. Show that $\mathfrak{p} \leq \mathfrak{t} \leq \mathfrak{b}$.

The main lemma here is “a sort of simultaneous diagonalization”.

Diagonalization Lemma. Suppose $\kappa < \mathfrak{p}$ and suppose that for every $s \in [\omega]^{<\omega}$ the family $\{A_{\alpha, s} : \alpha < \kappa\} \sse [\omega]^\omega$ has the finite intersection property. THEN there is a function $f \in \omega^\omega$ such that for each $\alpha < \kappa$ there is an $m$ such that for all $n \geq m$ we have $f(n) \in A_{\alpha, f \upharpoonright n}$.

proof. We start by finding a way to use $\mathfrak{p}$.

For each $s \in \omega^{<\omega}$ take $A_s \in [\omega]^{\omega}$ such that $A_s \sse^* A_{\alpha, s}$ for all $\alpha < \kappa$.

“Next a change of index set, countable, but different.”

For $\alpha < \kappa$, define $f_\alpha : \omega^{<\omega} \rightarrow \omega$ by letting $f_\alpha (s) := \min\{n \in A_s : A_s \setminus n \sse A_{\alpha, s}\}$.

Since $\kappa < \mathfrak{p} \leq \mathfrak{b}$, take $g : \omega^{< \omega} \rightarrow \omega$ such that (“now one must be careful…“) for all $\alpha < \kappa,  f_\alpha <^* g$.

“I put this strange notation. It means for all but finitely many coordinates $f_\alpha (n) < g(n)$.”

Assume also that for all $s \in \omega^{<\omega}$ we have $g(s) \in A_s$.

Define $f(n) := g(f \upharpoonright n)$.

“Clear definition by recursion. Properties, maybe not so clear.”

“What is going to be ‘m’? Before, I promised you an ‘m’.“

For $\alpha < \kappa$ let $m$ be such that $f\upharpoonright n \notin B(\alpha)$ for all $n \geq m$.

Here $B(\alpha) := \{s \in \omega^{<\omega} : f_\alpha (s) \not < g(s)\}$, and “the ‘B’ stands for the bad set”. [QED]

Martin’s Axiom for Separable Spaces

Proof that $\mathfrak{m}_{SEP} = \mathfrak{p}$.

We leave the direction $\mathfrak{m}_{SEP} \leq \mathfrak{p}$ as an exercise and only show $\mathfrak{m}_{SEP} \geq \mathfrak{p}$. Without loss of generality we assume that $\mathbb{P}$ has no atoms.

Let $\mathbb{P} = \bigcup_{n<\omega} \mathbb{P}_n$ be a given $\sigma$-centred poset and let $\{\mathcal{D}_\alpha : \alpha<\kappa\}$ be a collection of open dense subsets of $\mathbb{P}$ where $\kappa < \mathfrak{p}$.

We need to find a filter $G \sse \mathbb{P}$ such that $G \cap \mathcal{D}_\alpha \neq \emptyset$ for all $\alpha <\kappa$.

Without loss of generality (by a Löwenheim-Skolem argument), $\vert \mathbb{P} \vert \leq \kappa$. 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. $G \sse \mathbb{P}$ is a generic filter iff it is the upwards closure of a generic 2-linked set.

“This is actually quite natural.”

Build a sequence $p_{s \alpha}$ and families $A_{s \alpha}$ where $\alpha < \kappa$, and $s \in [\omega]^\omega$.

1. If $\vert s \vert = n+1$ then $p_{s \alpha} \in \mathbb{P}_{s(n)}$ for all $\alpha < \kappa$. (As Chris said: “The largest element of s knows about it.”)
2. $A_{s \alpha} := \{n < \omega : \exists q \in \mathbb{P}_n \cap \mathcal{D}_\alpha, q \leq p_{s \alpha}\}$ “is a very natural set, which is infinite (as $\mathbb{P}$ has no atoms).“
3. $n \in A_{s \alpha}$ implies that $p_{s n \alpha} \in \mathbb{P}_n \cap \mathcal{D}_\alpha$ and $p_{s n \alpha} \leq p_{s \alpha}$.

Check that for a fixed $s$, $\{A_{s \alpha} : \alpha < \kappa \}$ has the finite intersection property. This will allow us to apply the diagonalization lemma to find an $f: \omega \rightarrow \omega$ such that $\forall \alpha, \exists m_\alpha, \forall n \geq m_\alpha$ we get $f(n) \in A_{f \upharpoonright n \alpha}$.

Put $G := \{p_{f \upharpoonright (m_\alpha +1) \alpha} : \alpha <\kappa\}$. Note that by (2) it is generic, so all we need is the following exercise:

Exercise. Show that $G$ is linked.

[QED]

“Use classical things to help you in forcing.”

Exercise. If $\sigma$-linked posets have size at most continuum.

Question: Can a Boolean Algebra supporting a continuous submeasure algebra distinguish $\sigma$-finite-cc and $\sigma$-bounded-cc? (Compare with the chart at the end of class 1 notes.)

Martin’s Number for SH

We now attempt to show that $\mathfrak{m} \geq \mathfrak{m}_{SH}$ which, by a previous exercise, will give us $\mathfrak{m} = \mathfrak{m}_{SH}$. We first show the existence of a nice space, which will show $\mathfrak{m}_{SH} \leq \mathfrak{p}$. Stevo claimed in class that a further corollary is that $\mathfrak{m} < \mathfrak{m}_{SH}$ is impossible, but I can’t see how this follows.

Theorem. There is a $\sigma$-linked poset $\mathbb{P}$ of size $\mathfrak{t}$ without a centred subset of size $\mathfrak{t}$. Topologically, this corresponds to finding a compact, ccc, non-separable space of cardinality $\mathfrak{t}$.

“The key in forcing is your ability to make partially ordered sets.”

Corollary. $\mathfrak{m}_{SH} \leq \mathfrak{t}$

(For ease, in this proof we will be assuming that $\mathfrak{p} = \mathfrak{t}$, but you can make the obvious adjustments.)

proof.

Let $\{a_\xi : \xi < \mathfrak{t}\} \sse [\omega]^\omega$ be a non-extendible tower. For ease of notation, for $F \sse \mathfrak{t}$ we denote $a_F := \bigcap_{\xi \in F} a_\xi$.

“Most important is the metric structure. The distance set: ” $\Delta_F := \{\Delta(a_\xi, a_\eta) : \xi, \eta \in F. \xi \neq \eta\}$ where $\Delta (a,b) = \min (a \Delta b)$.

“Here I don’t have the options to meet dense sets.“

Let $\mathbb{P} := \{ F \in [\mathfrak{t}]^{<\omega} : \forall k < \omega, \vert a_F \cap k \vert \geq \vert \Delta_F \cap k \vert\}$.

“This doesn’t look promising. The order is key. It is reverse-inclusion.“

Claim 1. $\mathbb{P}$ is $\sigma$-k-linked for all $k \geq 2$.

Claim 2 (the promise that your generic object fills a tower). If $\mathcal{C} \sse \mathbb{P}$ is centred in $\mathcal{P}$ then $a_{\bigcup\mathcal{C}}$ is an infinite subset of $\omega$.

Thus if $\vert\mathcal{C} \vert = \mathfrak{t}$ then $a_{\bigcup\mathcal{C}}$ 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 $\sigma$-centred.”

proof of claim 2. Suppose $a_{\Gamma}$ is finite (where $\Gamma := \bigcup\mathcal{C}$). So there is a $k$ such that $a_\Gamma \sse k$. Also, since $a_\xi$ form a tower, it must be that $\Gamma$ is infinite. Thus there is a $k_1$ such that $\vert \Delta_\Gamma \cap k_1 \vert > \vert a_\Gamma \vert$.

“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 $F \in [\mathfrak{t}]^{<\omega}$ such that:

1. $\Delta_F \cap k_1 = \Delta_\Gamma \cap k$; and
2. $a_F \cap k_1 = a_\Gamma \cap k_1$.

We can do this because we only need finitely many branches to do the witnessing of finitely much. Thus we can conclude that $F \notin \mathbb{P}$ (by chasing the inequalities). But this violates the fact that $\mathcal{C}$ is centred. [QED]