# Stevo’s Forcing Class Fall 2012 – Class 15

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

## Finally, we are ready to prove the Baire-Class-1 lemma!

Some of these other proofs using forcing have classical proofs.

Remember we are aiming for the following theorem

##### Theorem. Every compact set of Baire-class-1 functions has a dense metrizable subspace.

Proved using a series of lemmas:

##### Lemma 1: If $K \subseteq \mathcal{B}_1 (\N^\N)$, compact, and $\mathcal{B} := \textrm{ro}(K)$ then $\mathcal{B}$ forces that the generic ultrafilter is countably generated.

We showed this is class 14.

This would be hard to pull the countably many generators out of the blue. The generic will give it to us.

definition 1. A regular pair of sets $(F,G)$ in some regular space is a pair where $F \subseteq G$ with $F$ closed, $G$ open.

definition 2
. A sequence of regular pairs $\{(F_\alpha, G_\alpha): \alpha < \theta\}$ is free iff
$\displaystyle \bigcap_{\alpha \in A} F_\alpha \cap \bigcap_{\beta \in B} (X \setminus G_\beta) \neq \emptyset$
for all pairs $A < B$ of finite subsets of $\theta$, (where each element of $A$ is below each element of $B$).

For more information on this you can see Stevo’s paper “On a $\Sigma^2$ functor.”

Lemma 2
. Let $X$ be a (non-trivial) compact space and let $\pi = \pi$-weight $(X)$, then there is a sequence of regular pairs $\{(F_\alpha, G_\alpha) : \alpha <\pi\}$ such that

• $\{ \textrm{int}(F_\alpha) : \alpha < \pi\}$ is a $\pi$-basis of $X$;
• If $\Gamma \subseteq \pi$ then $\{F_\gamma : \gamma \in \Gamma\}$ is centred iff $\{(F_\gamma, G_\gamma) \gamma \in \Gamma\}$ is free.

This shows up in Arhangel’skii’s famous theorem that first countable spaces have size $\leq \c$.

##### Topologically: $\{x_\xi : \xi < \theta\}$ is free iff $\displaystyle \forall \alpha < \theta, \overline{\{x_\xi : \xi < \alpha\}} \cap \overline{\{x_\xi : \alpha \leq \xi\}} = \emptyset$

The trick is seeing that this is true; the proof [of lemma 2] is obvious.

proof of lemma 2. Construct the sequences recursively assuming additionally that the $F_\alpha$‘s are $G_\delta$ sets and the $G_\alpha$‘s are $F_\sigma$. Then by compactness you are finished.

Back to our case.

Let $K \subseteq \mathcal{B}_1 (\N^\N)$ be compact, with $\pi = \pi$-weight $(X)$ and $\{(F_\alpha, G_\alpha) : \alpha <\pi\}$ as in lemma 2. Moreover, assume that the $F_\alpha$‘s and $G_\alpha$‘s are of the following form:
$\displaystyle F_\alpha = \{f \in K : \forall i \leq n_\alpha, f(x_i^\alpha) \leq \delta_i^\alpha\}$
and
$\displaystyle G_\alpha = \{f \in K : \forall i \leq n_\alpha, f(x_i^\alpha) \leq \epsilon_i^\alpha\}$
where:

• $x_0^\alpha, ..., x_{n_\alpha}^\alpha, \delta_0^\alpha, ..., \delta_{n_\alpha}^\alpha, \epsilon_0^\alpha, ..., \epsilon_{n_\alpha}^\alpha \in \mathbb{Q}$; and
• $\delta_i^\alpha < \epsilon_i^\alpha$ for each $i \leq n_\alpha$

Let $\dot{\mathcal{G}}$ be the $\mathcal{B} := \textrm{ro}(K)$-name for the generic filter.

##### Claim 1. $\Vdash_\mathcal{B} \{\alpha < ^\smile \pi : \textrm{int}(F_\alpha) \in \dot{\mathcal{G}}\}$ is countable.

Suppose that in some forcing extension $V^\mathcal{B}$, $\dot \Gamma$ is uncountable. Let $\Gamma_0$ be the first $\omega_1$ members of $\Gamma$.

I don’t care. Just take the first $\omega_1$ many.

Recall that $\hat K$ is relatively compact, so
$\displaystyle \overline{\hat K} = \overline{\{\hat f : f \in K\}} \subseteq \mathcal{B}_1 (\N^\N)^{V^\mathcal{B}}$

(“Do the $F_\alpha$ definition in the extension.“)

Now there is some funny notation. We could spend a long time chasing the truth.

All we want is that $\{(\tilde F_\alpha, \tilde G_\alpha) : \alpha \in \Gamma_\alpha\}$ is a free sequence. (“This is finitary, so it isn’t a big deal.“)

For $\alpha \in \Gamma$, by compactness (!!) pick
$\displaystyle \alpha \in \bigcap_{\xi \leq \alpha} \tilde F_\xi \cap \bigcap_{\xi > \alpha} (\overline{\hat K} \setminus \tilde G_\xi)$

Let $x \in K$ be a complete accumulation point of $\{x_\xi : \xi \in \Gamma_0\}$, which must exist by compactness.

Then [SOMETHING]. You should consult the $\Sigma^2$ paper for details.

“[After having some difficulty proving this technical lemma]. The notation of ‘free’ is self-dual. This causes problems in (my) understanding.”

Now comes the trick: the Baire Characterization Theorem.

By this theorem, $\overline{\hat K}$ is countably tight, which is a contradiction. [End of Claim 1]

Let $\Gamma := \{\alpha < ^\smile \pi : \textrm{int}(F_\alpha) \in \dot{G}\}$, which is countable by claim 1.

##### Claim 2. $\Vdash_{\mathcal{B}} \{\textrm{int}(F_\alpha) : \alpha \in \dot \Gamma\}$ is a base of $\dot{\mathcal{G}}$.

Take $O \subseteq \mathcal{B}^+$ and $O^\prime \in \mathcal{B}^\prime$.

So $O \Vdash O^\prime \in \mathcal{G}^\prime$. i.e. $O \subseteq O^\prime$.

Find $O_1$ such that $O_1 \Vdash \textrm{int}(F_\alpha) \in \mathcal{G}^\prime, \textrm{int}(F_\alpha) \subseteq O^\prime$.

(“Why not $F_\alpha \subseteq O^\prime$?“) [End of Claim 2]

##### Corollary: $K$ has a $\sigma$-disjoint $\pi$-base. (For $\sigma$-discrete, this is metric space.)

proof. Pick a $\mathcal{B}$-name $\tau$ such that $\Vdash_\mathcal{B} \tau : \omega \rightarrow \dot{\mathcal{G}}$ is a decreasing sequence which generates $\dot{\mathcal{G}}$.

Define a refining sequence $P_n$ for $n<\omega$ of maximal antichains of $\mathcal{B}^+$ recursively as follows:

(“How do they relate to $\tau$?“)

Let $P_0$ be a maximal antichain in $\mathcal{B}^+$ deciding $\tau(0)$.

inductive step.

For each $O \in P_n$ let $P_{n+1} (O)$ be a maximal antichain of regular open subsets of $O$ deciding the values $\tau(m)$ for $m \leq n+1$.

This is what you had to do. Now we hope for $\pi$-base.

Let $P := \bigcup_{n < \omega} P_n$.

##### Claim 3. $P$ is a $\pi$-base of $K$.

This is more or less obvious, but I am now suspicious of myself [becuase of the difficulty with the technical proofs.]

Now where is the dense metrizable subspace? It is the first countability.

Recall: $K^\prime := \{x \in K : x \textrm{ is a } G_\delta \textrm{ point }\}$ is dense in $K$.

(Question: Is $K^\prime$ a $G_\delta$ set itself? Note $\aleph_1$ sized compact sets contain $G_\delta$‘s and we are not adding reals.)

So $K^\prime$ is a first countable space with a $\sigma$-disjoint $\pi$-base. [End of Corollary]

##### Lemma from General Topology. Every first-countable space with a $\sigma$-disjoint $\pi$-base has a dense metrizable subspace.

hint at proof. Basically you have to avoid boundaries.

## We spend the rest of the time talking about measure.

Let $K \subseteq \mathcal{B}_1 (\N^\N)$ be compact and $\mu$ a Radon probability measure.

##### Theorem (Bourgain’s Thesis). For every compact $K$ of Baire-Class-1 functions, and every Radon probability measure $\mu$ on $K$, then $L_\mu (K)$ is separable.

(“In human language: the measure algebra of $\mu$ is separable; there are no wild measures; the space behaves like a measure space.“)

The proof is interesting from the point of view of the course. Remember when you suffered through all that Martin’s Axiom stuff at the beginning of the course? Now it will be of use.”

proof. Otherwise there is a sequence $\{A_\alpha : \alpha < \omega_1\}$ of $\mu$-independent $F_\sigma$ subsets of $K$ each of measure $1/2$, (by Maharam).

You want them to be topologically independent, not just measure independent.

You really want a sequence (“of regular pairs, it is always the same“) $\{(F_\alpha, G_\alpha) : \alpha < \omega_1\}$ of regular pairs that are independent, i.e. :
$\displaystyle \bigcap_{\alpha \in A} F_\alpha \cap \bigcap_{\beta \in B} (X \setminus G_\beta) \neq \emptyset$
for all finite, disjoint $A, B \subseteq \omega_1$. (There is no assumption that $A < B$. That would make the sequence free.)

We also need a theorem of Fremlin:

##### Lemma (Fremlin). Assume a compact space $K$ supports a non-separable Radon measure. There is a ccc poset $\mathbb{P}$ which introduces an uncountable independent sequence $\{(F_\alpha, G_\alpha) : \alpha < \omega_1\}$ of regular pairs of $K$.

(“It was rather surprising that is was ccc, and not just proper.“)

Take such a ccc poset as in Fremlin’s Lemma.

Work in $V^\PP$ and reinterpret $K$ to $\overline{\hat K}$ and $\{(\tilde F_\alpha, \tilde G_\alpha) : \alpha < \omega_1\}$. For each $\alpha$ pick $f : \overline{\hat K} \rightarrow [0,1]$ such that $f_\alpha \upharpoonright \tilde F_\alpha \equiv 0$ and $f_\alpha \upharpoonright (\overline{\hat K} \setminus \tilde G_\alpha) \equiv 1$.

Define $f: \overline{\hat K} \rightarrow [0,1]^{\omega_1}$ by $f(x) = \{f_\alpha (x) : \alpha < \omega_1\}$. The range of $f$ contains $\{0,1\}^{\omega_1} \subseteq [0,1]^{\omega_1}$.

Recall that $\omega_1 + 1$ embeds into $\{0,1\}^{\omega_1}$, so $\overline{\hat K}$ cannot be countably tight! [QED]