# Stevo’s Forcing Class Fall 2012 – Class 14

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

## Continuing from last class

##### Lemma: $\Vdash_\mathcal{B} \hat F = \{\hat f : f \in F\}$ is relatively compact

Restatement. $\Vdash_\mathcal{B} \overline{\hat F} \subseteq \mathcal{B}_1 (\N^\N)$.

## Proof of Lemma

Suppose we have $p \in \mathbb{P}$ and $\dot g$ a $\mathbb{P}$-name for a function from $\mathbb{N}^\N$ into $\mathbb{R}$ such that $p \Vdash \dot g \in \overline{\hat F} \setminus \mathcal{B}_1 ({(\N^\N)}^{V^\mathcal{B}})$.

So basically we try to reflect $g$.

By Baire’s Characterization Theorem, there exist $\mathbb{P}$-names $\dot p , \{\dot{x}_n : n < \omega\}, \{\dot{y}_n : n < \omega\}$ and $\epsilon < \delta$ and $p_0 \leq p$ such that
$\displaystyle p_0 \Vdash \overline{\{\dot{x}_n : n < \omega\}} = \overline{\{\dot{y}_n : n < \omega\}} = \dot p$
and
$\displaystyle \forall n, p_0 \Vdash \dot{g}(\dot{x}_n) < \epsilon \textrm{ and } \dot{g}(\dot{y}_n) > \delta$

This is what discontinuity means.

There will be dense sets that get separated ($\epsilon < \delta$).

$p_0$ just decides the rationals $\epsilon$ and $\delta$.

Now we approximate.

I think I wrote that there were seven inductive hypothesis [in my notes], so let us write them.

Starting with $p_0$ we construct:

• $\{f_n\} \subseteq F, \{f_n^k\}_{k < \omega} \rightarrow f_n$ where $\{f_n^k\} \subseteq \mathcal{C} (\N^\N)$;

What else? …

• $(p_n) \subseteq \mathbb{P}$ is decreasing;
• $\{k_n : n < \omega\}, \{i_n : n<\omega\} \subseteq \mathbb{N}$ are increasing;
• $T(n) \subseteq \mathbb{N}^{i_n}$ (“the forking level of some perfect tree“) is finite;
• $m_t \subseteq \mathbb{N}$, for $t \in T(n)$;
• $n_t \subseteq \mathbb{N}$, for $t \in T(n)$;

So that something happens…

1. $\forall t \in T(n), \exists t_0 \neq t_1 \in T(n+1)$ where $t \subseteq t_0 \cap t_1$;
2. $p_{n+1} \Vdash T(n) \subseteq \dot{p} \upharpoonright i_n$. (“These are branches through some tree“);
3. $\forall t \in T(n), p_{n+1} \Vdash \dot{x}_{m_t} \upharpoonright i_n = t = \dot{y}_{n_t} \upharpoonright i_n$. (“This is making sure we reflect that $\{\dot{x}_n : n < \omega\}, \{\dot{y}_n : n < \omega\}$ are dense“);
4. $p_{n+1} \Vdash \dot{x}_0 \upharpoonright i_n = s_0^n, ..., \dot{x}_n \upharpoonright i_n = s_n^n$. (“Here $p_{n+1}$ decides the value of all $x$‘s up to a point.” [To himself:] “But why do I care about all $x$‘s?“)
5. $f_m^{k_n} [s_l^n] \subseteq (-\infty, \epsilon)$ for all $l \leq m \leq n$;
6. $p_{n+1} \Vdash \dot{y}_0 \upharpoonright i_n = t_0^n, ..., \dot{y}_n \upharpoonright i_n = t_n^n$;
7. $f_m^{k_n} [t_l^n] \subseteq (\delta, \infty)$ for all $l \leq m \leq n$;

[At this point someone walks in and just stares at the board.]

Let us see if we can survive a step.”

## $n$ to $n+1$

Since $p_{n+1}$ forces $\dot p$ is perfect, find $q \leq p_{n+1}$, $i_{n+1}^\prime > i_n$ and for each $t \in T(n)$ two extensions $t^0 \neq t^1 \in \mathbb{N}^{i_{n+1}}$ such that
$\displaystyle q \Vdash [t^0 ] \cap \dot p \neq \emptyset \neq [t^1 ] \cap \dot p$
and
$\displaystyle T^\prime (n+1) = \{t^0, t^1 : t \in T(n)\} \subseteq \textrm{decisions of } q_0 \textrm{ about } \dot{x}_{m_t} \upharpoonright i_{n+1}^\prime, \dot{y}_{n_t} \upharpoonright i_{n+1}^\prime \textrm{ where } t \in T(n)$

(Aside: this will turn out to almost work. And the second part is addressing (3).)

Now I use the fact that these sets $\{\dot{x}_n : n<\omega\}, \{\dot{y}_n : n<\omega\}$ are forced to be be dense in $\dot{p}$.

We can find an extension $q_1 \leq q_0$ so $\{m_t, n_t : t \in T(n+1)\} \subseteq \mathbb{N}$ and
$\displaystyle q \Vdash \dot{x}_{m_t} \upharpoonright i_{n+1}^\prime = t, \dot{y}_{n_t} \upharpoonright i_{n+1}^\prime = t$

Remember the limit function is not continuous.”

Now we work. What is $f_{n+1}?$

Find $q_2 \leq q$ and $f_{n+1} \in F$ such that:

• $q \Vdash \hat{f}_{n+1} (\dot{x}_l) < \epsilon, \forall l \leq n$;
• $q \Vdash \hat{f}_{n+1} (\dot{y}_l) > \delta, \forall l \leq n$

(“They agree with $g$ up to a finite stage.$g$ is a pointwise limit of continuous functions. Plug in $\{\dot{x}_l : l \leq n\}, \{\dot{y}_l : l \leq n\}$ to get the error.)

Find $\{f^k_{n+1} : k < \omega\} \subseteq \mathcal{C}_1 (\N^\N)$ such that $f^k_{n+1} \rightarrow_k f_{n+1}$.

Find $q_3 \leq q$ and $k_{n+1} > k$ such that
$\displaystyle q \Vdash \forall k \geq k_{n+1}, \forall l \leq n, \hat{f}^k_{n+1} (\dot{x}_l) < \epsilon, \hat{f}^k_{n+1} (\dot{y}_l) > \delta$

This is pointwise convergence. We have finitely many conditions.

Find $i_{n+1} > i_{n+1}^\prime$ and for each $m \leq n$ and $l \leq n, s_l^{n+1} \in \mathbb{N}^{i_{n+1}}, t_l^{n+1} \in \mathbb{N}^{i_{n+1}}$ and $q_4 \leq q$ such that (5) and (7) are satisfied. (“I want to use continuity of all functions accumulated so far.“)

You might be worried about smaller indices, but that is the Inductive hypothesis.

Anything that doesn’t satisfy this [following condition] is out.

$\displaystyle T(n+1) \supseteq \{\textrm{extensions of } T^\prime(n+1)\} \supseteq \{s_l^{n+1}, t_l^{n+1} : l \leq m \leq n\} \cap [T^\prime(n+1)]$

(The second set is because “I’m building a tree!” and the last set is “things that project down to $T^\prime(n+1)$“)

Now, let $T := \bigcup_{n \in \omega} T(n)$ (or really the downward closure), and observe $Q := [T]$ is perfect.

Now reflection of the points will be the ‘u’s“.

Let
$\displaystyle u_l := \bigcup_{l \leq n < \omega} s_l^n \in Q$
and let
$\displaystyle v_l := \bigcup_{l \leq n < \omega} t_l^n \in Q$
and now
$\displaystyle \overline{\{ u_l : l < \omega \}} = Q = \overline{\{ v_l : l < \omega \}}$

Now do we have a contradiction?

Fix an $m$. By the inequalities,
$\displaystyle f_m^k (u_l) \rightarrow_k f_m(u_l) \leq \epsilon$
and
$\displaystyle f_m^k (v_l) \rightarrow_k f_m(v_l) \geq \delta$

## “Now comes the trick…“

Find a subsequence $f_{m_k}$ which converges to a Baire-Class-1 function!

So
$\displaystyle f \upharpoonright \{u_l : l < \omega\} \leq \epsilon$
and
$\displaystyle f \upharpoonright \{v_l : l < \omega\} \geq \delta$
a contradiction. [QED]

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