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) .

Recall Baire’s Characterization Theorem (“A great theorem“.): f \in \mathcal{B}_1 (\N^\N) iff for all closed K \subseteq \mathbb{N}^\N , f \upharpoonright K has a point of continuity. (Moreover, this K can be assumed to be perfect.)

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
\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
\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.

Extend only when you can.

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“.

\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
\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!

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

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