(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.)

**Summary**

- We spend the entire lecture proving a theorem about Baire-Class-1 functions.
- The proof is interesting because it uses forcing.
- The entire proof is in the inductive step.

## Continuing from last class

**Lemma**: is relatively compact

**Restatement**. .

##### Recall **Baire’s Characterization Theorem** (“A great theorem“.): iff for all closed , has a point of continuity. (Moreover, this can be assumed to be perfect.)

## Proof of Lemma

Suppose we have and a -name for a function from into such that .

“So basically we try to reflect .“

By Baire’s Characterization Theorem, there exist -names and and such that

and

“This is what discontinuity means.”

“There will be dense sets that get separated ().”

“ just decides the rationals and .”

“Now we approximate.”

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

Starting with we construct:

- where ;

“What else? …“

- is decreasing;
- are increasing;
- (“the forking level of some perfect tree“) is finite;
- , for ;
- , for ;

“So that something happens…“

- where ;
- . (“These are branches through some tree“);
- . (“This is making sure we reflect that are dense“);
- . (“Here decides the value of all ‘s up to a point.” [To himself:] “But why do I care about all ‘s?“)
- for all ;
- ;
- for all ;

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

“Let us see if we can survive a step.”

## to

Since forces is perfect, find , and for each two extensions such that

and

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

“Now I use the fact that these sets are forced to be be dense in .“

We can find an extension so and

“Remember the limit function is not continuous.”

“Now we work. What is “

Find and such that:

- ;

(“They agree with up to a finite stage.” is a pointwise limit of continuous functions. Plug in to get the error.)

Find such that .

Find and such that

“This is pointwise convergence. We have finitely many conditions.“

Find and for each and and 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.“

**the thing I mysteriously boxed off in my notes**

(notably, , )

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

(The second set is because “I’m building a tree!” and the last set is “things that project down to “)

Now, let (or really the downward closure), and observe is perfect.

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

Let

and let

and now

“Now do we have a contradiction?“

Fix an . By the inequalities,

and

## “Now comes the trick…“

Find a subsequence which converges to a Baire-Class-1 function!

So

and

a contradiction. [**QED**]

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