(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.)
- We introduce free sequences.
- A lemma (2) relating free sequences and -bases is proved.
- We prove that the generic ultrafilter is countably generated.
- We construct nice -disjoint -bases.
- A realted theorem about compact spaces supporting a radon measure is proved as a corollary.
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 , compact, and then 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 in some regular space is a pair where with closed, open.
definition 2. A sequence of regular pairs is free iff
for all pairs of finite subsets of , (where each element of is below each element of ).
For more information on this you can see Stevo’s paper “On a functor.”
Lemma 2. Let be a (non-trivial) compact space and let -weight , then there is a sequence of regular pairs such that
- is a -basis of ;
- If then is centred iff is free.
This shows up in Arhangel’skii’s famous theorem that first countable spaces have size .
Topologically: is free iff
“Why do we care about this? You will see in a moment.”
“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 ‘s are sets and the ‘s are . Then by compactness you are finished.
“Back to our case.“
Let be compact, with -weight and as in lemma 2. Moreover, assume that the ‘s and ‘s are of the following form:
- ; and
- for each
Let be the -name for the generic filter.
Claim 1. is countable.
Suppose that in some forcing extension , is uncountable. Let be the first members of .
“I don’t care. Just take the first many.“
Recall that is relatively compact, so
(“Do the definition in the extension.“)
“Now there is some funny notation. We could spend a long time chasing the truth.“
We might try
or something like it.
All we want is that is a free sequence. (“This is finitary, so it isn’t a big deal.“)
For , by compactness (!!) pick
Let be a complete accumulation point of , which must exist by compactness.
Then [SOMETHING]. You should consult the 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, is countably tight, which is a contradiction. [End of Claim 1]
Let , which is countable by claim 1.
Claim 2. is a base of .
Take and .
So . i.e. .
Find such that .
(“Why not ?“) [End of Claim 2]
Corollary: has a -disjoint -base. (For -discrete, this is metric space.)
proof. Pick a -name such that is a decreasing sequence which generates .
Define a refining sequence for of maximal antichains of recursively as follows:
(“How do they relate to ?“)
Let be a maximal antichain in deciding .
For each let be a maximal antichain of regular open subsets of deciding the values for .
“This is what you had to do. Now we hope for -base.“
Claim 3. is a -base of .
“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: is dense in .
(Question: Is a set itself? Note sized compact sets contain ‘s and we are not adding reals.)
So is a first countable space with a -disjoint -base. [End of Corollary]
Lemma from General Topology. Every first-countable space with a -disjoint -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 be compact and a Radon probability measure.
Theorem (Bourgain’s Thesis). For every compact of Baire-Class-1 functions, and every Radon probability measure on , then is separable.
(“In human language: the measure algebra of 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 of -independent subsets of each of measure , (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“) of regular pairs that are independent, i.e. :
for all finite, disjoint . (There is no assumption that . That would make the sequence free.)
We also need a theorem of Fremlin:
Lemma (Fremlin). Assume a compact space supports a non-separable Radon measure. There is a ccc poset which introduces an uncountable independent sequence of regular pairs of .
(“It was rather surprising that is was ccc, and not just proper.“)
Take such a ccc poset as in Fremlin’s Lemma.
Work in and reinterpret to and . For each pick such that and .
Define by . The range of contains .
Recall that embeds into , so cannot be countably tight! [QED]
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