On ADS (Part 3)February 9, 2012 at 11:40 | Posted in Uncategorized | Leave a comment
Continuing the construction, let us assume that is singular of countable cofinality and . For each , let be a sequence of sets such that
- for all , and
By induction on we choose sets such that for no and is a subset of .
Notice that each set of the form is of cardinality less than , and there are sets of this form. Our assumption that is greater than tells us that a suitable can always be found.
Let be a bijection, and let . We will show that the collection will witness .
What this amounts to is that for each , we need a function so that
for all , where
Lemma 1 For each , the family has a transversal (i.e., a one-to-one choice function) .
Proof: We define by induction on . This is trivial for and limit. If , then is not a subset of so can be defined easily.
Armed with the preceding lemma, we define a function as follows:
Given , we know for some unique . We define to be the unique such that .
In English, answers the question “when does enumerate the value of ?”.
Our goal is to show that the function almost works, and then show how to modify to a function that actually does the job.