I want to use this post to work through some material which solidifies the relationship between square-like principles and scales. We begin with a theorem due to Shelah.
Theorem: Let be regular cardinals with . Then there is a stationary set , and a sequence such that:
- Each is a closed subset of ;
- If , then ;
- If , then is a club subset of with order-type .
The above sequence is referred to as a continuity condition for by Shelah in Cardinal Arithmetic. We can think of these continuity conditions as weak fragments of square, which always hold in ZFC. The interesting thing is that continuity conditions allow us to produce exact upper bounds.
Definition: Let be a continuity condition for some stationary set as above, and let be a sequence of functions from to the ordinals. We say weakly obeys if:
If , then for each , we have for each .
One thing to note is that despite having been published in 1994, Cardinal Arithmetic has no new material past 1989. In particular, the existence of continuity conditions over stationary sets was not known in the case that for singular when cardinal arithmetic was sealed. However, Todd Eisworth’s chapter in the handbook has a nice exposition about the above theorem. On the other hand, the following is included in Cardinal Arithmetic (appearing as Claim 2.6A on page 16):
Theorem: Let be an ideal on a regular cardinal . Assume that:
- is regular such that there is some stationary which has a continuity condition ;
- is a sequence of functions from to the ordinals;
- obeys .
Then has a -exact upper bound.
Proof: We first produce a -least upper bound, and then show that this bound must be exact. In order to produce the desired lub, we inductively produce better and better upper bounds which are better and better approximations to a lub as follows:
Stage : Let be defined by .
Stage : Suppose that has been defined, and is a -upper bound for . If is a least-upper bound, then we can terminate the induction. Otherwise, there is some which is an upper bound for such that .
Stage limit: Suppose that has been defined for each . For each , define the set , and for each , define ,the projection of to , by setting .
Note that each is defined -almost everywhere, and that the sequence is -increasing. We claim that there exists some such that for each , we have , and we set . To see this, assume otherwise.
As the sequence never stabilizes we can find a club set such that
and implies .
Let , then is club in of order-type . We then inductively define an increasing sequence of ordinals in as follows:
We let . If has been defined, then we can find be such that . Just pick such that and (which we can do since is an accumulation point of ), and let be in . At limit stages, let , and let be in . Then our collection has the following properties:
Now for each , define the set
By condition 2, each , and so for each we can pick . Since we have -many of these sets, and ranges through , there must be some unbounded subset of for which is constant. Call this constant value . Now suppose are such that , then (in fact, it’s in . Since obeys , we have that , and so . Finally, since are both in and respectively, we get the following string of inequalities:
But then, the sequence is strictly increasing along , which has size . Since happens -many times, this is absurd. Thus, the induction can be carried through limit stages.
Next, we claim that this induction cannot have been carried through -many stages. Otherwise, assume that has been defined for each , and let . Next note since the sequence is -increasing for each , that the sequence must be decreasing and hence stabilize. Thus, we see that must also be eventually constant. On the other hand, for each , contradicting the fact that for large enough.
Therefore, has a -upper bound . If we can show that is exact, then we are done. So suppose otherwise and let witness this. That is, is a function from to the ordinals such that , but there is no such that . So for each define , which is -positive by assumption. Further, note that cannot stabilize modulo else if it does stabilize at some , then define by
if and otherwise.
It is straightforward to check that would then be an upper bound of with and , contradicting that is a lub. So as before, we can find a club such that for all in , we have that with . Letting , we can again find such that
Let , and again find an unbounded subset of such that . If we let be such that , we then get that
This is silly, and so is an exact upper bound of .