So the point of this post is to work through one of Shelah’s results that (roughly speaking) allows us to create a complicated coloring if we have a scale in hand. We begin with a few definitions.
Def: Let be cardinals with . We say that holds if there is a coloring of the pairs of in -many colors such that:
Given a sequence of pairwise disjoint elements of , and an ordinal , there are such that for every and , we have .
We can think of as a strong form of the assertion that . For example, if we let be a singular cardinal, then says that there is a coloring such that: Given any set and any ordinal , there are such that . In other words, this family of coloring principles assert that Ramsey’s theorem fails miserably, and in particular tells us that fails to be Jónsson in an incredibly spectacular way.
What I want to do in this post is work through a result of Shelah’s which allows us to produce witnesses to for any singular cardinal by using a scale to color pairs of . First recall what a scale is.
Def: Let be singular. We say that the pair is a scale on if:
- is a set of regular cardinals cofinal in ;
- is strictly increasing and cofinal in where is the ideal of bounded subsets of .
A theorem of Shelah’s tells us that scales exist. Scales allow us to show that some of the combinatorial information on can be pulled down to or even . In particular, scales are extremely useful for elementary submodel arguments because they allow us to smuggle information about into our submodel without needing all of or even present. This will be made apparent in the proof of the following (which can be found on page 67 of Cardinal Arithmetic):
Lemma 4.1B: (Shelah) Let be a scale on . Then, letting the principle holds. In particular, always holds.
The coloring used in the proof of the above lemma basically takes two ordinals to the first point past which and diverge completely. Much in the same way that coloring using Skolem functions is a useful technique, it’s nice to have this coloring lying around since scales always exist.
Proof: We begin by fixing a scale as in the hypotheses, and let . Let , and index so we may think of a function as a function in . Now partition into -many pieces by way of such that is unbounded in for each . For ease of notation, let . Now we define two colorings, on pairs of as follows:
Let be defined by for .
Let be defined by . So takes a pair to the first point where their corresponding functions diverge, and takes the associated cardinal in to which piece of the partition that cardinal lies in. We claim that witnesses . With that said, fix a sequence of pairwise disjoint elements of and note that there is some such that, for -many , we have . So, we may as well assume that each has this property by thinning out the collection we were originally handed, and reindexing. Further, let be a given color.
We want to show that there are such that for each we have . There are going to be a lot of indices to keep track of, but keeping the goal in mind (no matter how horrible it looks) will help. Our main tool for proving that this coloring is appropriate will be elementary submodels, so let be regular, and large enough with such that:
It won’t quite be apparent why we’re utilizing this machinery until a little bit later. Now since the collection is disjoint, we can assume that for each by reindexing the sets , and perhaps chopping off initial segments. Next let and define the characteristic function of with respect to by for every and everywhere else. Since , it follows that is non-zero almost everywhere (mod bounded) and so is in . Next since is cofinal in , there is some such that:
So not only does get above almost everywhere, we have that . This will allow us to take an appropriate Skolem Hull of and know that these things still happen. Next, since , we can fix for each an index such that: For every , we have that . Let since , so then for every and every , we have . This give us a pair of canonical witnesses to the fact that is thin, and gives us one of the indices we will use to witness the desired coloring property. So our next goal is to find our second (smaller) witness, which we will want to see enough of.
For each , define a function by:
Note that since , we have that . So above each , we have -many functions above it, and we are letting be the closest approximation to we can get by way of these ladders of functions. Next let , which we claim is -equivalent to . Otherwise, suppose that , and define a function by if and otherwise. As before, we can find some such that which is absurd. Now we pick an index such that:
We now make our first approximation to our desired companion to . We let be such that
We note that such a choice is possible, as implies that whereas since . Really, we only wanted to find an index such that we can get above each , and the important thing here is . With that said, let , and note that we have . So by elementarity we have
by our requirements on , and the fact that . So let be such that . Thus:
We claim that and are as desired. That is, we need to show that and for every pair , we have (so also that ). So if we can show that , then we are done since . But by the above inequality, we only have to show the following three things:
- For every pair , we have ;
- For every , we get that .
By a lemma due to Baumgartner, we have that (by our choice of , for every regular , if , then
Now since , it follows since , that as well. But since , we have that , and similarly for each since each is definable from parameters in . Now that we’ve shown 1. and 2. above, we only need to show 3.
Now for every , and we have that since it’s definable from parameters, and hence . On the other hand, our choice of tells us that , and so the result follows.