1. Colorings of pairs. I
There are several possible ways in which one can try to generalize Ramsey’s theorem to larger cardinalities. We will discuss some of these generalizations in upcoming lectures. For now, let’s highlight some obstacles.
Theorem 1 (-Kakutani) In fact,
Proof: Let Let be given by
Then, if are distinct, it is impossible that
Proof: With as above, let be given as follows: Let be a well-order of in order type Let be the lexicographic ordering on Set
Proof: Let be a counterexample. Let be least such that has size and let be such that if then To simplify notation, we will identify and For let be such that but By regularity of there is such that for many
But if and then iff so It follows that has size contradicting the minimality of
The lemma implies the result: If has size and is -homogeneous, then contradicts Lemma 3.
Now I want to present some significant strengthenings of the results above. The results from last lecture exploit the fact that a great deal of coding can be carried out with infinitely many coordinates. Perhaps surprisingly, strong anti-Ramsey results are possible, even if we restrict ourselves to colorings of pairs.
Recall our convention that if for some subset of a linearly ordered set, then
proved the following theorem in 1987. This is one of the gems of modern combinatorial set theory. Previous versions of this result were known to Hajnal and Rado in the presence of The argument below is due to Velleman, from 1990; thanks to Assaf Rinot for pointing me to this argument.
Proof: As in the proof of Theorem 4 from last lecture, it is enough to find a function such that whenever then contains a club. In effect, let be such that is stationary for all . and consider If then for all so
To define begin with an injective sequence of functions Also, for each fix an injection
For countable ordinals let
To see that is as wanted, fix an uncountable For let
If is countable, let
so is club. Now we show that
To see this, pick and with We will find an such that and
and so or Since and we have that so is uncountable, and
For in let and
Since and, by the definition of but so
Since is uncountable, there is an uncountable an and an such that for all () and so and it follows that Then is uncountable and, since Note that
Note that is finite, since is injective. We can then find an with Note that
Since then Since and then Since and then
Finally, if then, since we must have It follows that so This shows that Since was arbitrary, it follows that and we are done.
The result also holds for many larger cardinals. For example, we have the following generalization obtained independently by Shelah and The proof below is Shelah’s.
Proof: We use the method of walks.
Fix stationary and nonreflecting; we may assume that all the ordinals in are limit ordinals. For each nonzero choose so that:
- If is limit, is club in and disjoint from and any point of is a successor ordinal.
Fix a partition of into stationary sets.
Given in define by induction on
- If is defined and strictly larger than let
- Let be the least such that At this point, the construction stops.
We can think of the sequence of as describing a walk from down to through the ladders Note that is well-defined, since the form a decreasing sequence of ordinals.
Note that if are in and then For we have and equality occurs only if
Suppose are in Let Set
Then and, if then
- and for all and
Now define as follows: Given in let be largest such that, letting for all
- for all and
if such an exists, and in that case let be the unique such that Otherwise let
(The strange looking condition 2 will be used at the end of the argument. For now, notice that if does not satisfy conditions 1 and 2, then no larger can satisfy them either.)
Let now and be given. We need in with
Consider the model where holds iff Let be a continuous increasing sequence of elementary substructures of with and for all
Note that there is a club of ordinals such that This is because both and are filtrations of
Definition 6 A filtration of is a strictly increasing continuous sequence with union such that for all
(The above definition is somewhat more restrictive than the general notion of filtration one finds in algebra.) Suppose and are filtrations of By regularity of given any we can find such that and similarly there is some such that By an obvious back and forth argument we can find a club of ordinals such that
In particular, for club many We can then find a limit ordinal such that
Choose so and is defined and larger than
Let We claim that
Note that if then and therefore since is limit.
If then Since then is a successor ordinal, and must therefore be Hence This is the only place in the argument where we use the assumption that is nonreflecting.
Thus for all Therefore as claimed.
Let be the first order formula in the language of with parameters for all asserting that, letting and for all we have that:
- is limit,
- for all and
(That there is such a formula takes but a moment’s thought and is best argued by induction on The strange looking condition 4 is the counterpart of condition 2 in the definition of and will be used towards the end of the argument.)
Note all the parameters in belong to and since
Recall that and Let Then
so, by elementarity, the same holds in Since was arbitrary,
Now let Then
By elementarity, the same holds in and, since was arbitrary, then
Therefore there are in such that and
For all since we have that and
In particular, for all and and therefore
We claim that
To see this, notice that if then:
- for all since and Here, denotes for all
This means that satisfies the requirements of in the definition of
On the other hand, and, since it is larger than
But then no satisfies the requirements of in the definition of To see this, note that
for all in particular for
However, we must also have so
This means that in the definition of and Since we are done.
Corollary 7 If is regular,
Proof: The set is stationary and nonreflecting. This is because consists of ordinals of cofinality for any If is a club set of minimal order type, such that every ordinal in is a successor, then no ordinal in has cofinality and therefore avoids
This implies that for all nonzero
The situation for is different, and we turn to it before moving to larger cardinals.
The argument of Theorem 5 makes use of (basic) ideas from mathematical logic. This is a very useful tool in combinatorial set theory, and more substantial applications will be encountered in future lectures.
2. Canonical partitions
Definition 8 Let let be a set, and let A function is -canonical on iff whenever and are subsets of then iff for all
(Compare with the discussion of canonical relations in of lecture III.1.)
Here are some examples; suppose that and that is -canonical on Then:
- iff is constant on
- iff is injective on
- iff is (strictly) min-homogeneous, i.e., whenever then iff
- iff is (strictly) end-homogeneous, i.e., iff
- iff is (strictly) max-homogeneous, i.e., iff
(The nonstrict versions are obtained by replacing equivalence with implication: is min-homogeneous iff whenever etc.)
Note that if is finite, then necessarily so this generalizes Ramsey’s theorem.
Theorem 10 For all and all there is an -canonical partition of
Proof: The obvious attempt is to define by
where if and then means that
Clearly, if then
Conversely, suppose Let Then so Hence, Conversely,
Recall now the ultrafilter proof of Ramsey’s theorem from of lecture III.1. We showed there that if is a nonprincipal ultrafilter over and for some finite then there is some infinite set and a color such that and Here,
Definition 11 An ultrafilter over is Ramsey iff it is nonprincipal and for all is generated by the sets for
Whether there are Ramsey ultrafilters is independent of and follows from
It follows from the definition and the argument in of lecture III.1 that a nonprincipal ultrafilter over is Ramsey iff for any and any finite coloring of there is some such that is monochromatic.
For we have a “Ramsey-ultrafilter” version of Theoren 9:
Homework problem 14. Let be a nonprincipal ultrafilter over Show that is Ramsey iff for any there is some such that is 1-1 or constant.
Now we proceed to the proof of the canonization Theorem 9:
Proof: Let be given. For let be the set of all tuples where and
(Note that, against convention, we are not requiring that )
Then defines a coloring of with finitely many colors, and Ramsey’s theorem guarantees that there is some such that is constant. Let be the enumeration of
Let and let if and then
We claim that is -canonical on
Step 1: Suppose and We need to prove that
For this, we define a transformation as follows:
- If then
- Otherwise, let be least such that (so .) Let be such that and Then, by definition of and We then set
Iterating we see that for some such that and we are done.
Step 2: Conversely, suppose that that and that We need to prove that
Assume otherwise. Define an equivalence relation on pairs of elements of by setting for iff there is an order isomorphism such that and
The following is immediate from the fact that is homogeneous for
For let For let be such that
Then there is:
- An such that
- An such that
- An such that
Say that enumerates Then, by definition of and recalling that we have that
Since we can then find with such that for Let be such that
Define by setting and
Note that since then
However, by choice of and the definition of we have that and and therefore (since is homogeneous for ) and (by Lemma 12) contradiction.
It follows immediately that Let We can then find and such that is -canonical on If then is constant. For by passing to an appropriate proper subset of we necessarily make the range of smaller. In either case, we find an infinite subset of such that
We can extract a bit of additional information from the proof of Theorem 9 without too much effort. I state the result without proof.
Definition 13 Let and consider a coloring
Say that is invariant iff whenever and with as in the proof of Theorem 9 above, then iff
Say that is isomorphic to iff whenever and is the unique order isomorphism, then iff
Say that is stationary iff for all
Then we have:
Theorem 14 (Rado) The following are equivalent for and
- is invariant.
- is stationary.
- is -canonical on for some
3. Colorings of pairs. II
Magidor has shown that, consistently, every stationary subset of may reflect. Therefore we cannot apply Theorem 5 to conclude that Fortunately, this case is covered by another result of
Recall from Theorem 20 in lecture II.12 that for every singular cardinal there is an increasing sequence of regular cardinals cofinal in such that
i.e., there is a sequence where each and we have that whenever and that for any there is some such that where iff for all but boundedly many we have that
We call such a tuple a scale for
I omit the proof of Theorem 15 (which also uses an elementary substructures argument); see for example the paper by Burke and Magidor cited at the end. Instead, I briefly describe the witnessing coloring:
For each let witness Consider the map where is the largest such that if such exists, and otherwise.
Now let be the coloring given by where if and otherwise.
One then proceeds to show that witnesses To conclude, the following “stepping up” argument can now be invoked:
Lemma 16 Suppose that is an infinite cardinal and that Then also
Proof: Given witnessing let be an injection for all and set by if and otherwise.
To see that works, let and fix Let so but It follows that there is some such that
But now we can conclude that there are in such that and Thus and we are done.
Notice that the result applies to all singular cardinals in a large initial segment of the sequence of cardinals, for example, it applies to all below the first weakly inaccessible cardinal.
An analysis of the argument showing that indeed works led Shelah to isolate a family of combinatorial principles from which the following result follows:
Theorem 17 (Shelah) If is singular, then
Shelah has also shown that for many inaccessible cardinals (for example, for any cardinal that is inaccessible but not Mahlo.) Further results have been obtained by Eisworth and Shelah.
It is still open whether there can be a successor cardinal (necessarily, the successor of a singular) such that
Before transformed the area with his results, some particular cases were known. The following is from 1965:
Proof: Using that let enumerate the bounded subsets of in such a manner that
We claim that there is a function such that for all in with and all there is such that
Assume that is such a function, let and let Find some such that and let be larger than and than Then for some we have and we are done.
To prove the claim, for each let and list as Now define recursively so that for all Now set for all
The following extension of Theorem 18 shows that one can require a stronger property of the witnessing function
Proof: Let enumerate all partial functions with and
Given let and fix a bijection where
Now, by transfinite recursion, define so that
for all Notice that this is possible, since for all
To define we set for all and all and let for all not specified by this rule.
We claim that is as wanted. Suppose otherwise, and pick counterexamples with and Define by picking
Let be such that and pick such that and Let for the unique ordinal such that Then, by construction, But this contradicts the definition of and we are done.
One sometimes writes for and and and for and and The conclusion of the theorem above can then be stated as saying that or even that for all suitable
Notice that this is obvious if we just require that there is some pairwise disjoint collection in of size consisting of stationary subsets of all of which belong to This weaker version is an obvious consequence of the existence of Ulam matrices in does not require to hold in and only needs that
Now define for all We claim that for some all the sets are stationary. Clearly, they are disjoint, and the result follows.
To see the claim, assume otherwise. Then for each we can find and a club disjoint from Now set This is an uncountable (in fact, a club) set, so there is some such that
In particular, there is some such that But and so and we are forced to conclude that contradiction.
On the other hand, Paul Larson has shown that the conclusion of Corollary 20 (that there is such a sequence of length ) fails in this generality. Larson’s argument builds a forcing extension of in which reals are added and is collapsed. It is open whether Corollary 20 holds without the assumption of but assuming that no reals are added and no cardinals are collapsed.
In this regard, I should mention here the following recent result of Justin Moore. Moore’s construction allowed him to solve the -space problem and settle several famous problems in set theory and set theoretic topology. Moreover, just as ‘s original proof, Moore’s argument is sufficiently absolute that the function he builds satisfies the conclusion of Theorem 22 in any outer model with the same
I find it impossible to resist mentioning that Magidor has reproved Corollary 20 with a remarkably short argument. I proceed to include it below; once again, thanks to Assaf Rinot for showing me this result.
Theorem 23 (Magidor) Let be a cardinal, assume that and let be an outer model such that Then for every stationary in that remains stationary in there exists in a silly diamond sequence for over that is, a sequence such that, in is stationary for all
Proof: Working in let enumerate Given let enumerate
for all It is enough to check that for some is as wanted.
To see this, assume otherwise. Working in pick and for all so that for all Now set Since there is some such that Let be larger than and let be such that Then However, contradiction.
Corollary 24 Assume and let be an outer model with Then there is in a partition of into stationary sets, all of which remain stationary in
Proof: Note that the assumption guarantees that
Let be a silly diamond sequence for over For let
Then is stationary in Since lies in we are done.
To conclude, I want to mention briefly what the situation is at singular cardinals.
Theorem 25 Let be a singular cardinal and let be an integer. Then
Proof: Let be a partition of into many sets of size smaller than Given let be the ordinals such that and set
Note that the numbers in the sequence are nonzero and add up to
Lemma 26 For any integer let be the set of all sequences where is an integer, and and let Then
Proof: We may as well define noting that Note also that For and let be the set of all sequences in with Then and
The result follows now by an obvious induction.
It follows that there are possible values for and we can then define by
If there are at least different ordinals such that and it follows that
On the other hand, a version of the Canonization theorem for singular cardinals allows one to show the following:
i.e., for any coloring there is an such that
As we will see in future lectures, other than the cardinals such that (the weakly compact cardinals) are strongly inaccessible. I suspect that stronger versions of Theorem 27 should be known (with weaker large cardinal restrictions on or without the strong limit assumption on ), and would be glad to hear of any references.
Here are some references consulted while preparing this note:
- Maxim Burke, Menachem Magidor, Shelah’s pcf theory and its applications, Annals of Pure and Applied Logic, 50 (1990), 207–254.
- Todd Eisworth, Successors of singular cardinals, in Handbook of set theory, Matthew Foreman, Akihiro Kanamori, eds., forthcoming.
- Paul András Hajnal, Attila Máté, Richard Rado, Combinatorial set theory: partition relations for cardinals, North-Holland (1984).
- Akihiro Kanamori, The higher infinite, Springer (1994).
- Richard Rado, Note on canonical partitions, Bulletin of the London Mathematical Society, 18 (1986), 123-126.
- Assaf Rinot. Surprisingly short, unpublished note; version of April 3, 2009. Available (as of this writing) at his webpage.
- Saharon Shelah, Was Sierpinski right? I, Israel J. Math, 62 (3) (1988), 355–380.
- Stevo , Walks on ordinals and their characteristics, Birkhäuser (2007).
- Dan Velleman, Partitioning pairs of countable sets of ordinals, The Journal of Symbolic Logic, 55 (3) (Sep., 1990), 1019–1021.
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