**1. Larger cardinalities **

We have seen that (Ramsey) and (-Rado) for any On the other hand, we also have that () and (-Kakutani) for any infinite

Positive results can be obtained for larger cardinals than if we relax the requirements in some of the colors. A different extension, the -Rado theorem, will be discussed later.

This was originally shown by Dushnik and Miller in 1941 for regular, with providing the singular case. For regular one can in fact show something stronger:

Theorem 2 (-Rado)Suppose is regular and uncountable. Then

which means: If then either there is a stationary that is -homogeneous for , or else there is a closed subset of of order type that is -homogeneous for .

(Above, *top* stands for “topological.”)

*Proof:* Assume that for no set with supremum it is the case that is -homogeneous. For each fix an increasing -sequence with limit and define and a sequence as follows:

- and is 1-homogeneous.
- and is 1-homogeneous.
- In general, and

is 1-homogeneous.

- For each this is done for as long as we can. Notice that since if we can continue the construction for many steps, then we obtain a set of order type with supremum such that is 1-homogeneous, contradiction. It follows that the construction must stop at some finite stage. Let be the length of the sequence of so built, thus for any is not 1-homogeneous.

By repeated application of Fodor’s lemma, find stationary, and ordinals and such that for all is limit, for and for

We claim that is 0-homogeneous: If are in then for so if then would be 1-homogeneous, and we could have defined (for example, as ), contradicting the definition of

Since every stationary subset of contains closed subsets of order type for all countable ordinals we immediately obtain:

Corollary 3for any

We will return to these matters later, when we discuss the Baumgartner-Hajnal theorem. Let us now present the proof of Theorem 1:

*Proof:* The argument in Theorem 2 gives a stronger result for regular. Now assume that is singular, let and let be an increasing sequence of regular cardinals cofinal in and such that Fix a coloring

For let

If for every there is some such that we can define a sequence as follows: Let and let be such that Given and let and let be such that It follows that is 1-homogeneous, and we are done.

We can then assume that there is some such that for all Fix a partition of into disjoint sets such that By regularity of either there is an infinite -homogeneous subset of for some and again we are done, or else for each we can find a 0-homogeneous set

Fix Since for each there is some such that it follows that there is some such that

has size

We can now easily build an increasing sequence of ordinals below such that implies But then we can define

for each so Note that being a subset of is 0-homogeneous. Finally, by definition of the

has size and is 0-homogeneous as well, completing the proof.

Corollary 4 ()Let be an infinite set and suppose that and are two well-orderings of Then there is a such that

*Proof:* We may as well assume that is an initial ordinal Work in This is a model of choice. By the -Dushnik-Miller theorem, either there is an infinite subset of where and never coincide, or else there is a as wanted. Since the first case is impossible because the are well-orderings, we are done.

It is somewhat challenging to find a direct argument for the corollary that does not involve passing to a substructure where choice holds. Note that even for there is some work involved, since could have cofinality so the straightforward inductive argument may fail.

I believe it is still open whether holds whenever is a cardinal of uncountable cofinality.

**2. Jónsson cardinals **

Recall from Definition 15 on lecture III.2 that a cardinal is **Jónsson** iff every algebra on admits a proper subalgebra of size Here, an **algebra** is a structure of the form where for each there is an such that An algebra without such proper substructures is called a **Jónsson algebra**.

The following is immediate:

Lemma 5is not Jónsson.

*Proof:* Consider

From the Shelah- results from last lecture, we have:

Theorem 6 (Shelah, )If is the successor of a regular cardinal (or simply if is regular and admits a nonreflecting stationary subset), then is not Jónsson, since in fact

Similarly:

Theorem 7 ()If all regular cardinals below the singular cardinal satisfy then so does In particular, is not Jónsson. It suffices that there is a scale forsupportedby such cardinals

A few additional cases are established by the following theorems. First, an obvious observation:

*Proof:* Let be given by

We begin with several nice characterizations of Jónsson cardinals. Item 2 is due to -Hajnal, item 3 to Keisler-Rowbottom, items 4, 5 are due to Tryba, and item 6 is due to Kleinberg.

Theorem 9The following are equivalent:

is Jónsson.Any structure for a countable first order language with domain of size admits a proper elementary substructure with domain of sizeFor every there exists an elementary embedding (for some transitive ) such that andFor some ordinal there exists an elementary embedding such that andThere is some cardinal such thatFor every sufficiently large regular and any there is an elementary substructure of where is a well-ordering of such that and yet

Of course, items 5 and 6 can also be stated in terms of the structures, and item 7 also has a version in which we just require that some has the stated property.

Recall that is the collection of sets whose transitive closure has size strictly less than This is a model of The well-ordering is mostly for convenience, and provides us with definable Skolem functions.

*Proof:* *1 implies 2.* Given consider the algebra Since is Jónsson, there is a proper subalgebra of of size In particular, if is the universe of this subalgebra,

*2 implies 3.* By means of appropriate coding, one can represent any function by finitely many functions For example, if we can set and Note that here means, according to our conventions, where and lists the elements of in increasing order.

Given a structure with universe in a countable language, we can add Skolem functions to it, and still obtain a countable language. Apply the coding mentioned above, and let list all the resulting colorings in such a way that is -ary for some (Add “dummy” functions if necessary to achieve this.)

Now define by If is of size and closed under then is closed under the and therefore under the Skolem functions for the original structure. But then is the universe of a proper elementary substructure of

*3 implies 4.* Let By the Löwenheim-Skolem theorem there is an

with and here, denotes the elementary substructure relation.

Consider the structure where is treated as a constant and is a bijection. Since is Jónsson, admits a proper elementary substructure

of size

Let be the Mostowski collapsing map, which is defined since is transitive. Since we necessarily have Otherwise, since is a bijection, we could define and it would follow that Also, since we have that (again, because is a bijection), and therefore but is not the identity.

We now obtain the desired embedding by considering

*4 implies 5.* This is obvious.

*5 implies 6.* Assume is elementary, where and It is easy to check that is regular and uncountable in

We claim that holds in (It is in order to have the relevant functions around that we require that ) Recall that means that whenever there is some such that

Assume otherwise, and let be a counterexample in By elementarity, is an uncountable regular cardinal and witnesses in and therefore in In particular, if then must have size However, has size contradiction.

By elementarity, it follows that holds in

*6 implies 1.* This is clear.

*7 implies 2.* If is not Jónsson and is as in 7 (for any ), there is in a witness to the failure of 2. But is closed under contradiction.

*2 implies 7.* Given let where is treated as a constant, and and as relations, so any substructure of contains and as elements.

Let be a complete set of Skolem functions for with of arity for all As in the proof that 2 implies 3, we may assume that rather that

Set by

Here, for and is the subset of consisting of its first many elements in increasing order.

By assumption, there is such that Let

so and is the universe of an elementary substructure of Note that so It follows that is as required.

By a standard Löwenheim-Skolem type of argument, we may also assume in 7 that for any with

Theorem 10 (-Hajnal)If is not Jónsson, neither is

*Proof:* For each let witness Since is not Jónsson, these functions exist. Now set by

for if this maximum is at least Set otherwise.

It is straightforward to check that witnesses that is not Jónsson. For suppose Let and let be such that and There is some such that Then and we are done.

Lemma 11 (Kleinberg)If is Jónsson, then the least cardinal such that is regular and uncountable, and we actually have

*Proof:* That there is some such follows from Theorem 9. Letting be least such that Lemma 8 shows that and it suffices to check that

For this, assume otherwise, and let be a counterexample. Let be increasing and cofinal. For each let witness

Now set by

whenever for some and whenever is not of this form.

It is straightforward to check that witnesses This is because if then for each we can find some such that There is some such that Let and let Then so there is some such that Now let have the form where is chosen so that Then Since was arbitrary, this shows that contradiction.

In light of characterization 6 in Theorem 9, it is natural to wonder whether one can have This is impossible for regular:

*Proof:* Let be strictly increasing and cofinal with Define

by iff

Assume that and Then Thus and

This shows that witnesses

Recall that if is a cardinal, and then is **-Rowbottom** iff for any and it is Rowbottom iff it is -Rowbottom.

Obviously, if is -Rowbottom for some then it is Jónsson. We want to show that, in fact, the least Jónsson cardinal must be -Rowbottom for some First, an easy extension of the techniques showing Theorem 9 shows:

Lemma 13 (Rowbottom)Suppose that and Then the following are equivalent:

Every structure of size in a countable language with a distinguished unary relation of size admits an elementary substructure of size where the relation has size strictly smaller than

*Proof:* That 2 implies 1 is obvious: Given let

where is treated as a relation, and we are using that straightforward coding allows us to consider colorings (with domain for some ) as elements of the language (rather than functions whose domain is for some ).

There is an elementary substructure of of size such that Since is closed under then witnesses for

That 1 implies 2 is essentially the proof that 2 implies 7 in Theorem 9: Given

a structure in a countable language, where is treated as a relation, let be a complete set of Skolem functions for with of arity for all and define by setting

Here, for and is the subset of consisting of its first many elements in increasing order.

By assumption, there is such that Let

so and is the universe of an elementary substructure of Note that so It follows that is as required.

*Proof:* Let be any function extending a witness to Given a structure

in a countable language, extend to

using Skolem functions for define as before. Let be such that Let be the elementary substructure of generated by so is obtained by closing under the Skolem functions for As before, This implies that, in fact, since is closed under The result now follows from Lemma 13.

Corollary 15 (Kleinberg)Let be the least Jónsson cardinal, and let be least such that Then is -Rowbottom.

*Proof:* We need to show that for all Towards a contradiction, suppose is least such that this fails, as witnessed by Note that since and By Lemma 14, there is some and an such that Then, up to reindexing, By minimality of there must be some such that This is a contradiction.

Tryba has shown the following result, closely related to Corollary 15:

Theorem 16 (Tryba)Assume that is Jónsson and strongly inaccessible. Then is -Rowbottom for some .

Apparently it is still open whether every weakly inaccessible Jónsson cardinal satisfies the conclusion of Theorem 16.

From Lemmas 12 and 14, it follows that if then must be a Jónsson cardinal.

Theorem 17 (Prikry)Suppose that is a singular limit of measurable cardinals. Then is -Rowbottom.

As shown in Theorems 25 and 26 below, any measurable cardinal is Jónsson. It is then natural to ask whether can ever hold in the context of Theorem 17. This was recently shown by Woodin.

Theorem 18 (Woodin)If is measurable, is a limit of measurable cardinals, and then

We postpone the proof until next Section.

The following is a counterpart to ‘s Theorem 7. It illustrates a very useful pcf trick, the use of “characteristic functions.”

Theorem 19 (Shelah)Suppose that where is a singular cardinal. Let be a scale for such that each is not Jónsson. Then neither is

*Proof:* We use characterization 7 of Theorem 9.

Recall that if is a scale for then is a strictly increasing sequence of regular cardinals cofinal in and where each the sequence is -increasing and cofinal, i.e., a witness to

Let be a sufficiently large regular cardinal and let be such that and We need to show that

Suppose first that for all large enough we have that Consider the **characteristic function of on **, defined by

By definition, Also, by our assumption, for all large enough. Since is -cofinal in and there must be some such that

for all large enough Note that and so for all This, of course, contradicts the displayed inequality.

It follows that for arbitrarily large values of For any such since and is not Jónsson, then This is because contains some witness to by elementarity, and so

It follows that If and then as must contain some surjection from onto and should therefore also contain its range. Since is unbounded in we can now conclude that

Theorem 20 (Rowbottom)The first Jónsson cardinal is either weakly inaccessible, or singular of cofinality

*Proof:* From the results above, the first Jónsson cardinal is a limit. Suppose

Fix a club consisting of cardinals, with For each fix also a witness to Finally, let extend any witness to Let be such that

for all

Now set by

Let It is enough to check that is a Jónsson algebra.

Towards a contradiction, let be the domain of a proper subalgebra of Since is in the language of we have By the presence of it follows that

Let Pick such that and, recursively, choose so for each

- and

Using that note that Since is club, is defined, and by choice of the we have that Recall that because It easily follows from the definition of and the choice of that Since then Since was arbitrary,

This contradicts that is Jónsson.

Theorem 20 can be strengthened as follows:

Theorem 21 (Tryba)If is a singular cardinal of uncountable cofinality, and is Jónsson, then

contains a club.

There is a companion result to the theorem above for successors of singulars of uncountable cofinality:

Theorem 22 (Shelah)Suppose that is singular of uncountable cofinality, and that is Jónsson. Then

is club in

The following is the main open problem in this area:

**Open question.** *Can be Jónsson?*

Kleinberg showed that the notion of Jónsson and Rowbottom are equiconsistent. Clearly, from characterization 6 in Theorem 9, say, if there is a Jónsson cardinal then exists. Building on results of Donder and Koepke, Peter Koepke showed that if is Jónsson, then for every there is an inner model with many measurable cardinals. I expect the assumption that is Jónsson to carry a much higher consistency strength.

**3. Large cardinals **

Recall from Definition 13 in lecture III.2 that a cardinal is **Ramsey** iff

Remember that this means that for any there is an that is homogeneous for for all Obviously, Ramsey cardinals are Jónsson. They are large cardinals as well:

Definition 23Aweakly compactcardinal is a cardinal such that

Clearly, Ramsey cardinals are weakly compact.

Lemma 24 ()If is weakly compact, then it is strongly inaccessible.

*Proof:* It follows from Sierpi\’nski’s Theorem that must be strong limit: Otherwise, let be such that Then so

We argue that must also be regular: Otherwise, for some sets of size less than Let be given by iff are in the same Then does not have a homogeneous set of size : It clearly does not admit a 0-homogeneous set, but if then (since there must be two elements of in the same so cannot be 1-homogeneous.

(For a different proof of regularity, see Lemma 7 in lecture III.2.)

One can show that if is weakly compact, then

is stationary in i.e., is Mahlo.

By induction, say that is 1-Mahlo iff it is Mahlo; say that it is -Mahlo iff it is strongly inaccessible and is -Mahlo is stationary in ; and say that it is -Mahlo, for limit, iff it is -Mahlo for all

Then one can in fact show that if is weakly compact, it is -Mahlo and limit of cardinals that are -Mahlo.

Many equivalent characterizations of weakly compact cardinals are known. For example, is weakly compact iff it is inaccessible and has the **tree property**. This means that whenever is a tree of height all of whose levels have size then there is a cofinal branch through i.e., the version of König’s lemma for holds.

Keisler proved that is weakly compact iff it has the **extension property**: for any there is a transitive and an such that

Theorem 25A Ramsey cardinal is Rowbottom. In fact, is Ramsey iff for all

*Proof:* Let be Ramsey and consider and a coloring Define by setting unless for some and in which case

If is homogeneous for then it must clearly be -homogeneous, and therefore it is also homogeneous for

*Proof:* We present an argument due to Rowbottom. Let be a normal -complete nonprincipal ultrafilter over Let and consider a coloring

By induction on we show that for any there is an that is homogeneous for

For this is clear from the -completeness of Assume the result for and consider a map For each let be the map

By -completeness, for each there is a color and a set such that By induction, there is a color and a set such that for all

For let so by -completeness. Let so It is enough to check that is -homogeneous for

For this, let and let and so

since and

It follows that for all there is a set homogeneous for Let Then and is homogeneous for

Let me now sketch the proof of Woodin’s Theorem 18.

*Proof:* Recall that Theorem 18 states that whenever is measurable, and is limit of measurable cardinals.

The argument requires some knowledge of directed systems. See, for example, the corresponding discussion in Chapter 0 of Kanamori’s book.

Let be strictly increasing and cofinal with and measurable but not a limit of measurables, for all nonzero

Suppose that there is an elementary embedding with and such that It follows that By the proof that 5 implies 6 in Theorem 9, it follows that

It therefore suffices to build such an embedding This is accomplished by building a directed system of embeddings

together with embeddings

for all such that the embeddings commute with the embeddings and, letting be the corresponding direct limit embedding, we have and

The construction is recursive. To begin with, arguing as in the proof that 3 implies 4 in Theorem 9, but using that measurable cardinals are Rowbottom, we may find

with such that but Set

Let be the preimage of Suppose that and

has been defined.

- If is limit, let be the direct limit embedding.
- If let and

Let be the natural embedding, and set and so is a measurable cardinal. Let be a normal -complete ultrafilter over with for some that, from the point of view of is a normal measure over

Let enumerate the members of so and

Let be in this intersection. Using one can define an embedding

by setting One easily checks that is indeed well-defined and elementary.

Let and in general iterate the above construction to obtain a directed system

together with embeddings

that commute with the

Let be the direct limit of and define for and in the natural way. One easily checks that and

This completes the construction of the directed system.

I close with a result about a strong version of Jónsson cardinals in the presence of a very strong axiom contradicting choice.

Recall that Kunen showed that there are no embeddings All the known proofs of this result (see for example lectures II.10, II.12, and III.3) use the axiom of choice in an essential way.

Theorem 27 (Sargsyan)In assume there is an elementary embedding

Let be the first fixed point of past its critical point. Then for all cardinals and all there is a cardinal such that for every cardinal we have that

In particular, all cardinals are Jónsson.

*Proof:* Let be elementary, and set Assume first that so Towards a contradiction, suppose that is the least cardinal such that for unboundedly many Since is fixed by and is definable from and then

Let and suppose that as witnessed by the function Then

and for all

However, and has size at most This is a contradiction.

It follows that for all This contradicts the definition of and we are done in the case that

Consider now an arbitrary Again, suppose that and that for unboundedly many

Recall that the **critical sequence** is defined recursively by and for all so

Let be the least fixed point of past and recursively define as the least fixed point of past Define and set

One easily checks that is elementary, and whenever

Notice that the contradiction for the case works for our current inside by considering where is chosen so that This completes the proof.

** Bibliography **

Here are some references consulted while preparing this note:

- Arthur Apter, Grigor Sargsyan,
*Jónsson-like partition relations and*, The Journal of Symbolic Logic,**69 (4)**(Dec., 2004), 1267–1281. - 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). - Peter Koepke,
*Some applications of short core models*, Annals of Pure and Applied Logic,**37**(1988), 179–204. - Grigor Sargsyan, unpublished notes.
- Jan Tryba,
*On Jónsson cardinals with uncountable cofinality*, Israel Journal of Mathematics,**49 (4)**(1984), 315–324. - Jan Tryba,
*Rowbottom-type properties and a cardinal arithmetic*, Proceedings of the American Mathematical Society,**96 (4)**(Apr., 1986), 661–667.

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