At the end of last lecture we defined club sets and showed that the diagonal intersection of club subsets of a regular cardinal is club.

**Definition 10.** Let be a limit ordinal of uncountable cofinality. The set is *stationary **in* iff for all club sets

For example, let be a regular cardinal strictly smaller than Then is a stationary set, since it contains the -th member of the increasing enumeration of any club in This shows that whenever there are disjoint stationary subsets of Below, we show a stronger result. The notion of stationarity is central to most of set theoretic combinatorics.

**Fact 11.** *Let be stationary in *

*is unbounded in**Let be club in Then is stationary.*

**Proof.** 1. must meet for all and is therefore unbounded.

2. Given any club sets and and it follows that is stationary.

Continuing with the intuition that a club subset of is large, we should think of stationary sets as being those that are not small. Thinking in terms of measure theory may be helpful: A club set is like a set of full measure, and a stationary set is like a set of positive measure. Clearly, we cannot have disjoint full measure sets, this corresponds to the fact that the intersection of two clubs is club. However, it is possible to have disjoint stationary sets. It is because of this fact that last lecture I made the comment that stationary sets are a manifestation of the axiom of choice: It is consistent with (for example, it is a consequence of determinacy) that every stationary subset of contains a club. However, under choice any stationary subset of an ordinal of cofinality can be split into many disjoint stationary subsets. This is a result of Solovay. For or, in general, for any successor cardinal, this was known prior to Solovay’s theorem; it is a consequence of a nice construction due to Ulam.

**Definition 12.** Let be a cardinal. An *Ulam matrix on* is defined as follows: For each fix an injection . For and define

For define

The following observations are immediate from the definition:

- If then for any
- If then for any since the are injective.
- for each

It follows from 3. and Theorem 7 that for each in fact, contains a club. It follows from the pigeonhole principle that for some the set has size

This shows that there are many disjoint stationary subsets of In fact, more is true: Given stationary, by replacing each with the same argument shows:

**Theorem 13. (Ulam). ***For each any stationary subset of can be split into many disjoint stationary sets. *

A little argument with the pigeonhole principle shows that, for every has size at most and therefore for all but boundedly many is unbounded in

**Question 14. **Are there many values of such that ?

Of course, the answer to Question 14 is yes if is regular.

We are now ready to provide a very useful characterization of stationary sets due to Fodor.

**Definition 13.** Let be a set of ordinals. A function is regressive iff for all

**Theorem 14. (Fodor). ***Suppose that is stationary in the regular cardinal and is a regressive function. Then there is some such that is stationary.*

**Proof. Otherwise, for each there is a club disjoint from Let so is club by Lemma 9. Let For all so Therefore, and is not regressive. Contradiction. **

**Corollary 15. Let be an uncountable regular cardinal and let The following are equivalent:**

*is stationary.**For every regressive function on there is some such that is unbounded.**For every regressive function on there is some such that is stationary.**For every function on there is some stationary subset of where is constant, or else there is a stationary subset of where is strictly increasing.*

**Proof. **Clearly 3. implies 2., and 1. implies 3. by Theorem 14. Assume that is not stationary, and let be a club set disjoint from For every larger than the set is nonempty and bounded below so it has a largest element. The map that assigns to this largest element is regressive on By mapping every other (if any) to the map extends to a regressive function on However, the preimage of every point in its range is bounded. This proves that 2. implies 1.

Obviously, 4. implies 1. Suppose that is stationary in that and that is not stationary. Let so is stationary and for all By induction, define a sequence such that

- for all
- for all and in fact
- whenever and
- whenever is a limit ordinal.

(Whenever condition 5 holds in a sequence, we say that the sequence is *continuous*. If condition 3 holds, this indeed corresponds to the enumeration map being continuous in the order topology.)

That we can satisfy requirements 1–5 follows from the regularity of By construction, is a club subset of and therefore is stationary. By construction, whenever and This shows that 4. holds.

**Remark 16. **If has uncountable cofinality but is not a regular cardinal, it is not necessarily true that every regressive function on every stationary subset of is constant on a stationary subset. However, a weak version of Fodor’s lemma still holds in this case, namely, for any stationary in and any regressive there is some stationary subset of in which is *bounded.* The proof is an easy extension of the argument for Theorem 14.

Note that the collection of subsets of that contain a club forms a -complete filter (the *club filter*), and the collection of *nonstationary* sets forms a -complete ideal (the *nonstationary ideal*).

Before stating Silver’s result, we need one additional result. Recall that was introduced last lecture in Definition 4, it is the statement that for all infinite cardinals We break the result into two parts, Lemmas 17 and 18, corresponding respectively to Theorems II.1.9 (describing the behavior of the exponential ) and II.1.10 (describing the computation of powers ) from lecture II.2.

**Lemma 17.** *Assume Let be singular and let Then *

**Proof.** Assume let be singular and let be as in the statement of the lemma. By the Bukovský-Hechler theorem, if is eventually constant below then and otherwise, By we have that since we are assuming that

**Lemma 18.** *Assume Let be infinite. Then *

**Proof. **Now we use Theorem II.1.10. Assume We argue by induction on that the formula holds for all If then as before.

Let If and then where the induction hypothesis is used in the last inequality.

Suppose now that and If is not eventually constant below then by Otherwise, is eventually constant below If is sufficiently large, then we have that and where the last inequality is by the induction hypothesis, and we are done.

**Corollary 19.** * holds iff for all infinite cardinals *

What the results say is that makes all powers as small as possible modulo the size of the exponential function Notice that the arguments we gave are *local*, meaning that the computations hold at some assuming only that holds up to

**Definition 20. (Shelah). **Let be a cardinal. A cardinal is *-inaccessible* iff for all cardinals

The following is a fairly general statement of Silver’s result (still some additional generality is possible). I’ll present some corollaries and explain how they follow from the theorem. Then I will prove in detail a (very) particular case which, however, contains all the required ingredients for the proof of the general result. The argument I present is due to Baumgartner and Prikry and is combinatorial in nature. Silver’s original proof used the technique of forcing and depended on a generic ultrapower argument. We will see a similar (forceless) argument in Section 4, in the context of large cardinals. For the original approach, see Jack Silver, *On the singular cardinals problem*, in **Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974)**, Vol. **1**, . Canad. Math. Congress, Montreal, Que., 1975, 265–268.

**Theorem 21. (Silver).**

*Let be a cardinal. Assume that and is -inaccessible. Let be a strictly increasing and continuous sequence of cardinals cofinal in Suppose that there is some such that is stationary in Then**In the situation of item 1., if in addition for all then**If is a strong limit singular cardinal of uncountable cofinality, and is stationary in then*

**Corollary 22. (Silver).**

*Let be a singular strong limit cardinal of uncountable cofinality Let be a strictly increasing and continuous sequence of cardinals cofinal in Suppose that and is stationary. Then*

*Let be a singular cardinal of uncountable cofinality and let If is stationary, then**Suppose that and is stationary. Then**Suppose that is strong limit, and is stationary. Then**Suppose that and is stationary. Then**The first counterexample to is not a singular cardinal of uncountable cofinality**The first counterexample to is not a singular cardinal of uncountable cofinality.*

**Proof.** 1. This follows immediately from the theorem, using that since is strong limit.

2. Note that is strong limit: If then since It follows that for unboundedly many, and therefore for all,

Let be a strictly increasing continuous sequence of cardinals cofinal in Then is stationary in

But for all Now the result follows from 1.

3. For all by Tarski’s formula, Homework problem 8 from lecture II.3. It follows that for stationarily many But, for any countable limit ordinal we have that The result now follows from the theorem.

4. By 3., since is strong limit.

5. and 6. By 2.

7. Suppose that holds below that is singular, and that is uncountable. Suppose first that Then in fact since We then have that and also holds at

Suppose now that We claim that in fact is -inaccessible. Assume that this is the case, and fix a strictly increasing and continuous sequence of cardinals converging to For any limit ordinal we have that By below as long as is sufficiently large to ensure that (We are using Corollary 19 here, more precisely, the fact that Corollary 19 has a local proof as explained above.)

It follows that Theorem 21.1 applies, with and therefore and also holds at as wanted.

All that remains is to argue that is indeed -inaccessible. For this, suppose that Using below by (the local nature of the proof of) Corollary 19 we have that and we are done.

Now I state and prove the best known particular case of Silver’s result.

**Corollary 23. (Silver). *** is not the first counterexample to *

**Proof.** Suppose that for all In what follows, by we mean the set product (the collection of functions) rather than its cardinality, and similarly for other products. Fix injections for all For let be the map Set

The following properties of are immediate:

- Whenever there is in fact an such that for all countable
- since the assignment is 1-1.

Whenever a collection of functions satisfies 1., we will say that is *almost disjoint*.

**Lemma 24.** *Suppose that is almost disjoint, and for all the set is stationary in Then *

**Proof.** Given define so that for all Then is regressive in a stationary set. By Fodor’s lemma (Theorem 14), there is some such that is stationary.

For and let Then either is empty, or else the map that assigns to its restriction is injective, since is almost disjoint and is stationary in and therefore unbounded.

By definition of if then But then it follows from that

The result follows, since and there are only many possible values of and many possible values of

**Corollary 25. ***Let and let be almost disjoint. Suppose that for all is stationary. Then *

**Proof.** This is immediate from the lemma since injects into

To complete the proof of the theorem, we now argue that can be split into many sets to which the corollary applies.

**Definition 26.** Let be the club filter on Given functions we say that iff contains a club.

Suppose that Then, by Corollary 22, at most many functions in are below on a stationary set, i.e., for all but at most many functions

By induction, we can therefore define a -increasing sequence of members of . But notice that for *any* it is the case that for some Otherwise, contradicts the previous paragraph. It follows that where and we are done by Corollary 22.

The arguments above (see the proof of Corollary 22.7, for example) suggest that guarantees that Tarski’s conjecture (see lecture II.3) holds. This easily follows from Lemma 18. We can in fact say a bit more:

**Homework problem 9. (Jech-Shelah). ***Assume that Tarski’s conjecture fails, so for some limit ordinal there is some strictly increasing sequence of cardinals with limit such that *

*Show that and there is some cardinal such that**Suppose that is least such that there is a counterexample as above. Then ( and)**If Tarski’s conjecture fails, then there is a cardinal of uncountable cofinality such that for all and*

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