2. Silver’s theorem.

From the results of the previous lectures, we know that any power can be computed from the cofinality and gimel functions (see the Remark at the end of lecture II.2). What we can say about the numbers varies greatly depending on whether is regular or not. If is regular, then As mentioned on lecture II.2, forcing provides us with a great deal of freedom to manipulate the exponential function at least for regular. In fact, the following holds:

**Theorem 1. (Easton).** *If holds, then for any definable function from the class of infinite cardinals to itself such that:*

*whenever and**for all*

*there is a class forcing that preserves cofinalities and such that in the extension by it holds that for all regular cardinals here, is the function as computed prior to the forcing extension. *

For example, it is consistent that for all regular cardinals (as mentioned last lecture, the same result is consistent for all cardinals, as shown by Foreman and Woodin, although their argument is significantly more elaborate that Easton’s). There is *almost* no limit to the combinations that the theorem allows: We could have whenever is regular and is an even ordinal, and whenever for some odd ordinal Or, if there is a proper class of weakly inaccessible cardinals (regular cardinals such that ) then we could have the third weakly inaccessible strictly larger than for all regular cardinals etc.

Morally, Easton’s theorem says that there is nothing else to say about the gimel function on regular cardinals, and all that is left to be explored is the behavior of for singular In this section we begin this exploration. However, it is perhaps sobering to point out that there are several weaknesses in Easton’s result.

**First****,** there is the assumption of in the ground model. When I began typing these notes, I was tempted to say that this is not a horribly serious problem, but then I actually thought about it. What I mean is this:

**Question 2.** Assume that is as above and also satisfies

3. for all regular cardinals

Does it follow that there is some cofinality-preserving forcing extension where for all regular cardinals ?

Years ago, Foreman, Magidor and Shelah asked whether the following is consistent: “Every nontrivial (set) forcing either adds a real or collapses a cardinal.”

This “maximality principle,” as they called it, implies that fails everywhere and there are no inaccessible cardinals (regular strong limit cardinals). Assuming that the principle is consistent gives us that nothing like Easton’s original approach would work to produce the cofinality preserving extensions we require. This is because Easton’s forcing is a product of class many nontrivial set forcings as factors. Of course, mutual genericity gives us that only set many of these factors can add reals, so at least one cardinal (probably a proper class) will be collapsed. This does not rule out that a completely different approach will give Easton’s result, but it makes the problem potentially more difficult than I originally imagined.

The work of Foreman, Magidor and Shelah on this “maximality principle” led to their discovery of Martin’s Maximum, arguably one of the most important events in modern set theory. A local version of the principle already follows from , the bounded proper forcing axiom:

**Theorem 3. (Todorčević).** * implies that every forcing that adds a subset of must either add a real or collapse *

For the proof, see Stevo Todorčević, *Some combinatorial properties of trees*, Bull. London Math. Soc., **14 (3)**,(1982), 213–217. All that is needed is that and that every tree of height and size is sealed, i.e., there is a function such that whenever and then in fact or This notion is due to Baumgartner, who called these trees special. That these statements are consequences of follows from work of Baumgartner and of Justin Moore. Todorčević requires that the forcing notion is a set, but one can dispense with this assumption.

An examination of the proof of Todorčević’s theorem actually shows that it answers Question 1 negatively. Very briefly, the argument is this: Assume and suppose that a forcing adds a subset of but no reals. Given a name for this set, we can then build a tree of height essentially the tree of attempts to decide the name. The assumption of gives that the tree has size at most and from it a cofinal map from to can be constructed.

Assume that is a forcing answering Question 1 affirmatively, where the function satisfies and We then have many distinct names for subsets of to which the above procedure can be applied. It follows that the tree construction must fail, which only happens because some proper initial segment of the new subset is not in the ground model, so a new real is added. A mutual genericity argument ensures that there must be many new reals added this way.

**Second**, Easton’s forcing preserves cofinalities and therefore cardinals. However, it does not preserve in general any significant large cardinals. For example, as we will see in Section 4, if is a measurable cardinal, then is not the first counterexample to so although Easton forcing preserves the fact that is weakly inaccessible, it may very easily destroy its measurability. I do not know of any general Easton-like result for definable maps that takes into account a significant amount of the large cardinal structure of the universe.

**Third**, of course, Easton’s result only applies to the regular cardinals. In fact, in the models obtained by Easton’s method we have no freedom whatsoever to manipulate the gimel function for singular cardinals. This can be made precise as follows:

**Definition 4.** The **singular cardinals hypothesis**, is the statement that for all cardinals we have that

Note that the equality holds if is regular, so only has content for singular cardinals, which explains its name. For such notice that the left hand side is always at least as large as the right hand side, so what says is that is as small as possible for singular.

is a trivial consequence of and it is easy to see that it holds in the models obtained by Easton’s method. One may then wonder whether it is actually provable. Magidor showed that this is not the case; for example, beginning with significantly large cardinals, Magidor obtained a model in which is a strong limit cardinal, and

In this model fails since There are simpler models of the failure of I mentioned this one because it also witnesses a failure of Tarski’s conjecture, stated last lecture, the claim that whenever is a limit ordinal and is a strictly increasing sequence of cardinals cofinal in

In effect, Tarski’s conjecture fails in Magidor model for as witnessed by the sequence

This is because where the last equality follows from Tarski’s formula, Homework problem 8 (or from Theorem 10 in lecture II.2).

Magidor’s counterexample occurs at a singular cardinal of cofinality Silver’s remarkable result is that this must always be the case:

**Theorem 5. (Silver). ***If holds for all singular cardinals of countable cofinality, then it holds everywhere. *

Next time we will state Silver’s theorem in full generality and prove a particular case that illustrates the combinatorial ideas that guided the proof of the general result and its many generalizations. It follows from Silver’s theorem that in an Easton-like result that also considers singular cardinals the function must satisfy additional restrictions. I do not know of any general results of this kind, although many particular examples of possible behaviors and a significant list of restrictions have been identified.

A key feature of Silver’s result is the use of the notion of stationarity. In a sense, stationary sets are the first nontrivial manifestation of the axiom of choice at the uncountable level. Their definition requires an additional, very important notion.

**Definition 6.** Let be an ordinal. A set is *club* in if and only if it is closed in the order topology of and it is cofinal in (Club stands for “closed and unbounded.”)

The definition above is unnecessarily general. The empty set is club in A closed subset of is club if and only if If has cofinality then, for any strictly increasing and cofinal function its range is club in and if is closed in then is club if and only if contains the range of such an

This means that the concept is not interesting if is a successor. Automatically, any intersection of clubs is club. It also means that the concept is not interesting if has cofinality as we can have disjoint club sets.

We want club sets to capture enough of the structure of that they can be used as surrogates for A good working intuition is that club sets provide a measure of “size” of subsets of A set is large if it contains a club. In order for this to be a reasonable notion of largeness, we need that no two large sets be disjoint. On the other hand, we don’t want that an arbitrary intersection of large sets be large.

This means that the notion of club set is only interesting (or at least, only captures our working intuition) for a limit ordinal of uncountable cofinality (as the following theorem verifies), and from now on when we mention a club subset of we (perhaps implicitly) assume that satisfies these non-triviality requirements.

It may be worth pointing out that the topological closure condition of a club set can be easily restated as follows: A set is closed in iff for all if and is unbounded in then

Examples of club subsets of include the set of all countable limit ordinals, the set of all countable indecomposable ordinals of the form for limit (ordinal exponentiation), and many other, much more elaborate, sets. At first sight, it may look like these sets are actually rather thin, compared with, for example, the set of all countable successor ordinals. However, the closure condition in fact gives us that a club set has many “accumulation” points, and it is this reason what makes them so useful (and large, in some intuitive sense).

It is easy to verify that if is club in then the set of limit points of (in the topological sense, as a subset of ) is also a club set, and we can iterate this procedure. In fact, we can iterate it for a long while:

**Theorem 7.** *Assume that Then for any the intersection of any many club subsets of is club.*

**Proof. **Let be a sequence of club subsets of where Let so is closed, and all we need to verify is that it is unbounded.

For this, let be given. By induction on we define a strictly increasing sequence so that for all Assume that has been defined, and Since the supremum of this sequence is strictly below Let be least such that Such an ordinal exists since is unbounded in

Let Again, Repeat the above construction with in place of to obtain a sequence Iterate this procedure times. The sequences we have produced then satisfy that is actually *independent* of Call this ordinal Since, by construction, for all it follows that because is closed in (and since ).

We have shown that and, by construction, This shows that is unbounded.

When is a regular cardinal, we can say even more.

**Definition 8. **Let be a regular cardinal and let be a sequence of club subsets of The *diagonal intersection* of the in symbols, is the set of all such that for all

For example, let Then while if then is the set of limit ordinals below What we are doing is “fattening up” the club sets a tiny bit before taking their intersection: Given let for all Then

**Lemma 9.** *If is regular, the diagonal intersection of any many club subsets of is club in *

This is as strong a result as we can hope for. Considering again the example of it is clear that we cannot strengthen the result to conclude that the intersection of many club subsets of is club in

**Proof.**Let be a sequence of club subsets of and let be their diagonal intersection. The characterization mentioned right before the statement of the lemma shows that is closed. Now we show that it is unbounded. For this, let be given. Set so is club. Let be strictly larger than Set so is also club. Let be strictly larger than

Continuing in this fashion we build a strictly increasing -sequence of elements of and a sequence of club sets such that for all Let so is club in and let so Note that, for any the ordinals all belong to so Therefore,

If then for some , so all belong to and so does It follows that Since we have that is unbounded, as we wanted to show.

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