580 -Cardinal arithmetic (4)

2. Silver’s theorem.

From the results of the previous lectures, we know that any power $\kappa^\lambda$ 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 $\gimel(\lambda)$ varies greatly depending on whether $\lambda$ is regular or not. If $\lambda$ is regular, then $\gimel(\lambda)=2^\lambda.$ As mentioned on lecture II.2, forcing provides us with a great deal of freedom to manipulate the exponential function $\kappa\mapsto 2^\kappa,$ at  least for $\kappa$ regular. In fact, the following holds:

Theorem 1. (Easton). If ${\sf GCH}$ holds, then for any definable function $F$ from the class of infinite cardinals to itself such that:

1. $F(\kappa)\le F(\lambda)$ whenever $\kappa\le\lambda,$ and
2. $\kappa<{\rm cf}(F(\kappa))$ for all $\kappa,$

there is a class forcing ${\mathbb P}$ that preserves cofinalities and such that in the extension by ${\mathbb P}$ it holds that $2^\kappa=F^V(\kappa)$ for all regular cardinals $\kappa;$ here, $F^V$ is the function $F$ as computed prior to the forcing extension. $\Box$

For example, it is consistent that $2^\kappa=\kappa^{++}$ for all regular cardinals $\kappa$ (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 $2^\kappa=\kappa^{+16}$ whenever $\kappa=\aleph_\tau$ is regular and $\tau$ is an even ordinal, and $2^\kappa=\kappa^{+17}$ whenever $\kappa=aleph_\tau$ for some odd ordinal $\tau.$ Or, if there is a proper class of weakly inaccessible cardinals (regular cardinals $\kappa$ such that $\kappa=\aleph_\kappa$) then we could have $2^\kappa=$ the third weakly inaccessible strictly larger than $\kappa,$ for all regular cardinals $\kappa,$ 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 $\gimel(\lambda)$ for singular $\lambda.$ 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 ${\sf GCH}$ 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 $F$ is as above and also satisfies

3. $F(\kappa)\ge 2^\kappa$ for all regular cardinals $\kappa.$

Does it follow that there is some cofinality-preserving forcing extension where $2^\kappa=F^V(\kappa)$ for all regular cardinals $\kappa$?

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 ${\sf GCH}$ 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, ${\sf MM},$ arguably one of the most important events in modern set theory. A local version of the principle already follows from ${\sf BPFA}$, the bounded proper forcing axiom:

Theorem 3. (Todorčević). ${\sf BPFA}$ implies that every forcing that adds a subset of $\omega_1$ must either add a real or collapse $\omega_2.$ $\Box$

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 $2^{\aleph_0}=\aleph_2$ and that every tree ${\mathcal T}$ of height and size $\omega_1$ is sealed, i.e., there is a function $f:{\mathcal T}\to\omega$ such that whenever $f(x)=f(y)=f(z)$ and $x\le_{\mathcal T}y,z,$ then in fact $y\le_{\mathcal T} z$ or $z\le_{\mathcal T} y.$ This notion is due to Baumgartner, who called these trees special. That these statements are consequences of ${\sf BPFA}$ 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 ${\sf BPFA}$ and suppose that a forcing adds a subset of $\omega_1$ but no reals. Given a name for this set, we can then build a tree of height $\omega_1,$ essentially the tree of attempts to decide the name. The assumption of ${\sf BPFA}$ gives that the tree has size at most $\omega_2$ and from it a cofinal map from $\omega_1$ to $\omega_2$ can be constructed.

Assume that ${\mathbb P}$ is a forcing answering Question 1 affirmatively, where the function $F$ satisfies $F(\aleph_0)=\aleph_2$ and $F(\aleph_1)=\aleph_3.$ We then have $\aleph_3$ many distinct names for subsets of $\omega_1$ 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 $\aleph_3$ 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 $\kappa$ is a measurable cardinal, then $\kappa$ is not the first counterexample to ${\sf GCH},$ so although Easton forcing preserves the fact that $\kappa$ is weakly inaccessible, it may very easily destroy its measurability. I do not know of any general Easton-like result for definable maps $F$ 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, ${\sf SCH},$ is the statement that for all cardinals $\lambda,$ we have that $\lambda^{{\rm cf}(\lambda)}=\lambda^++2^{{\rm cf}(\lambda)}.$

Note that the equality holds if $\lambda$ is regular, so ${\sf SCH}$ only has content for singular cardinals, which explains its name. For such $\lambda,$ notice that the left hand side is always at least as large as the right hand side, so what ${\sf SCH}$ says is that $\gimel(\lambda)$ is as small as possible for $\lambda$ singular.

${\sf SCH}$ is a trivial consequence of ${\sf GCH},$ 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 $\kappa=\aleph_{\omega_1+\omega_1}$ is a strong limit cardinal, $\gimel(\kappa)=\kappa^{+\omega+2}$ and $\gimel(\kappa^{+\omega})=\kappa^{+\omega+1}.$

In this model ${\sf SCH}$ fails since $2^{{\rm cf}(\kappa)}+\kappa^+=2^{\aleph_1}+\kappa^+=\kappa^+<\gimel(\kappa).$ There are simpler models of the failure of ${\sf SCH};$ I mentioned this one because it also witnesses a failure of Tarski’s conjecture, stated last lecture, the claim that $\prod_{\alpha<\beta}\lambda_\alpha=\lambda^{|\beta|}$ whenever $\beta$ is a limit ordinal and $(\lambda_\alpha:\alpha<\beta)$ is a strictly increasing sequence of cardinals cofinal in $\lambda.$

In effect, Tarski’s conjecture fails in Magidor model for $\beta=\omega_1+\omega,$ as witnessed by the sequence

$\displaystyle \lambda_\alpha=\left\{\begin{array}{cl}\aleph_\alpha &\mbox{ if }\alpha<\omega_1,\\\kappa^{+n}&\mbox{ if }\alpha=\omega_1+n.\end{array}\right.$

This is because $\prod_{\alpha<\beta}\lambda_\alpha =\prod_{\alpha<\omega_1}\aleph_\alpha\prod_{n<\omega}\kappa^{+n} =\gimel(\aleph_{\omega_1})\gimel(\kappa^{+\omega})<\kappa\gimel(\kappa)$ $=\gimel(\kappa)=(\kappa^{+\omega})^{|\omega_1+\omega|},$ 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 $\omega.$ Silver’s remarkable result is that this must always be the case:

Theorem 5. (Silver). If ${\sf SCH}$ holds for all singular cardinals of countable cofinality, then it holds everywhere. $\Box$

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 $F$ 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.

After proving the particular case of Silver’s theorem, I will present in detail a generalization in Section 3, and will state (without proof) several very general results that indicate the wide scope of the techniques involved. These techniques emphasize the role played by the combinatorics of infinitary products. In a sense, it is natural that the study of the gimel function requires of us some understanding of products. For example, if $\tau$ is a regular cardinal and $(\kappa_\alpha:\alpha<\tau)$ is a strictly increasing sequence of cardinals cofinal in $\kappa$, then it follows from the results from last lecture that $\gimel(\kappa)=\prod_\alpha\kappa_\alpha.$

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 $\alpha$ be an ordinal. A set $C\subseteq\alpha$ is club in $\alpha$ if and only if it is closed in the order topology of $\alpha,$ and it is cofinal in $\alpha.$ (Club stands for “closed and unbounded.”)

The definition above is unnecessarily general. The empty set is club in $\alpha=0.$ A closed subset $C$ of $\alpha+1$ is club if and only if $\alpha\in C.$ If $\alpha$ has cofinality $\omega$ then, for any strictly increasing and cofinal function $f:\omega\to\alpha,$ its range is club in $\alpha,$ and if $C\subseteq\alpha$ is closed in $\alpha,$ then $C$ is club if and only if $C$ contains the range of such an $f.$

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

We want club sets to capture enough of the structure of $\alpha$ that they can be used as surrogates for $\alpha.$ A good working intuition is that club sets provide a measure of “size” of subsets of $\alpha.$ 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 $\alpha$ a limit ordinal of uncountable cofinality (as the following theorem verifies), and from now on when we mention a club subset of $\alpha$ we (perhaps implicitly) assume that $\alpha$ 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 $C\subseteq\alpha$ is closed in $\alpha$ iff for all $\beta<\alpha,$ if $0<\beta$ and $C\cap\beta$ is unbounded in $\beta,$ then $\beta\in C.$

Examples of club subsets of $\omega_1$ include the set of all countable limit ordinals, the set of all countable indecomposable ordinals of the form $\omega^\delta$ for $\delta$ 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 $C$ is club in $\alpha,$ then $C'\subsetneq C,$ the set of limit points of $C$ (in the topological sense, as a subset of $\alpha$) 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 $\omega<\kappa={\rm cf}(\alpha).$ Then for any $\lambda<\kappa,$ the intersection of any $\lambda$ many club subsets of $\alpha$ is club.

Proof. Let $(C_\gamma:\gamma<\lambda)$ be a sequence of club subsets of $\alpha,$ where $\lambda<\kappa.$ Let $C=\bigcap_\gamma C_\gamma,$ so $C$ is closed, and all we need to verify is that it is unbounded.

For this, let $\beta<\alpha$ be given. By induction on $\gamma,$ we define a strictly increasing sequence $(\beta^0_\gamma:\gamma<\lambda)$ so that $\beta^0_i\in C_i$ for all  $i.$ Assume that $(\beta^0_\gamma:\gamma<\delta)$ has been defined, and $\delta<\lambda.$ Since $\delta<\lambda<\kappa,$ the supremum $\epsilon$ of this sequence is strictly below $\alpha.$ Let $\beta^0_\delta\in C_\delta$ be least such that $\epsilon,\beta<\beta^0_\delta.$ Such an ordinal $\beta^0_\delta$ exists since $C_\delta$ is unbounded in $\alpha.$

Let $\hat\beta=\sup_\gamma\beta^0_\gamma.$ Again, $\hat\beta<\alpha.$ Repeat the above construction with $\hat\beta$ in place of $\beta$ to obtain a sequence $(\beta^1_\gamma:\gamma<\lambda).$ Iterate this procedure $\omega$ times. The sequences we have produced then satisfy that $\theta_\gamma=\sup_n\beta^n_\gamma$ is actually independent of $\gamma.$ Call this ordinal $\theta.$ Since, by construction, $\beta^n_\gamma\in C_\gamma$ for all $n,$ it follows that $\theta\in C_\gamma,$ because $C_\gamma$ is closed in $\alpha$ (and $\theta<\alpha$ since $\kappa>\omega$).

We have shown that $\theta\in\bigcap_\gamma C_\gamma=C$ and, by construction, $\beta<\theta.$ This shows that $C$ is unbounded. ${\sf QED}$

When $\alpha=\kappa={\rm cf}(\kappa)$ is a regular cardinal, we can say even more.

Definition 8. Let $\kappa$ be a regular cardinal and let $(C_\alpha:\alpha<\kappa)$ be a sequence of club subsets of $\kappa.$ The diagonal intersection of the $C_\alpha,$ in symbols, $\bigtriangleup_\alpha C_\alpha,$ is the set of all $\beta<\kappa$ such that $\beta\in C_\alpha$ for all $\alpha<\beta.$

For example, let $C_\alpha=\kappa\setminus\alpha.$ Then $\bigtriangleup_\alpha C_\alpha=\kappa$ while if $C_\alpha=\kappa\setminus(\alpha+2),$ then $\bigtriangleup_\alpha C_\alpha$ is the set of limit ordinals below $\kappa.$ What we are doing is “fattening up” the club sets $C_\alpha$ a tiny bit before taking their intersection: Given $(C_\alpha:\alpha<\kappa),$ let $D_\alpha=C_\alpha\cup(\alpha+1)$ for all $\alpha+1.$ Then $\bigtriangleup_\alpha C_\alpha=\bigcap_\alpha D_\alpha.$

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

This is as strong a result as we can hope for. Considering again the example of $C_\alpha=\kappa\setminus\alpha,$ it is clear that we cannot strengthen the result to conclude that the intersection of $\kappa$ many club subsets of $\kappa$ is club in $\kappa.$

Proof. Let $(C_\alpha:\alpha<\kappa)$ be a sequence of club subsets of $\alpha,$ and let $C$ be their diagonal intersection. The characterization mentioned right before the statement of the lemma shows that $C$ is closed. Now we show that it is unbounded. For this, let $\alpha=\alpha_0<\kappa$ be given. Set $C^0=\bigcap_{\beta<\alpha_0}C_\beta,$ so $C^0$ is club. Let $\alpha_1\in C^0$ be strictly larger than $\alpha_0.$ Set $C^1=\bigcap_{\beta<\alpha_1}C_\beta,$ so $C^1$ is also club. Let $\alpha_2\in C^1$ be strictly larger than $\alpha_1.$

Continuing in this fashion we build a strictly increasing $\omega$-sequence $\alpha_0<\alpha_1<\dots$ of elements of $\kappa$ and a sequence of club sets $C^n=\bigcap_{\beta<\alpha_n}C_\beta$ such that $\alpha_{n+1}\in C^n$ for all $n.$ Let $C^\omega=\bigcap_n C_n,$ so $C^\omega$ is club in $\kappa,$ and let $\gamma=\sup_n\alpha_n,$ so $\gamma<\kappa.$ Note that, for any $n,$ the ordinals $\alpha_{n+1},\alpha_{n+2},\dots$ all belong to $C^n,$ so $\gamma\in C_n.$ Therefore, $\gamma\in C^\omega.$

If $\beta<\gamma,$ then $\beta<\alpha_n$ for some $n$, so $\alpha_{n+1},\alpha_{n+2},\dots$ all belong to $C_\beta$ and so does $\gamma.$ It follows that $\gamma\in C=\bigtriangleup_\beta C_\beta.$ Since $\alpha<\gamma,$ we have that $C$ is unbounded, as we wanted to show. ${\sf QED}$

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