## 580 -Cardinal arithmetic (5)

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 $\alpha$ be a limit ordinal of uncountable cofinality. The set $S\subseteq\alpha$ is stationary in $\alpha$ iff $S\cap C\ne\emptyset$ for all club sets $C\subseteq\alpha.$

For example, let $\lambda$ be a regular cardinal strictly smaller than ${\rm cf}(\alpha).$ Then $S^\alpha_\lambda:=\{\beta<\alpha : {\rm cf}(\beta)=\lambda\}$ is a stationary set, since it contains the $\lambda$-th member of the increasing enumeration of any club in $\alpha.$ This shows that whenever ${\rm cf}(\alpha)>\omega_1,$ there are disjoint stationary subsets of $\alpha.$ Below, we show a stronger result. The notion of stationarity is central to most of set theoretic combinatorics.

Fact 11. Let $S$ be stationary in $\alpha.$

1. $S$ is unbounded in $\alpha.$
2. Let $C$ be club in $\alpha.$ Then $S\cap C$ is stationary.

Proof. 1. $S$ must meet $\kappa\setminus\alpha$ for all $\alpha$ and is therefore unbounded.

2. Given any club sets $C$ and $D,$ $(S\cap C)\cap D=S\cap(C\cap D)\ne\emptyset,$ and it follows that $S\cap C$ is stationary. ${\sf QED}$

Continuing with the intuition that a club subset of $\alpha$ 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 ${\sf ZF}$ (for example, it is a consequence of determinacy) that every stationary subset of $\omega_1$ contains a club. However, under choice any stationary subset of an ordinal $\alpha$ of cofinality $\kappa$ can be split into $\kappa$ many disjoint stationary subsets. This is a result of Solovay. For $\omega_1$ 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 $\kappa$ be a cardinal. An Ulam matrix on $\kappa^+$ is defined as follows: For each $\beta<\kappa^+$ fix an injection $f_\beta:\beta\to\kappa$. For $\iota<\kappa$ and $\alpha<\kappa^+$ define $A^\iota_\alpha = \{ \beta:\alpha<\beta\mbox{ and }f_\beta(\alpha)=\iota \}.$
For $\alpha<\kappa^+$ define $supp(\alpha)=\{ \iota: A^\iota_\alpha \mbox{ is stationary }\}.$

The following observations are immediate from the definition:

1. If $\iota\ne\eta$ then $A^\iota_\alpha\cap A^\eta_\alpha=\emptyset$ for any $\alpha<\kappa^+.$
2. If $\alpha\ne\gamma$ then $A^\iota_\alpha\cap A^\iota_\gamma=\emptyset$ for any $\iota<\kappa,$ since the $f_\beta$ are injective.
3. $\bigcup_\iota A^\iota_\alpha=\kappa^+\setminus(\alpha+1)$ for each $\alpha<\kappa^+.$

It follows from 3. and Theorem 7 that $supp(\alpha)\ne\emptyset$ for each $\alpha,$ in fact, $\bigcup_{\iota\in supp(\alpha)}A^\iota_\alpha$ contains a club.  It follows from the pigeonhole principle that for some $\iota,$ the set $\{\alpha:\iota\in supp(\alpha)\}$ has size $\kappa^+.$

This shows that there are $\kappa^+$ many disjoint stationary subsets of $\kappa^+.$ In fact, more is true: Given $S$ stationary, by replacing each $A^\iota_\alpha$ with $S\cap A^\iota_\alpha,$ the same argument shows:

Theorem 13. (Ulam). For each $\kappa,$ any stationary subset of $\kappa^+$ can be split into $\kappa^+$ many disjoint stationary sets. $\Box$

A little argument with the pigeonhole principle shows that, for every $\eta<\kappa,$ $\{ \alpha<\kappa^+ : supp(\alpha)\subseteq\eta \}$ has size at most $|\eta|$ and therefore for all but boundedly many $\alpha,$ $supp(\alpha)$ is unbounded in $\kappa.$

Question 14. Are there $\kappa^+$ many values of $\alpha$ such that $|supp(\alpha)|=\kappa$?

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

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

Definition 13. Let $S$ be a set of ordinals. A function $f : S\to{\sf ORD}$ is regressive iff $f(\alpha)<\alpha$ for all $0<\alpha\in S.$

Theorem 14. (Fodor). Suppose that $S$ is stationary in the regular cardinal $\kappa,$ and $f:S\to\kappa$ is a regressive function. Then there is some $\alpha$ such that $f^{-1}\{\alpha\}$ is stationary.

Proof. Otherwise, for each $\alpha<\kappa$ there is a club $C_\alpha$ disjoint from $f^{-1}\{\alpha\}.$ Let $C=\bigtriangleup_\alpha C_\alpha,$ so $C$ is club by Lemma 9.  Let $\alpha\in C\cap S.$ For all $\beta<\alpha,$ $\alpha\in C_\beta,$ so $f(\alpha)\ne\beta.$ Therefore, $f(\alpha)\ge\alpha,$ and $f$ is not regressive. Contradiction. ${\sf QED}$

Corollary 15.Let $\kappa$ be an uncountable regular cardinal and let $S\subseteq \kappa.$ The following are equivalent:

1. $S$ is stationary.
2. For every regressive function $f$ on $S,$ there is some $\alpha$ such that $f^{-1}\{\alpha\}$ is unbounded.
3. For every regressive function $f$ on $S,$ there is some $\alpha$ such that $f^{-1}\{\alpha\}$ is stationary.
4. For every function $f$ on $S,$ there is some stationary subset of $S$ where $f$ is constant, or else there is a stationary subset of $S$ where $f$ is strictly increasing.

Proof. Clearly 3. implies 2., and 1. implies 3. by Theorem 14. Assume that $S$ is not stationary, and let $C$ be a club set disjoint from $S.$ For every $\gamma\in S$ larger than ${\rm min}(C),$ the set $C\cap\gamma$ is nonempty and bounded below $\gamma,$ so it has a largest element. The map that assigns to $\gamma$ this largest element is regressive on $S\setminus({\rm min}(C)+1).$ By mapping every other $\gamma\in S$ (if any) to ${}0,$ the map extends to a regressive function on $S.$ However, the preimage of every point in its range is bounded. This proves that 2. implies 1.

Obviously, 4. implies 1. Suppose that $S$ is stationary in $\kappa,$ that $f:S\to\kappa$ and that $T=\{\alpha\in S : f(\alpha)<\alpha\}$ is not stationary. Let $S'=S\setminus T,$ so $S'$ is stationary and $f(\alpha)\ge\alpha$ for all $\alpha\in S'.$ By induction, define a sequence $(a_\alpha:\alpha<\kappa)$ such that

1. $a_0\in S,$
2. $a_{\alpha+1}\in S$ for all $\alpha<\kappa,$
3. $a_{\alpha+1}>a_\alpha$ for all $\alpha<\kappa$ and in fact
4. $a_{\alpha+1}>f(a_\alpha)$ whenever $a_\alpha\in S,$ and
5. $a_\alpha=\sup\{a_\beta:\beta<\alpha\}$ whenever $\alpha$ 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 $\kappa.$ By construction, $C=\{a_\alpha:\alpha<\kappa\}$ is a club subset of $\kappa,$ and therefore $C\cap S'$ is stationary. By construction, $f(a_\alpha)\ge a_\alpha>f(a_\beta)$ whenever $a_\alpha,a_\beta\in S'\cap C$ and $\beta<\alpha.$ This shows that 4. holds. ${\sf QED}$

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

Note that the collection of subsets of $\kappa$ that contain a club forms a ${\rm cf}(\kappa)$-complete filter (the club filter), and the collection of nonstationary sets forms a ${\rm cf}(\kappa)$-complete ideal (the nonstationary ideal).

Before stating Silver’s result, we need one additional result. Recall that ${\sf SCH}$ was introduced last lecture in Definition 4, it is the statement that $\gimel(\kappa)=\kappa^++2^{{\rm cf}(\kappa)}$ for all infinite cardinals $\kappa.$ We break the result into two parts, Lemmas 17 and 18, corresponding respectively to Theorems II.1.9 (describing the behavior of the exponential $\lambda\mapsto2^\lambda$) and II.1.10 (describing the computation of powers $\kappa^\lambda$) from lecture II.2.

Lemma 17.Assume ${\sf SCH}.$ Let $\kappa$ be singular and let $\tau=\sup_{\rho<\kappa}2^{\rho}.$ Then $\displaystyle 2^\kappa=\left\{\begin{array}{cl}\tau&\mbox{if }\rho\mapsto2^\rho\mbox{ is eventually constant below }\kappa,\\ \tau^+&\mbox{otherwise.}\end{array}\right.$

Proof. Assume ${\sf SCH},$ let $\kappa$ be singular and let $\tau$ be as in the statement of the lemma. By the Bukovský-Hechler theorem, if $\rho\mapsto 2^{\rho}$ is eventually constant below $\kappa,$ then $2^\kappa=\tau,$ and otherwise, $2^\kappa=\gimel(\tau).$ By ${\sf SCH},$ we have that $\gimel(\tau)=\tau^+,$ since we are assuming that $2^{{\rm cf}(\kappa)}<\tau.$ ${\sf QED}$

Lemma 18.Assume ${\sf SCH}.$ Let $\kappa,\lambda$ be infinite. Then $\displaystyle \kappa^\lambda=\left\{\begin{array}{cl}2^\lambda&\mbox{if }\kappa\le2^\lambda,\\ \kappa&\mbox{if }\lambda<{\rm cf}(\kappa)\mbox{ and }2^\lambda<\kappa,\\ \kappa^+&\mbox{ if }{\rm cf}(\kappa)\le\lambda\mbox{ and }2^\lambda<\kappa.\end{array}\right.$

Proof. Now we use Theorem II.1.10. Assume ${\sf SCH}.$ We argue by induction on $\kappa$ that the formula holds for all $\lambda.$ If $\kappa\le2^\lambda,$ then $\kappa^\lambda=2^\lambda$ as before.

Let $\tau=\sup_{\rho<\kappa}|\rho|^\lambda.$ If $2^\lambda<\kappa$ and $\lambda<{\rm cf}(\kappa),$ then $\kappa^\lambda=\kappa\tau=\kappa\sup_{2^\lambda\le\rho<\kappa}|\rho|^\lambda\le\kappa\sup_{2^\lambda\le\rho<\kappa}|\rho|^+=\kappa,$ where the induction hypothesis is used in the last inequality.

Suppose now that $2^\lambda<\kappa$ and ${\rm cf}(\kappa)\le\lambda.$ If $\rho\mapsto\rho^\lambda$ is not eventually constant below $\kappa,$ then $\kappa^\lambda=\gimel(\kappa)=\kappa^++2^{{\rm cf}(\kappa)}=\kappa^+,$ by ${\sf SCH}.$ Otherwise, $\rho\mapsto\rho^\lambda$ is eventually constant below $\kappa.$ If $\rho<\kappa$ is sufficiently large, then we have that $2^\lambda<\rho,$ $\tau=\rho^\lambda,$ and $\kappa^\lambda=\tau=\rho^\lambda\le\rho^+\le\lambda,$ where the last inequality is by the induction hypothesis, and we are done. ${\sf QED}$

Corollary 19.${\sf SCH}$ holds iff $\kappa^\lambda\le2^\lambda+\kappa^+$ for all infinite cardinals $\kappa,\lambda.$ $\Box$

What the results say is that ${\sf SCH}$ makes all powers as small as possible modulo the size of the exponential function $\lambda\mapsto2^\lambda.$ Notice that the arguments we gave are local, meaning that the computations hold at some $\kappa$ assuming only that ${\sf SCH}$ holds up to $\kappa.$

Definition 20. (Shelah). Let $\kappa$ be a cardinal. A cardinal $\lambda$ is $\kappa$-inaccessible iff $\mu^\kappa<\lambda$ for all cardinals $\mu<\lambda.$

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).

1.  Let $\lambda$ be a cardinal. Assume that ${\rm cf}(\lambda)=\kappa>\omega$ and $\lambda$ is $\kappa$-inaccessible. Let $(\lambda_\alpha:\alpha<\kappa)$ be a strictly increasing and continuous sequence of cardinals cofinal in $\lambda.$ Suppose that there is some $\mu<\kappa$ such that $\{\alpha<\kappa:\prod_{\beta<\alpha}\lambda_\beta\le\lambda_\alpha^{+\mu}\}$ is stationary in $\kappa.$ Then $\lambda^\kappa\le\lambda^{+\mu}.$
2. In the situation of item 1., if in addition $2^\tau\le\lambda^\kappa$ for all $\tau<\lambda,$ then $2^\lambda\le\lambda^{+\mu}.$
3. If $\aleph_\lambda$ is a strong limit singular cardinal of uncountable cofinality, and $\{\alpha<\lambda:\gimel(\aleph_\alpha)\le\aleph_{\alpha\cdot2}\}$ is stationary in $\lambda,$ then $2^{\aleph_\lambda}<\aleph_{\lambda\cdot2}.$ $\Box$

Corollary 22. (Silver).

1. Let $\lambda$ be a singular strong limit cardinal of uncountable cofinality $\kappa.$ Let $(\lambda_\alpha:\alpha<\kappa)$ be a strictly increasing and continuous sequence of cardinals cofinal in $\lambda.$ Suppose that $\mu<\kappa$ and $\{\alpha<\kappa:\prod_{\beta<\alpha}\lambda_\beta\le\lambda_\alpha^{+\mu}\}$ is stationary. Then $2^\lambda\le\lambda^{+\mu}.$
2. Let $\lambda$ be a singular cardinal of uncountable cofinality $\kappa,$ and let $\mu<\kappa.$ If $\{\delta<\lambda:2^\delta\le\delta^{+\mu}\}$ is stationary, then $2^\lambda\le\lambda^{+\mu}.$
3. Suppose that $\mu<\omega_1,$ $2^{\aleph_1}<\aleph_{\omega_1},$ and $\{\alpha<\omega_1:\aleph_\alpha^{\aleph_0}\le\aleph_{\alpha+\mu}\}$ is stationary. Then $\aleph_{\omega_1}^{\aleph_1}\le\aleph_{\omega_1+\mu}.$
4. Suppose that $\mu<\omega_1,$ $\aleph_{\omega_1}$ is strong limit, and $\{\alpha<\omega_1:\aleph_\alpha^{\aleph_0}\le\aleph_{\alpha+\mu}\}$ is stationary. Then $2^{\aleph_{\omega_1}}\le\aleph_{\omega_1+\mu}.$
5. Suppose that $\mu<\omega_1$ and $\{\alpha<\omega_1 : 2^{\aleph_\alpha}\le\aleph_{\alpha+\mu}\}$ is stationary. Then $2^{\aleph_{\omega_1}}\le\aleph_{\omega_1+\mu}.$
6. The first counterexample to ${\sf GCH}$ is not a singular cardinal of uncountable cofinality
7. The first counterexample to ${\sf SCH}$ is not a singular cardinal of uncountable cofinality.

Proof. 1. This follows immediately from the theorem, using that $2^\lambda=\lambda^\kappa$ since $\lambda$ is strong limit.

2. Note that $\lambda$ is strong limit: If $\delta<\lambda$ then $\delta^{+\mu}<\lambda,$ since $\mu<\kappa.$ It follows that $2^\delta<\lambda$ for unboundedly many, and therefore for all, $\delta<\lambda.$

Let $(\lambda_\alpha:\alpha<\kappa)$ be a strictly increasing continuous sequence of cardinals cofinal in $\lambda.$ Then $\{\alpha:2^{\lambda_\alpha}\le\lambda_\alpha^{+\mu}\}$ is stationary in $\kappa.$

But $\prod_{\beta<\alpha}\lambda_\beta\le2^{\lambda_\alpha}$ for all $\alpha.$ Now the result follows from 1.

3. For all $\alpha<\omega_1,$ $\aleph_\alpha^{\aleph_1}=2^{\aleph_1}\aleph_\alpha^{\aleph_0}$ by Tarski’s formula, Homework problem 8 from lecture II.3. It follows that $\aleph_\alpha^{\aleph_1}\le\aleph_{\alpha+\mu}$ for stationarily many $\alpha.$ But, for any countable limit ordinal $\alpha,$ we have that $\prod_{\beta<\alpha}\aleph_\beta=\aleph_\alpha^{\aleph_0}.$ The result now follows from the theorem.

4. By 3., since $\aleph_{\omega_1}$ is strong limit.

5. and 6. By 2.

7. Suppose that ${\sf SCH}$ holds below $\lambda,$ that $\lambda$ is singular, and that $\kappa={\rm cf}(\lambda)$ is uncountable. Suppose first that $2^\kappa\ge\lambda.$ Then in fact $2^\kappa>\lambda,$ since ${\rm cf}(2^\kappa)>\kappa={\rm cf}(\lambda).$ We then have that $\gimel(\lambda)=2^\kappa=2^\kappa+\lambda^+,$ and ${\sf SCH}$ also holds at $\lambda.$

Suppose now that $2^\kappa<\lambda.$ We claim that in fact $\lambda$ is $\kappa$-inaccessible. Assume that this is the case, and fix a strictly increasing and continuous sequence of cardinals $(\lambda_\alpha:\alpha<\kappa)$ converging to $\lambda.$ For any limit ordinal $\alpha<\kappa,$ we have that $\prod_{\beta<\alpha}\lambda_\beta\le\lambda_\alpha^{|\alpha|}.$ By ${\sf SCH}$ below $\lambda,$ $\lambda_\alpha^{|\alpha|}\le\lambda_\alpha^+$ as long as $\alpha$ is sufficiently large to ensure that $2^{|\alpha|}\le2^\kappa<\lambda_\alpha.$ (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 $\mu=1,$ and therefore $2^\kappa + \lambda^+ = \lambda^+ \le \gimel(\lambda) \le \lambda^+,$ and ${\sf SCH}$ also holds at $\lambda,$ as wanted.

All that remains is to argue that $\lambda$ is indeed $\kappa$-inaccessible. For this, suppose that $\delta<\lambda.$ Using ${\sf SCH}$ below $\lambda,$ by (the local nature of the proof of) Corollary 19 we have that $\delta^\kappa\le\delta^++2^\kappa<\lambda,$ and we are done. ${\sf QED}$

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

Corollary 23. (Silver). $\aleph_{\omega_1}$ is not the first counterexample to ${\sf GCH}.$

Proof. Suppose that $2^\kappa=\kappa^+$ for all $\kappa<\aleph_{\omega_1}.$ In what follows, by $\prod_{\alpha<\omega_1}\aleph_{\alpha+1}$ we mean the set product (the collection of functions) rather than its cardinality, and similarly for other products. Fix injections $\pi_\alpha : {\mathcal P}(\aleph_\alpha)\to\aleph_{\alpha+1}$ for all $\alpha<\omega_1.$ For $X\subseteq\aleph_{\omega_1}$ let $f_X\in\prod_\alpha\aleph_{\alpha+1}$ be the map $f_X(\alpha)=\pi_\alpha(X\cap\aleph_\alpha).$ Set ${\mathcal F}=\{f_X : X\subseteq\aleph_{\omega_1}\}.$

The following properties of ${\mathcal F}$ are immediate:

1. Whenever $f\ne g\in{\mathcal F},$ there is in fact an $\alpha$ such that $f(\beta)\ne g(\beta)$ for all countable $\beta\ge\alpha.$
2. $|{\mathcal F}|=2^{\aleph_{\omega_1}},$ since the assignment $X\mapsto f_X$ is 1-1.

Whenever a collection ${\mathcal G}$ of functions $f:\omega_1\to{\sf ORD}$ satisfies 1., we will say that ${\mathcal G}$ is almost disjoint.

Lemma 24. Suppose that ${\mathcal G}\subseteq\prod_{\alpha<\omega_1}\aleph_{\alpha+1}$ is almost disjoint, and for all $f\in {\mathcal G},$ the set $\{\alpha : f(\alpha)<\aleph_\alpha\}$ is stationary in $\omega_1.$ Then $|{\mathcal G}|\le\aleph_{\omega_1}.$

Proof. Given $f\in{\mathcal G}$ define $\hat f:\omega_1\to\omega_1$ so that $\aleph_0+|f(\alpha)|=\aleph_{\hat f(\alpha)}$ for all $\alpha.$ Then $\hat f$ is regressive in a stationary set. By Fodor’s lemma (Theorem 14), there is some $\gamma_f\in\omega_1$ such that $S_f=\{\alpha : \hat f(\alpha)=\gamma_f\}$ is stationary.

For $\gamma\in\omega_1$ and $S\subseteq\omega_1,$ let ${\mathcal A}_{\gamma,S}=\{f\in{\mathcal G}:\gamma_f=\gamma\mbox{ and }S_f=S\}.$ Then either ${\mathcal A}_{\gamma,S}$ is empty, or else the map that assigns to $f\in {\mathcal A}_{\gamma,S}$ its restriction $f\upharpoonright S$ is injective, since ${\mathcal G}$ is almost disjoint and $S$ is stationary in $\omega_1$ and therefore unbounded.

By definition of $\gamma_f,$ if $f\in{\mathcal A}_{\gamma,S}$ then $f\upharpoonright S:\omega_1\to\aleph_{\gamma+1}.$ But then it follows from ${\sf GCH}$ that $|{\mathcal A}_{\gamma,S}|\le\aleph_{\gamma+2}.$

The result follows, since ${\mathcal G}=\bigcup_{\gamma,S}{\mathcal A}_{\gamma,S},$ and there are only $\aleph_1$ many possible values of $\gamma$ and $\aleph_2$ many possible values of $S.$ ${\sf QED}$

Corollary 25. Let $g\in\prod_{\alpha<\omega_1}\aleph_{\alpha+1}$ and let ${\mathcal G}$ be almost disjoint. Suppose that for all $f\in{\mathcal G},$ $\{\alpha<\omega_1 : f(\alpha) is stationary. Then $|{\mathcal G}|\le\aleph_{\omega_1}.$

Proof. This is immediate from the lemma since $\prod_\alpha g(\alpha)$ injects into $\prod_\alpha\aleph_\alpha.$ ${\sf QED}$

To complete the proof of the theorem, we now argue that ${\mathcal F}$ can be split into $\aleph_{\omega_1 +1}$ many sets to which the corollary applies.

Definition 26. Let ${\mathcal C}_{\omega_1}$ be the club filter on $\omega_1.$ Given functions $f,g:\omega_1\to{\sf ORD},$ we say that $f<_* g$ iff $\{\alpha:f(\alpha) contains a club.

Suppose that $g\in{\mathcal F}.$ Then, by Corollary 22, at most $\aleph_{\omega_1}$ many functions in ${\mathcal F}$ are below $g$ on a stationary set, i.e., $g<_*f$ for all but at most $\aleph_{\omega_1}$ many functions $f\in{\mathcal F}.$

By induction, we can therefore define a $<_*$-increasing sequence $(f_\eta : \eta<\aleph_{\omega+1})$ of members of ${\mathcal F}$. But notice that for any $f\in{\mathcal F},$ it is the case that $f<_* f_\eta$ for some $\eta.$ Otherwise, $f$ contradicts the previous paragraph. It follows that ${\mathcal F}=\bigcup_\eta{\mathcal F}_\eta,$ where ${\mathcal F}_\eta=\{f\in{\mathcal F} : f<_* f_\eta\},$ and we are done by Corollary 22. ${\sf QED}$

The arguments above (see the proof of Corollary 22.7, for example) suggest that ${\sf SCH}$ 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 $\beta$ there is some strictly increasing sequence of cardinals $(\kappa_\alpha:\alpha<\beta)$ with limit $\kappa$ such that $\prod_{\alpha<\beta}\kappa_\alpha<\kappa^{|\beta|}.$

1. Show that ${\rm cf}(\beta)<|\beta|<\beta$ and there is some cardinal $\lambda<\kappa$ such that $\lambda^{|\beta|}>\kappa.$
2. Suppose that $\beta$ is least such that there is a counterexample as above. Then ($\aleph_0<|\beta|$ and) $\beta=|\beta|+\omega.$
3. If Tarski’s conjecture fails, then there is a cardinal $\aleph_\gamma$ of uncountable cofinality $\tau$ such that $\gamma>\tau,$ $\aleph_\nu^\tau<\aleph_\gamma$ for all $\nu<\gamma,$ and $\gimel(\aleph_{\gamma+\omega})<\gimel(\aleph_\gamma).$

### 9 Responses to 580 -Cardinal arithmetic (5)

1. […] (or everywhere) result, that gives us information beyond the pointwise theorems from last lecture, like Corollary 23. Then I state a result where the hypotheses, as in Silver’s theorem, are […]

2. […] Recall the very general statement of Silver’s theorem, Theorem 21 in lecture II.5: […]

3. […] this, proceed by contradiction, assuming that and consider a Ulam matrix see Definition 12 in lecture II.5. Arguing just as in the proof of Theorem 13 there, notice that has measure 1 for all and by […]

4. […] of a successor cardinal can be split into many disjoint stationary sets, this is Theorem 13 in lecture II.5. Consequently, we can partition into many stationary pieces. Let be such a partition. By […]

5. […] and Corollary 23, since for all by induction (by the local nature of the proof of Lemma 18 in lecture II.5). The result follows […]

6. […] with a result of Solovay. We need a preliminary lemma, generalizing Ulam’s Theorem 13 from lecture II.5 that stationary subsets of can be split into many stationary subsets: Theorem 3 (Solovay) Any […]

7. […] with a result of Solovay. We need a preliminary lemma, generalizing Ulam’s Theorem 13 from lecture II.5 that stationary subsets of can be split into many stationary […]

8. […] of a successor cardinal can be split into many disjoint stationary sets, this is Theorem 13 in lecture II.5. Consequently, we can partition into many stationary pieces. Let be such a partition. By […]

9. […] this, proceed by contradiction, assuming that and consider a Ulam matrix see Definition 12 in lecture II.5. Arguing just as in the proof of Theorem 13 there, notice that has measure 1 for all and by […]