## 580 -Cardinal arithmetic (11)

4. Strongly compact cardinals and ${{\sf SCH}}$

Definition 1 A cardinal ${\kappa}$ is strongly compact iff it is uncountable, and any ${\kappa}$-complete filter (over any set ${I}$) can be extended to a ${\kappa}$-complete ultrafilter over ${I.}$

The notion of strong compactness has its origin in infinitary logic, and was formulated by Tarski as a natural generalization of the compactness of first order logic. Many distinct characterizations have been found.

Let’s begin by defining the (first order) infinitary logics ${{\mathcal L}_{\kappa\lambda}.}$

Definition 2 Let ${\kappa}$ and ${\lambda}$ be nonzero cardinals or ${\infty.}$ Given a set of relation, function, and constant symbols (a language), the ${{\mathcal L}_{\kappa,\lambda}}$ formulas of the language are defined inductively by closing under the following clauses:

1. Atomic formulas are ${{\mathcal L}_{\kappa,\lambda}}$ formulas.
2. If ${\phi}$ is an ${{\mathcal L}_{\kappa,\lambda}}$ formula, so is ${\lnot\phi.}$
3. If ${X}$ is a set of fewer than ${\kappa}$ many ${{\mathcal L}_{\kappa,\lambda}}$ formulas, then ${\bigwedge X}$ is also an ${{\mathcal L}_{\kappa,\lambda}}$ formula.
4. If ${X}$ is a set of fewer than ${\lambda}$ many variables, and ${\phi}$ is an ${{\mathcal L}_{\kappa,\lambda}}$ formula, then so is ${\exists X\,\phi.}$

One can formalize the above even further by, for example, starting with variables ${x_\alpha=(0,\alpha)}$ for each ordinal ${\alpha,}$ ${n}$-ary relation symbols ${R_{\alpha,n}=(1,\alpha,n)}$ for all ${n<\omega}$ and all ordinals ${\alpha,}$ and ${n}$-ary function symbols ${f_{\alpha,n}=(2,\alpha,n)}$ for all ${n<\omega}$ and all ordinals ${\alpha,}$ and assigning a (well-founded) tree decomposition to each formula.

One then defines the corresponding semantic naturally. The well-foundedness of the tree associated to each formula is used to guarantee that the obvious inductive definition takes into account all formulas. For example, given a model ${M}$ for the language under consideration, an assignment ${\pi:x_\alpha\mapsto t_\alpha}$ of elements of ${M}$ to variables, a set ${X}$ of fewer than ${\lambda}$ many variables, and an ${{\mathcal L}_{\kappa,\lambda}}$ formula ${\phi,}$ we say that ${M\models\exists X\,\phi[\pi]}$ holds iff there is a function ${\tau:X\rightarrow M}$ such that if ${\hat\pi}$ is the map that to ${x_\alpha}$ assigns ${t_\alpha}$ unless ${x_\alpha\in X,}$ in which case ${\hat\pi(x_\alpha)=\tau(x_\alpha),}$ then ${M\models\phi[\hat\pi].}$

One says that ${M\models\psi}$ iff ${M\models\psi[\pi]}$ for all assignments ${\pi.}$

One defines free and bound variables as usual, and the standard notions carry through: A sentence is a formula without free variables, a theory is a collection of sentences, and a theory ${\Sigma}$ is satisfiable iff it has a model, i.e., a structure ${M}$ in the language of ${\Sigma}$ such that ${M\models\phi}$ for all ${\phi\in\Sigma.}$

Definition 3 An ${{\mathcal L}_{\kappa,\lambda}}$ theory ${\Sigma}$ is ${\rho}$-satisfiable iff whenever ${T\subseteq\Sigma}$ and ${|T|<\rho,}$ then ${T}$ is satisfiable.

Hence the compactness theorem for first order logic (i.e., for ${{\mathcal L}_{\omega,\omega}}$) is the claim that if ${\Sigma}$ is an ${{\mathcal L}_{\omega,\omega}}$ theory that is finitely (i.e., ${\omega}$-) satisfiable, then ${\Sigma}$ itself is satisfiable.

A result of Keisler and Tarski showed that the corresponding notion of compactness for ${{\mathcal L}_{\kappa,\kappa}}$ with ${\kappa>\omega}$ is of large cardinal character in that it at least implies measurability of ${\kappa.}$ The argument naturally leads to the notion of a fine measure.

Definition 4 Given a cardinal ${\kappa}$ and a set ${X,}$ denote by ${{\mathcal P}_\kappa(X)=[X]^{<\kappa}}$ the collection of subsets of ${X}$ of size strictly smaller than ${\kappa.}$

An ultrafilter ${{\mathcal U}}$ over ${{\mathcal P}_\kappa(X)}$ is fine iff it is ${\kappa}$-complete, nonprincipal, and for any ${x\in X,}$ ${\{\sigma\in {\mathcal P}_\kappa(X):x\in \sigma\}\in {\mathcal U}.}$

If ${\gamma\ge\kappa}$ say that ${\kappa}$ is ${\gamma}$-compact iff it is uncountable and there is a fine ultrafilter over ${{\mathcal P}_\kappa(\gamma).}$

Actually, the notion originally considered was not that of a fine measure, but a closely related one:

Definition 5 Let ${\kappa\le\lambda}$ be cardinals. An ultrafilter ${{\mathcal U}}$ over a set ${X}$ is ${(\kappa,\lambda)}$-regular iff there is a family ${(X_\alpha:\alpha<\lambda)}$ of elements of ${{\mathcal U}}$ such that for every ${S\subseteq \lambda}$ of size ${\kappa,}$ ${\bigcap_{\alpha\in S}X_\alpha=\emptyset.}$

Lemma 6 Suppose that ${\kappa\le\lambda.}$ Then there is a fine ultrafilter over ${{\mathcal P}_\kappa(\lambda)}$ iff there is a ${\kappa}$-complete ${(\kappa,\lambda)}$-regular ultrafilter.

Proof: Let ${{\mathcal U}}$ be a fine ultrafilter over ${{\mathcal P}_\kappa(\lambda)}$ and let ${X_\alpha=\{\sigma\in {\mathcal P}_\kappa(\lambda):\alpha\in\sigma\}}$ for all ${\alpha<\lambda.}$ Then ${(X_\alpha:\alpha<\lambda)}$ witnesses that ${{\mathcal U}}$ is ${(\kappa,\lambda)}$-regular.

Conversely, let ${{\mathcal D}}$ be a ${\kappa}$-complete ${(\kappa,\lambda)}$-regular ultrafilter as witnessed by the sequence ${(X_\alpha:\alpha<\lambda).}$ Consider the projection ${f:\bigcup_\alpha X_\alpha\rightarrow{\mathcal P}_\kappa(\lambda)}$ given by ${f(x)=\{\alpha<\lambda:x\in X_\alpha\}.}$ Then ${f_*({\mathcal D}\upharpoonright\bigcup_\alpha X_\alpha)}$ is a fine ultrafilter over ${{\mathcal P}_\kappa(\lambda).}$ $\Box$

Lemma 7 If ${\kappa}$ is strongly compact, then it is regular, and therefore measurable.

Proof: Recall that for any nonprincipal ultrafilter ${{\mathcal U},}$ its additivity ${{\rm add}({\mathcal U}),}$ the first ${\tau}$ such that ${{\mathcal U}}$ is not ${\tau}$-complete, is either ${\omega}$ or a measurable cardinal. Let ${{\mathcal F}}$ be the filter of cobounded subsets of ${\kappa^+.}$ This is a ${\kappa^+}$-complete filter so, by assumption, it can be extended to a ${\kappa}$-complete nonprincipal ultrafilter over ${\kappa^+.}$ If ${\kappa}$ is singular, the additivity of this ultrafilter cannot be ${\kappa,}$ but it is not smaller than ${\kappa,}$ and it is not larger than ${\kappa^+}$, so it would have to be ${\kappa^+.}$ But ${\kappa^+}$ is not measurable, being a successor, contradiction. $\Box$

Theorem 8 The following are equivalent about the cardinal ${\kappa:}$

1. ${\kappa}$ is strongly compact.
2. ${\kappa}$ is ${\lambda}$-compact for all ${\lambda\ge\kappa.}$
3. ${\kappa}$ is uncountable, and for all regular ${\lambda\ge\kappa}$ there is a uniform ${\kappa}$-complete ultrafilter over ${\lambda.}$
4. For all ${\lambda}$ there is an elementary embedding ${j:V\rightarrow M}$ with ${{\rm cp}(j)=\kappa}$ such that whenever ${X\subseteq M}$ and ${|X|\le\lambda}$ then there is some ${Y\in M}$ such that ${X\subseteq Y}$ and ${M\models|Y|
5. For all ${\lambda}$ there is an elementary embedding ${j:V\rightarrow M}$ with ${{\rm cp}(j)=\kappa}$ such that ${{}^\kappa M\subset M}$ and there is a set ${X\in M}$ such that ${j[\lambda]\subseteq X}$ and ${M\models|X|
6. ${\kappa}$ is uncountable, and the compactness theorem holds for ${{\mathcal L}_{\kappa\kappa}:}$ Any ${\kappa}$-satisfiable ${{\mathcal L}_{\kappa,\kappa}}$-theory is satisfiable.
7. ${\kappa}$ is uncountable, and for every ${\lambda\ge\kappa}$ there is a ${\kappa}$-complete, ${(\kappa,\lambda)}$-regular ultrafilter.

Proof: 1 implies 2. Let ${\lambda\ge\kappa.}$ The collection ${\{\{x\in {\mathcal P}_\kappa(\lambda):\sigma\subseteq x\}:\sigma\in {\mathcal P}_\kappa(\lambda)\}}$ generates a ${\kappa}$-complete filter by closing under supersets (because ${\kappa}$ is regular, by Lemma 7). By 1, there is a ${\kappa}$-complete ultrafilter over ${{\mathcal P}_\kappa(\lambda)}$ that extends it. This ultrafilter is clearly fine.

2 implies 5. We may assume that ${\lambda\ge\kappa.}$ Let ${{\mathcal U}}$ be a fine ultrafilter over ${{\mathcal P}_\kappa(\lambda)}$ and form the ultrapower embedding ${j:V\rightarrow M={\rm Ult}(V,{\mathcal U}).}$ Obviously (by considering the sets ${\{x\in {\mathcal P}_\kappa(\lambda):\alpha\in x\}}$ for ${\alpha\in\lambda}$), ${{\mathcal U}}$ is not ${\kappa^+}$-complete. It follows that ${{\rm cp}(j)=\kappa.}$ From, say, Theorem 6 and preceeding remarks from last lecture, it follows that ${{}^\kappa{\sf ORD}\subset M.}$ By choice, this implies that ${{}^\kappa M\subset M.}$

The result follows, since we can take ${X=\langle{\rm id}\rangle:}$ That ${j[\lambda]\subseteq X}$ simply means that for all ${\alpha<\lambda,}$ ${\{x\in {\mathcal P}_\kappa(\lambda):\alpha\in x\}\in {\mathcal U}.}$ That ${M\models|X| simply means that ${{\mathcal P}_\kappa(\lambda)=\{x\in {\mathcal P}_\kappa(\lambda):|x|<\kappa\}\in {\mathcal U}.}$

5 implies 2. If ${j:V\rightarrow M}$ and ${X}$ are as in 5, without loss, ${X\in {\mathcal P}_{j(\kappa)}^M(j(\lambda));}$ otherwise, replace ${X}$ with ${X\cap j(\lambda).}$

Define

$\displaystyle {\mathcal U}=\{A\subseteq{\mathcal P}_\kappa(\lambda):X\in j(A)\}.$

Then ${{\mathcal U}}$ is a fine ultrafilter over ${{\mathcal P}_\kappa(\lambda).}$

2 implies 4. Again, take ${j:V\rightarrow M={\rm Ult}(V,{\mathcal U})}$ where ${{\mathcal U}}$ is a fine measure over ${{\mathcal P}_\kappa(\lambda).}$ Given ${X\subseteq M}$ with ${|X|\le\lambda,}$ say ${X=\{\langle f_\alpha\rangle:\alpha<\lambda\},}$ let ${F:{\mathcal P}_\kappa(\lambda)\rightarrow V}$ be given by ${F(\sigma)=\{f_\alpha(\sigma):\alpha\in\sigma\}.}$ Then ${Y=\langle F\rangle}$ is as required.

4 implies 1. Let ${{\mathcal F}}$ be a ${\kappa}$-complete filter over a set ${I}$ and let ${\lambda=|{\mathcal F}|.}$ Take ${j:V\rightarrow M}$ a witness of 4 for ${\lambda.}$ Let ${Y\in M}$ extend ${j[{\mathcal F}]}$ and have size smaller than ${j(\kappa)}$ in ${M.}$ Then ${\bigcap(j({\mathcal F})\cap Y)\ne\emptyset,}$ since in ${M}$ it is an intersection of fewer than ${j(\kappa)}$ members of the ${j(\kappa)}$-complete filter ${j({\mathcal F}).}$ Let ${c}$ be in this intersection, and define an ultrafilter by

$\displaystyle {\mathcal U}=\{X\subseteq I: c\in j(X)\}.$

Then ${{\mathcal U}}$ is a ${\kappa}$-complete ultrafilter extending ${{\mathcal F}.}$

(Notice we have shown that 4 and 5 are equivalent, although neither property seems to imply the other directly.)

2 implies 6. Let ${\Sigma=\{\phi_\alpha:\alpha<\lambda\}}$ be a ${\kappa}$-satisfiable ${{\mathcal L}_{\kappa,\kappa}}$ theory. We may assume that ${\lambda\ge\kappa,}$ or there is nothing to show. For any ${\sigma\in {\mathcal P}_\kappa(\lambda)}$ let ${M_\sigma\models\bigwedge\{\phi_\alpha:\alpha\in\sigma\}.}$ Let ${{\mathcal U}}$ be a fine ultrafilter over ${{\mathcal P}_\kappa(\lambda),}$ and consider the ultraproduct ${\prod_{\sigma\in {\mathcal P}_\kappa(\lambda)}M_\sigma/{\mathcal U}.}$

This is defined by a straightforward generalization of Definition 1 from lecture II.9: For ${f,g\in\prod_\sigma M_\sigma}$ set ${f=_{\mathcal U}g}$ iff ${\{\sigma:f(\sigma)=g(\sigma)\}\in {\mathcal U}.}$ Let ${[f]}$ be the equivalence class of ${f}$ under this equivalence relation. Now we do not need Scott’s trick to make sense of ${[f],}$ since there are only set-many such functions ${f}$ to begin with. Let

$\displaystyle \prod_{\sigma\in {\mathcal P}_\kappa(\lambda)}M_\sigma/{\mathcal U}$

be the collection of these equivalence classes. Define ${R_{\mathcal U},}$ for ${R}$ an ${n}$-ary relation symbol in the language of ${\Sigma,}$ by

$\displaystyle ([f_1],\dots,[f_n])\in R_{\mathcal U}\mbox{ iff }\{\sigma:M_\sigma\models (f_1(\sigma),\dots,f_n(\sigma))\in R\}\in {\mathcal U}.$

Define ${f_{\mathcal U},}$ for ${f}$ an ${n}$-ary function symbol in the language of ${\Sigma,}$ by

$\displaystyle f_{\mathcal U}([g_1],\dots,[g_n])=[g]$

iff

$\displaystyle \{\sigma:M_\sigma\models f(g_1(\sigma),\dots,g_n(\sigma))=g(\sigma)\}\in {\mathcal U}.$

Finally, define ${c_{\mathcal U},}$ for ${c}$ a constant symbol in the language of ${\Sigma,}$ by ${c=[f],}$ where ${M_\sigma\models f(\sigma)=c}$ for all ${\sigma.}$

It is easy to check that these definitions make sense (i.e., they do not depend on the representatives of the equivalence classes under consideration), and they turn ${\prod_{\sigma\in {\mathcal P}_\kappa(\lambda)}M_\sigma/{\mathcal U}}$ into a structure in the language of ${\Sigma.}$

The proof of ${\mbox{\L o\'s}}$‘s lemma generalizes to this setting, since ${{\mathcal U}}$ is ${\kappa}$-complete and, since ${{\mathcal U}}$ is fine, this implies that ${\prod_{\sigma\in {\mathcal P}_\kappa(\lambda)}M_\sigma/{\mathcal U}\models\phi_\alpha}$ for all ${\alpha<\lambda.}$

(Notice that the same argument gives a proof of the compactness of ${{\mathcal L}_{\omega,\omega}}$ and shows that, over ${{\sf ZF},}$ the existence of nonprincipal ultrafilters over any infinite set guarantees the compactness of ${{\mathcal L}_{\omega,\omega}.}$)

6 implies 1. Let ${{\mathcal F}}$ be a ${\kappa}$-complete filter over a set ${I.}$ Consider the language that has a constant symbol ${\hat X}$ for all ${X\subseteq I,}$ a different constant symbol ${c,}$ and a binary relation symbol ${\hat\in.}$ Let ${\Sigma}$ be the theory in this language consisting of the union of the ${{\mathcal L}_{\kappa,\kappa}}$-theory of ${(I\cup{\mathcal P}(I),\in,X)_{X\subseteq I}}$ and ${\{c\hat\in\hat X:X\in {\mathcal F}\}.}$ The ${\kappa}$-completeness of ${{\mathcal F}}$ ensures that ${\Sigma}$ is ${\kappa}$-satisfiable, and 6 implies that it is satisfiable. Let ${M}$ be a model of ${\Sigma}$ and define an ultrafilter ${{\mathcal U}}$ over ${I}$ by

$\displaystyle {\mathcal U}=\{X\subseteq I:M\models c\hat\in \hat X\}.$

It is straightforward to check that ${{\mathcal U}}$ is ${\kappa}$-complete and extends ${{\mathcal F}.}$

(Notice that, over ${{\sf ZF},}$ the same argument shows that compactness for ${{\mathcal L}_{\omega,\omega}}$ implies (and is therefore equivalent to) the existence of nonprincipal ultrafilters over any infinite set.)

1 implies 3. Use 1 to extend the filter of cobounded subsets of ${\lambda}$ to a ${\kappa}$-complete ultrafilter over ${\kappa,}$ and notice that it is necessarily uniform.

2 is equivalent to 7. By Lemma 6.

3 implies 7. This is a theorem of Ketonen. Notice first that ${\kappa}$ is regular. By induction on ${\lambda,}$ one shows that if ${\lambda\ge\kappa}$ is regular then there is a ${\kappa}$-complete ${(\kappa,\lambda)}$-regular ultrafilter ${{\mathcal D}_\lambda}$ over ${\lambda.}$

Notice that the base of the induction (${\lambda=\kappa}$) is clear.

Definition 9 Let ${\lambda}$ be regular. An ultrafilter ${{\mathcal D}}$ over ${\lambda}$ is weakly normal iff whenever ${f:\lambda\setminus\{0\}\rightarrow\lambda}$ is regressive, there is some ${\alpha<\lambda}$ such that ${\{\beta<\lambda:f(\beta)<\alpha\}\in {\mathcal D}.}$

First we check that we may strengthen the assumption on the ultrafilters given by 3:

Lemma 10 Let ${\kappa\le\lambda}$ be uncountable regular cardinals. If ${{\mathcal D}}$ is a uniform ${\kappa}$-complete ultrafilter over ${\lambda}$ then there is a uniform ${\kappa}$-complete weakly normal ultrafilter ${{\mathcal U}}$ over ${\lambda}$ such that ${{\mathcal U}\le_{RK}{\mathcal D}.}$

The following notion will prove useful:

Definition 11 Given a ${\sigma}$-complete ultrafilter ${{\mathcal V}}$ and an ordinal ${\alpha,}$ let

$\displaystyle \delta^{\mathcal V}_\alpha=\sup j_{\mathcal V}[\alpha]$

where, as usual, ${j_{\mathcal V}:V\rightarrow {\rm Ult}(V,{\mathcal V})}$ is the ultrapower embedding.

If ${{\mathcal V}}$ is a ${\sigma}$-complete ultrafilter over ${\lambda}$ and ${\langle f\rangle=\delta^{\mathcal V}_\lambda,}$ we say that ${f}$ is the first function of ${{\mathcal V}.}$

Proof: Let ${f}$ be the first function of ${{\mathcal D}.}$ Then ${{\mathcal U}=f_*({\mathcal D})}$ works. $\Box$

It follows that for all regular ${\lambda\ge\kappa}$ there is a weakly normal uniform ${\kappa}$-complete ultrafilter ${{\mathcal U}_\lambda}$ over ${\lambda.}$

Definition 12 Let ${{\mathcal U}}$ and ${{\mathcal V}}$ be ultrafilters over sets ${X}$ and ${Y,}$ respectively. Then ${{\mathcal U}\times{\mathcal V}}$ is the ultrafilter over ${X\times Y}$ given by

$\displaystyle A\in {\mathcal U}\times{\mathcal V}\mbox{ iff }\{x\in X:\{y\in Y:(x,y)\in A\}\in {\mathcal V}\}\in {\mathcal U}.$

It is straightforward to check that ${{\mathcal U}\times{\mathcal V}}$ in Definition 12 is indeed an ultrafilter.

Assuming that ${\mu\ge\kappa}$ is regular and ${{\mathcal D}_\mu}$ exists, we claim that we can take ${{\mathcal D}_{\mu^+}={\mathcal U}_{\mu^+}\times{\mathcal D}_\mu}$ (more carefully, this is an ultrafilter over ${\mu^+\times\mu,}$ but we obtain the desired ultrafilter over ${\mu^+}$ via a bijection.)

In effect, it is easy to check that this product is ${\kappa}$-complete and uniform. We need to verify that it is ${(\kappa,\mu^+)}$-regular. This follows from a general result.

Lemma 13 Suppose ${\lambda}$ is regular and ${{\mathcal D}}$ is a uniform ${\sigma}$-complete ultrafilter over ${\lambda.}$ Suppose that ${{\rm Ult}(V,{\mathcal D})\models{\rm cf}(\delta^{\mathcal D}_\lambda) Then ${{\mathcal D}}$ is ${(\kappa,\lambda)}$-regular.

Proof: Let ${f}$ be the first function of ${{\mathcal D},}$ so ${Y=\{\alpha<\lambda:{\rm cf}(f(\alpha))<\kappa\}\in {\mathcal D}.}$ Let ${M={\rm Ult}(V,{\mathcal D}).}$

For each ${\alpha\in Y}$ let ${A_\alpha\subseteq f(\alpha)}$ be cofinal and of size smaller than ${\kappa,}$ and let ${F:\lambda\rightarrow V}$ be a function such that ${F(\alpha)=A_\alpha}$ for all ${\alpha\in Y,}$ so ${\langle F\rangle}$ is a cofinal subset of ${\delta^{\mathcal D}_\lambda}$ that, in ${M,}$ has size smaller than ${j(\kappa).}$

By induction define disjoint intervals ${I_\eta=(\xi_\eta,\mu_\eta)}$ for ${\eta<\lambda}$ as follows: Given ${\{I_\eta:\eta<\eta'\}}$ for some ${\eta'<\lambda,}$ let ${\xi_{\eta'}=(\sup_{\eta<\eta'}\mu_\eta)+1,}$ and let ${\mu_{\eta'}}$ be the least ordinal ${\mu<\lambda}$ such that ${\langle F\rangle\cap(j(\xi_{\eta'}),j(\mu))\ne\emptyset.}$

Note that ${\langle F\rangle\cap j(I_\eta)\ne\emptyset}$ for all ${\eta<\lambda,}$ by construction, and therefore the sets ${X_\eta=\{\alpha\in Y:A_\alpha\cap I_\eta\ne\emptyset\}}$ are in ${{\mathcal D}}$ for all ${\eta<\lambda.}$ However, if ${S\subseteq\lambda}$ has size ${\kappa,}$ then

$\displaystyle \bigcap_{\eta\in S}X_\eta =\{\alpha\in Y:\forall \eta\in S\,(A_\alpha\cap I_\eta\ne\emptyset)\}=\emptyset,$

because all the ${A_\alpha}$ have size smaller than ${\kappa}$ and there are ${\kappa}$ disjoint intervals ${I_\eta}$ for ${\eta\in S.}$ Hence, ${(X_\eta:\eta<\lambda)}$ witnesses that ${{\mathcal D}}$ is ${(\kappa,\lambda)}$-regular. $\Box$

Notice that Lemma 13 is in fact an equivalence. In order to make use of this result, we need to identify the first function for ${{\mathcal U}_{\mu^+}\times{\mathcal D}_\mu.}$

Definition 14 Let ${\kappa\le\lambda}$ be regular uncountable cardinals. Let ${{\mathcal U}}$ be a weakly normal ${\kappa}$-complete uniform ultrafilter over ${\lambda.}$ For each ${\alpha<\lambda}$ let ${{\mathcal V}_\alpha}$ be a ${\kappa}$-complete ultrafilter over some set ${X_\alpha.}$ The ${{\mathcal U}}$-sum of the ultrafilters ${{\mathcal V}_\alpha}$ is the ultrafilter ${{\mathcal D}}$ over ${\Gamma=\{(\alpha,\beta):\alpha<\lambda,\beta\in X_\alpha\}}$ given by

$\displaystyle X\in {\mathcal D}\mbox{ iff }\{\alpha: \{\beta:(\alpha,\beta)\in X\}\in {\mathcal V}_\alpha\}\in {\mathcal U}.$

In particular, if ${{\mathcal V}_\alpha={\mathcal V}}$ for all ${\alpha,}$ the above reduces to the product ${{\mathcal U}\times{\mathcal V}.}$

Definition 15 In the setting of Definition 14, let ${\langle g_\alpha\rangle_{{\mathcal V}_\alpha}=\delta^{{\mathcal V}_\alpha}_\alpha}$ for all ${\alpha<\lambda,}$ and let ${f_\lambda:\Gamma\rightarrow V}$ be the function

$\displaystyle f_\lambda(\alpha,\beta)=g_\alpha(\beta).$

Lemma 16 In the setting of Definition 15, ${\langle f_\lambda\rangle_{\mathcal D}=\delta^{\mathcal D}_\lambda.}$

In particular, if ${{\mathcal U}={\mathcal U}_{\mu^+}}$ and ${{\mathcal V}_\alpha={\mathcal D}_\mu}$ for all ${\alpha<\mu^+,}$ then ${{\mathcal D}={\mathcal D}_{\mu^+}}$ (up to a bijection), and ${f_{\mu^+}}$ is the first function of ${{\mathcal D}_{\mu^+}.}$

Proof: By uniformity of ${{\mathcal U},}$ ${\langle f_\lambda\rangle_{\mathcal D}\ge\delta^{\mathcal D}_\lambda.}$ Suppose that ${\langle g\rangle_{\mathcal D}<\langle f_\lambda\rangle_{\mathcal D},}$ so

$\displaystyle S=\{\alpha:\{\beta: g(\alpha,\beta)

By definition of ${g_\alpha,}$ for each ${\alpha\in S}$ there is ${t_\alpha<\alpha}$ such that

$\displaystyle \{\beta: g(\alpha,\beta)

By weak normality of ${{\mathcal U},}$ there is a fixed ${\xi<\lambda}$ such that ${\{\alpha\in S: t_\alpha\le\xi\}\in {\mathcal U}.}$ But then ${\langle g\rangle_{\mathcal D}\le j(\xi)<\delta^{\mathcal D}_\lambda.}$ $\Box$

By a straightforward generalization of Lemma 13, we get:

Corollary 17 For ${{\mathcal D}}$ as in Definition 14, ${\{\alpha:{\mathcal V}_\alpha}$ is ${(\kappa,{\rm cf}(\alpha))}$-regular${\}\in {\mathcal U}}$ iff ${\{(\alpha,\beta):{\rm cf}(g_\alpha(\beta))<\kappa\}\in {\mathcal D}}$ iff ${{\mathcal D}}$ is ${(\kappa,\lambda)}$-regular. ${\Box}$

In particular, ${{\mathcal D}_{\mu^+}}$ is indeed ${(\kappa,\mu^+)}$-regular.

More generally, if the inductive assumption holds for all ${\alpha<\lambda}$ and ${\lambda}$ is regular, we can choose ${{\mathcal V}_\alpha}$ to be a ${\kappa}$-complete ${(\kappa,{\rm cf}(\alpha))}$-regular ultrafilter over some set ${X_\alpha,}$ and define ${{\mathcal D}_\lambda}$ to be the ${{\mathcal U}_\lambda}$-sum of the ultrafilters ${{\mathcal V}_\alpha.}$ Then ${{\mathcal D}_\lambda}$ is ${(\kappa,\lambda)}$-regular, and we are done. $\Box$

Corollary 18 Let ${\kappa}$ be ${\gamma}$-compact for all ${\gamma\in[\kappa,\tau],}$ where ${\tau}$ is strongly compact. Then ${\kappa}$ is strongly compact.

Proof: By the local nature of the proof of the equivalence of conditions 2 and 3 in Theorem 8. $\Box$

In terms of cardinal arithmetic, the goal of this lecture is to prove the following theorem of Solovay:

Theorem 19 (Solovay) Assume that ${\kappa}$ is strongly compact. Then ${{\sf SCH}}$ holds above ${\kappa,}$ i.e., for all ${\lambda\ge\kappa,}$ ${\lambda^{{\rm cf}(\lambda)}=2^{{\rm cf}(\lambda)}+\lambda^+.}$

The proof below is due to Matteo Viale and uses the notion of a covering matrix, that has proved very useful as well in the presence of forcing axioms like ${{\sf PFA}}$ or ${{\sf MM}.}$ I have had occasion of using both the theory developed in the course of the proof of Theorem 8, and Viale’s machinery, in my own research.

Definition 20 Let ${\kappa}$ be a cardinal. A precovering matrix for ${\kappa}$ is an array ${{\mathcal D}=(K(n,\alpha):n<\omega,\alpha<\kappa)}$ of subsets of ${\kappa}$ such that

$\displaystyle \alpha\subseteq\bigcup_n K(n,\alpha)$

for all ${\alpha<\kappa.}$

We say that a precovering matrix is nontrivial iff there is some ${\beta<\kappa}$ such that ${{\rm ot}(K(n,\alpha))<\beta}$ for all ${n<\omega}$ and ${\alpha<\kappa.}$ The least such ${\beta}$ is denoted ${\beta_{\mathcal D}.}$

Viale’s notion of a covering matrix satisfies several additional requirements. For our application, precovering suffices.

Lemma 21 Suppose that ${\kappa}$ is a singular cardinal of cofinality ${\omega.}$ Then there is a nontrivial precovering matrix ${{\mathcal D}}$ for ${\kappa^+}$ with ${\beta_{\mathcal D}=\kappa.}$

Proof: Let ${(\kappa_n:n<\omega)}$ be a strictly increasing sequence of regular cardinals cofinal in ${\kappa,}$ and for each nonzero ${\eta<\kappa^+}$ let ${\phi_\eta:\kappa\rightarrow\eta}$ be surjective. Define the entries of ${{\mathcal D}}$ inductively by

$\displaystyle K(n,\alpha)=\bigcup\{K(n,\beta)\cup\{\beta\}:\beta\in\phi_\alpha[\kappa_n]\}.$

Clearly this works. In fact, for any ${n}$ and all ${\alpha,}$ ${{\rm ot}(K(n,\alpha))<\kappa_n^+.}$ $\Box$

Definition 22 The covering property holds at ${\kappa,}$ ${{\sf CP}(\kappa),}$ iff for any precovering matrix ${{\mathcal D}}$ for ${\kappa}$ there is an unbounded set ${A\subseteq\kappa}$ such that

$\displaystyle [A]^{\aleph_0}\subseteq\bigcup_{n,\alpha}[K(n,\alpha)]^{\aleph_0}.$

Viale’s definition of the condition ${{\sf CP}}$ is more restrictive, but for our application, this more general version suffices. Notice that ${{\sf CP}(\kappa)}$ says something about cardinal arithmetic at ${\kappa,}$ at least if there are nontrivial precovering matrices for ${\kappa.}$

Corollary 23 If ${{\sf CP}(\kappa)}$ holds then

$\displaystyle {\rm cf}(\kappa)^{\aleph_0}\le\kappa\sup_{\rho<\beta_{\mathcal D}}\rho^{\aleph_0}$

for any precovering matrix ${{\mathcal D}}$ for ${\kappa.}$

Proof: This is immediate from the displayed equation in Definition 22. $\Box$

Theorem 24 (Viale) Suppose that ${\kappa}$ is strongly compact. Then ${{\sf CP}(\lambda)}$ holds for all regular ${\lambda\ge\kappa.}$

Proof: Let ${\lambda\ge\kappa}$ be regular, and let ${{\mathcal D}}$ be a precovering matrix for ${\lambda.}$ Let ${{\mathcal U}}$ be a ${\kappa}$-complete uniform ultrafilter over ${\lambda.}$ For each ${\alpha<\lambda}$ and each ${n<\omega}$ let

$\displaystyle A^\alpha_n=\{\beta>\alpha:\alpha\in K(n,\beta)\}.$

Since ${\bigcup_n A^\alpha_n=(\alpha,\lambda)\in {\mathcal U},}$ there is some ${n_\alpha}$ such that ${A^\alpha_{n_\alpha}\in {\mathcal U}}$ for all ${\alpha.}$

For each ${n<\omega}$ let

$\displaystyle A_n=\{\alpha: A^\alpha_n\in {\mathcal U}\}.$

Since ${\bigcup_n A_n=\lambda}$ by the above, there is some ${n}$ such that ${A_n\in {\mathcal U}.}$

In particular, ${A_n}$ is unbounded in ${\lambda,}$ and we claim that ${A_n}$ witnesses the covering property for ${{\mathcal D}:}$ Let ${X\subset A_n}$ be countable. Then (by definition of ${A_n}$) ${A^\alpha_n\in {\mathcal U}}$ for all ${\alpha\in X,}$ so ${\bigcap_{\alpha\in X}A^\alpha_n\in {\mathcal U}.}$ In particular, this set is nonempty. Let ${\beta\in\bigcap_{\alpha\in X}A^\alpha_n.}$ Then (by definition of ${A^\alpha_n}$) ${\beta\ge\sup(X)}$ and ${\alpha\in K(n,\beta)}$ for all ${\alpha\in X.}$ Hence, ${X\subseteq K(n,\alpha).}$ $\Box$

Proof of Theorem 19: Let ${\kappa}$ be strongly compact. We show that ${\lambda^{{\rm cf}(\lambda)}=2^{{\rm cf}(\lambda)}+\lambda^+}$ for all ${\lambda\ge\kappa}$ by induction on ${\lambda.}$ The result is clear if ${\lambda}$ is regular, and follows by induction if ${{\rm cf}(\lambda)>\omega,}$ by Silver’s theorem. Assume now that ${{\rm cf}(\lambda)=\omega.}$ Let ${{\mathcal D}}$ be a nontrivial precovering matrix for ${\lambda^+}$ with ${\beta_{\mathcal D}=\lambda.}$

By ${{\sf CP}(\lambda^+)}$ and Corollary 23, ${\lambda^{\aleph_0}\le(\lambda^+)^{\aleph_0}\le\lambda^+\sup_{\tau<\lambda}\tau^{\aleph_0}=\lambda^+}$ since ${\tau^{\aleph_0}\le\tau^+}$ for all ${\tau\in[\kappa,\lambda),}$ by induction (by the local nature of the proof of Lemma 18 in lecture II.5). The result follows immediately. ${\Box}$

Strongly compact cardinals are a bit of an oddity in the large cardinal hierarchy in that they are of really high consistency strength (for example, if ${\kappa}$ is ${\kappa^+}$-compact, then there are inner models of ${{\sf ZF}+{\sf AD}_{\mathbb R}}$ that contain all the reals; this is significantly above in consistency strength than, say, the existence of a proper class of strong cardinals). However, in terms of size they need not be too large; in fact, Magidor showed that it is consistent that the first strongly compact is also the first measurable cardinal.

In contrast with the situation from strong cardinals shown last lecture, the following is still open:

Question 25 (Woodin) Assume that ${\kappa}$ is strongly compact and that ${{\sf GCH}}$ holds below ${\kappa.}$ Does it follow that it holds everywhere?

For more on infinitary logic, see Jon Barwise, Solomon Feferman, eds., Model-theoretic logics, Springer (1985). Of particular interest is Part C (Infinitary languages), 269-441, and even more specifically, the two articles Mark Nadel, ${{\mathcal L}_{\omega_1,\omega}}$ and admissible fragments, 271–316, and Max Dickmann, Large infinitary languages, 317–363.

For the general theory of strongly compact cardinals, see Akihiro Kanamori, The Higher Infinite, Springer (1994) and Jussi Ketonen, Strong compactness and other cardinal sins, Annals of Mathematical Logic 5 (1972), 47–76. For more on the covering property, see Matteo Viale, A family of covering properties, Mathematical Research Letters, 15 (2) (2008), 221–238.

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### One Response to 580 -Cardinal arithmetic (11)

1. […] One can also prove Rado’s selection principle as a consequence of the compactness of first order logic; for example, it is easy to give a proof using the ultrapower construction as in the proof of 2 implies 6 of Theorem 8 in lecture II.11. […]