## 580 -Cardinal arithmetic (8)

4. Large cardinals and cardinal arithmetic

In section 3 we saw how the powers of singular cardinals (or, at least, of singulars of uncountable cofinality) satisfy strong restrictions. Here I show that similar restrictions hold at large cardinals. There is much more than one could say about this topic, and the results I present should be seen much more like an invitation than a full story. Also, for lack of time, I won’t motivate the large cardinals we will discuss. (In the ideal world, one should probably say a few words about one’s beliefs in large cardinals, since their existence and even their consistency goes beyond what can be done in the standard system ${{\sf ZFC}.}$ I’ll however take their existence for granted, and proceed from there.)

1. Measurable cardinals

Definition 1 ${\kappa}$ is a measurable cardinal iff ${\kappa>\omega}$ and there is a nonprincipal ${\kappa}$-complete ultrafilter over ${\kappa.}$

Recall that ${{\mathcal U}\subseteq{\mathcal P}(X)}$ is an ultrafilter (over ${X,}$ or on ${{\mathcal P}(X)}$) iff the following hold:

1. ${\forall Y\subseteq Z\subseteq X,}$ if ${y\in{\mathcal U},}$ then ${Z\in{\mathcal U}.}$
2. If ${Y,Z\in{\mathcal U},}$ then ${Y\cap Z\in{\mathcal U}.}$
3. ${\emptyset\notin{\mathcal U}.}$
4. For all ${Y\subseteq X,}$ either ${Y\in{\mathcal U}}$ or else ${X\setminus Y\in{\mathcal U}.}$

Conditions 1–3 state that ${{\mathcal U}}$ is a filter. Notice that by 2 and 3, ${{\mathcal U}}$ has the finite intersection property. Conversely, any collection of subsets of ${X}$ with the finite intersection property generates in a natural way a filter.

Condition 4 is the ultrafilter condition. Equivalently, one can request that ${{\mathcal U}}$ is maximal under containment (subject to being a filter). The equivalence is an easy consequence of Zorn’s lemma.

An ultrafilter ${{\mathcal U}}$ over ${X}$ is nonprincipal iff ${\{x\}\notin{\mathcal U}}$ for any ${x\in X.}$ Otherwise, it is principal. Clearly, any ultrafilter over a finite set is principal. On the other hand, any infinite set admits a nonprincipal ultrafilter, in fact, since the collection of subsets of ${X}$ whose complements have size strictly smaller than ${X}$ has the finite intersection property, there are uniform ultrafilters over ${X}$, i.e., ultrafilters all of whose elements have the same size as ${X.}$

As before, an ultrafilter ${{\mathcal U}}$ is ${\lambda}$-complete iff any intersection of fewer than ${\lambda}$ many members of ${{\mathcal U}}$ is in ${{\mathcal U}.}$ If ${\lambda=\omega_1}$ we also say that ${{\mathcal U}}$ is ${\sigma}$-complete. Equivalently, by considering complements, ${{\mathcal U}}$ is ${\lambda}$-complete iff whenever a union of fewer than ${\lambda}$ many sets is in ${{\mathcal U},}$ then at least one of the sets in the union is in ${{\mathcal U}.}$

Clearly, no nonprincipal ultrafilter over ${\kappa}$ can be ${\kappa^+}$-complete. On the other hand, normality is a stronger requirement than ${\kappa}$-completeness.

Homework problem 10. Assume that ${{\mathcal U}}$ is a ${\sigma}$-complete nonprincipal ultrafilter over some set ${X.}$ Let ${\kappa}$ be the completeness of ${{\mathcal U},}$ i.e., ${\kappa}$ is smallest such that ${{\mathcal U}}$ is not ${\kappa^+}$-complete. Show that ${\kappa}$ is measurable.

Ulam introduced measurable cardinals by working on a question on Banach. The problem is whether there could be a nontrivial measure space ${(X,{\mathcal P}(X),\mu).}$ In detail, ${\mu:{\mathcal P}(X)\rightarrow[0,\infty],}$ and ${\mu(\bigcup_{n\in\omega}A_n)=\sum_n\mu(A_n)}$ for any pairwise disjoint sequence of subsets of ${X.}$

The nontriviality condition means several things. First of all, ${\mu(\emptyset)=0}$ (otherwise, we must have ${\mu(E)=\infty}$ for all ${E\subseteq X.}$)

But this is not enough. For consider any set ${X,}$ any function ${f:X\rightarrow[0,\infty)}$ and any (possibly empty, or perhaps nonproper) ${\sigma}$-ideal ${{\mathcal I}}$ of subsets of ${X.}$ Then we can set ${\mu(E)=\sum_{x\in E}f(x)}$ if ${E\in{\mathcal I}}$ and ${\mu(E)=\infty}$ if ${E\notin{\mathcal I},}$ and certainly ${(X,{\mathcal P}(X),\mu)}$ is a measure space. We then say that a measure space ${(X,{\mathcal P}(X),\mu)}$ is nontrivial iff it does not arise in this fashion.

Assume that ${(X,{\mathcal P}(X),\mu)}$ is nontrivial. Let ${f(x)=\mu(\{x\})}$ if ${\mu(\{x\})<\infty}$ and ${f(x)=0}$ otherwise. Let ${{\mathcal I}}$ be the ${\sigma}$-ideal generated by the subsets of ${X}$ of finite measure, and let ${\mu'}$ be the trivial measure over ${X}$ generated by ${f}$ and ${{\mathcal I}.}$ Then ${\mu'(A)\le\mu(A)}$ for all ${A\subseteq X}$ but ${\mu'\ne\mu,}$ since we are assuming that ${\mu}$ is nontrivial. It follows that there is some ${A}$ such that ${\mu'(A)<\mu(A).}$ It must then be the case that ${A\in{\mathcal I},}$ and therefore (by definition of ${{\mathcal I}}$) there is an increasing sequence ${(A_n:n<\omega)}$ of subsets of ${X}$ such that ${A=\bigcup_n A_n}$ and ${\mu(A_n)<\infty}$ for all ${n.}$ Necessarily, ${\mu'(A_n)<\mu(A_n)}$ for some ${n,}$ and we can define a measure ${\mu'':{\mathcal P}(A_n)\rightarrow[0,\infty]}$ by setting ${\mu''(E)=\mu(E)-\mu'(E).}$

Notice that in fact ${0<\mu''(A_n)<\infty}$ and ${\mu''(\{x\})=0}$ for all ${x\in A_n.}$ By normalizing, we may as well assume that ${\mu''(A_n)=1.}$

This shows that the measure problem has a solution iff there is a nontrivial probability space ${(Y,{\mathcal P}(Y),\lambda)}$ (so in particular ${\lambda(\emptyset)=0}$ and ${\lambda(Y)=1}$) such that ${\lambda(\{y\})=0}$ for all ${y\in Y,}$ and this is the way in which the question is most commonly posed in the literature.

Let ${{\rm add}(\lambda)}$ be the additivity of the measure ${\lambda,}$ i.e., the least cardinal ${\kappa}$ such that the measure of any disjoint union of fewer than ${\kappa}$ many disjoint subsets of ${Y}$ is the sum of the measures of the sets in the union, understanding that ${{\rm add}(\lambda)=\infty}$ if there is no such cardinal.

Let ${{\mathcal N}_\lambda}$ be the collection of subsets of ${Y}$ of measure zero. Then ${{\mathcal N}_\lambda}$ is an ideal. For any ideal ${{\mathcal I}}$ over a set ${Z}$ we can define ${{\rm add}({\mathcal I})}$ as the least cardinality of a collection of sets in ${{\mathcal I}}$ whose union is not in ${{\mathcal I},}$ understanding that ${{\rm add}({\mathcal I})=\infty}$ if there is no such cardinal.

If ${(Y,{\mathcal P}(Y),\lambda)}$ is a probability space, then either ${{\rm add}(\lambda)={\rm add}({\mathcal N}_\lambda)=\infty}$ and the space is trivial, or else ${{\rm add}(\lambda)={\rm add}({\mathcal N}_\lambda)}$ is some cardinal ${\kappa.}$ In this latter case, there must be a sequence ${(E_\alpha:\alpha<\kappa)}$ of nonempty pairwise disjoint sets in ${{\mathcal N}_\lambda}$ whose union ${E}$ is not in ${{\mathcal N}_\lambda.}$

In this case we can define a measure ${\mu}$ over ${\kappa}$ as follows: For ${x\in E}$ let ${\pi(x)\in\kappa}$ be the ordinal such that ${x\in E_{\pi(x)}.}$ Let ${X\subseteq\kappa.}$ Then set ${\displaystyle \mu(X)=\frac{\lambda(\{x\in E:\pi(x)\in X\})}{\lambda(E)}.}$ Then ${(\kappa,{\mathcal P}(\kappa),\mu)}$ is a probability space, ${\mu(\{\alpha\})=0}$ for all ${\alpha\in\kappa,}$ and ${\mu}$ is ${\kappa}$-additive.

Ulam noticed that in this case an interesting dichotomy happens: Either ${\mu}$ is nonatomic, meaning that for for all ${X\subseteq\kappa,}$ if ${\mu(X)>0}$ then there is some ${Y\subseteq X}$ such that ${0<\mu(Y)<\mu(X),}$ in which case ${\kappa\le{\mathfrak c}}$ and we say that ${\kappa}$ is (atomlessly) real-valued measurable, or else ${\mu}$ admits an atom ${X}$, that necessarily must have size ${\kappa.}$ In this case, for all ${Y\subseteq X,}$ either ${\mu(Y)=0}$ or ${\mu(Y)=\mu(X).}$ By renormalizing and composing with a bijection, this gives us a probability space ${(\kappa,{\mathcal P}(\kappa),\nu)}$ where ${\nu:{\mathcal P}(\kappa)\rightarrow\{0,1\},}$ ${\nu(\{\alpha\})=0}$ for all ${\alpha<\kappa,}$ and ${\nu}$ is ${\kappa}$-complete. Letting ${{\mathcal U}=\{Y\subseteq \kappa:\nu(Y)=1\},}$ ${{\mathcal U}}$ is a ${\kappa}$-complete nonprincipal ultrafilter over ${\kappa,}$ i.e., ${\kappa}$ is measurable.

Conversely, if ${\kappa}$ is measurable and ${{\mathcal U}}$ is a ${\kappa}$-complete nonprincipal ultrafilter over ${\kappa,}$ then ${\chi_{\mathcal U},}$ the characteristic function of ${{\mathcal U},}$ is an example of such a probability measure ${\nu.}$

Theorem 2 (Tarski-Ulam) If ${\kappa}$ is measurable or real-valued measurable, then it is weakly inaccessible. If ${\kappa}$ is in fact measurable, then it is strong limit and therefore (strongly) inaccessible.

Proof: Fix a witnessing probability ${\nu,}$ so ${\nu}$ is ${\kappa}$-additive, singletons are null, and ${\nu}$ is either nonatomic or bivalued.

${\kappa}$ is regular. Because, by ${\kappa}$-completeness, every bounded subset of ${\kappa,}$ in particular every ordinal below ${\kappa,}$ is null, and therefore ${\kappa}$ cannot be the union of fewer than ${\kappa}$ many ordinals below ${\kappa,}$ since any such union has measure zero.

${\kappa}$ is limit. For this, proceed by contradiction, assuming that ${\kappa=\lambda^+,}$ and consider a ${\lambda\times\lambda^+}$ Ulam matrix ${(A^\iota_\alpha:\iota<\lambda,\alpha<\kappa),}$ see Definition 12 in lecture II.5. Arguing just as in the proof of Theorem 13 there, notice that ${\bigcup_{\iota<\lambda}A^\iota_\alpha=(\alpha,\kappa)}$ has measure 1 for all ${\alpha<\kappa,}$ and by ${\kappa}$-completeness of ${\nu}$ there must be some ${\iota_\alpha}$ such that ${A^{\iota_\alpha}_\alpha}$ has positive measure.

By the pigeonhole principle, there is some fixed ${\iota}$ such that ${\iota_\alpha=\iota}$ for ${\kappa}$ many distinct values of ${\alpha.}$ Recall now that ${A^\iota_\alpha\cap A^\iota_\beta=\emptyset}$ whenever ${\alpha\ne\beta.}$ By the pigeonhole principle again, since ${\kappa>\omega,}$ there is some nonzero ${n\in\omega}$ such that ${\nu(A^\iota_\alpha)>1/n}$ for uncountably many ordinals ${\alpha.}$ This is clearly a contradiction.

Assume now that ${\kappa}$ is measurable. Then ${\kappa}$ is strong limit. Let ${{\mathcal U}}$ be the ultrafilter corresponding to ${\nu.}$ Let ${\rho<\kappa}$ and suppose that ${\{X_f:f\in{}^\rho2\}\subseteq{\mathcal P}(\kappa)}$ and ${\bigcup_f X_f\in{\mathcal U}.}$ It is enough to see that ${X_f\in{\mathcal U}}$ for some ${f,}$ since this shows that ${{\mathcal U}}$ is ${(2^\rho)^+}$-complete, so ${2^\rho<\kappa.}$ Notice that for each ${\alpha<\rho}$ there is a (least) ${\epsilon\in 2}$ such that ${\bigcup_{f(\alpha)=\epsilon}X_f\in{\mathcal U}.}$ Let ${h:\rho\rightarrow2}$ be the function that to each ${\alpha<\rho}$ assigns the corresponding ${\epsilon.}$ Then ${X_h=\bigcap_{\alpha<\rho}\bigcup_ {f(\alpha)=h(\alpha)} X_f\in{\mathcal U},}$ since ${{\mathcal U}}$ is at least ${\rho^+}$-complete. $\Box$

Homework problem 11. Show that if ${\kappa}$ is real-valued measurable then ${\kappa\le{\mathfrak c},}$ by showing that ${{\rm add}(\nu)\le{\mathfrak c},}$ where ${\nu}$ is as above. For this, begin by showing that for any ${X\subseteq\kappa}$ there is some ${Y\subseteq X}$ such that ${\nu(Y)=\nu(X)/2.}$

Of course, Theorem 2 implies that the existence of measurable cardinals is not provable in ${{\sf ZFC},}$ since if ${\kappa}$ is strongly inaccessible then ${V_\kappa\models{\sf ZFC}.}$

Whether all inaccessible cardinals are measurable is a different story. In fact, if ${\kappa}$ is measurable, then it is the ${\kappa}$-th inaccessible cardinal, and one can show much stronger results showing how vast the increment in strength is between both notions. This we will do by means of the ultrapower construction, a key idea in the study of large cardinals.

The reduction indicated above of the measure problem to the question of the existence of a ${\kappa}$-complete probability space ${(\kappa,{\mathcal P}(\kappa),\nu)}$ where singletons are null is probably classical. I followed the approach of David Fremlin, Real-valued measurable cardinals, in Set Theory of the reals, Haim Judah, ed., Israel Mathematical Conference Proceedings 6, Bar-Ilan University (1993), 151–304.

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### One Response to 580 -Cardinal arithmetic (8)

1. […] The study of ultrapowers originates in model theory, although it has found applications both in algebra and in analysis. However, it is accurate to say that it is mainly exploited in set theory. Here I present the basic idea, showing its close connection to the study of measurable cardinals, defined last lecture. […]