## 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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