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 I’ll however take their existence for granted, and proceed from there.)
1. Measurable cardinals
Definition 1 is a measurable cardinal iff and there is a nonprincipal -complete ultrafilter over
Recall that is an ultrafilter (over or on ) iff the following hold:
- if then
- If then
- For all either or else
Conditions 1–3 state that is a filter. Notice that by 2 and 3, has the finite intersection property. Conversely, any collection of subsets of with the finite intersection property generates in a natural way a filter.
Condition 4 is the ultrafilter condition. Equivalently, one can request that is maximal under containment (subject to being a filter). The equivalence is an easy consequence of Zorn’s lemma.
An ultrafilter over is nonprincipal iff for any 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 whose complements have size strictly smaller than has the finite intersection property, there are uniform ultrafilters over , i.e., ultrafilters all of whose elements have the same size as
As before, an ultrafilter is -complete iff any intersection of fewer than many members of is in If we also say that is -complete. Equivalently, by considering complements, is -complete iff whenever a union of fewer than many sets is in then at least one of the sets in the union is in
Clearly, no nonprincipal ultrafilter over can be -complete. On the other hand, normality is a stronger requirement than -completeness.
Homework problem 10. Assume that is a -complete nonprincipal ultrafilter over some set Let be the completeness of i.e., is smallest such that is not -complete. Show that is measurable.
Ulam introduced measurable cardinals by working on a question on Banach. The problem is whether there could be a nontrivial measure space In detail, and for any pairwise disjoint sequence of subsets of
The nontriviality condition means several things. First of all, (otherwise, we must have for all )
But this is not enough. For consider any set any function and any (possibly empty, or perhaps nonproper) -ideal of subsets of Then we can set if and if and certainly is a measure space. We then say that a measure space is nontrivial iff it does not arise in this fashion.
Assume that is nontrivial. Let if and otherwise. Let be the -ideal generated by the subsets of of finite measure, and let be the trivial measure over generated by and Then for all but since we are assuming that is nontrivial. It follows that there is some such that It must then be the case that and therefore (by definition of ) there is an increasing sequence of subsets of such that and for all Necessarily, for some and we can define a measure by setting
Notice that in fact and for all By normalizing, we may as well assume that
This shows that the measure problem has a solution iff there is a nontrivial probability space (so in particular and ) such that for all and this is the way in which the question is most commonly posed in the literature.
Let be the additivity of the measure i.e., the least cardinal such that the measure of any disjoint union of fewer than many disjoint subsets of is the sum of the measures of the sets in the union, understanding that if there is no such cardinal.
Let be the collection of subsets of of measure zero. Then is an ideal. For any ideal over a set we can define as the least cardinality of a collection of sets in whose union is not in understanding that if there is no such cardinal.
If is a probability space, then either and the space is trivial, or else is some cardinal In this latter case, there must be a sequence of nonempty pairwise disjoint sets in whose union is not in
In this case we can define a measure over as follows: For let be the ordinal such that Let Then set Then is a probability space, for all and is -additive.
Ulam noticed that in this case an interesting dichotomy happens: Either is nonatomic, meaning that for for all if then there is some such that in which case and we say that is (atomlessly) real-valued measurable, or else admits an atom , that necessarily must have size In this case, for all either or By renormalizing and composing with a bijection, this gives us a probability space where for all and is -complete. Letting is a -complete nonprincipal ultrafilter over i.e., is measurable.
Conversely, if is measurable and is a -complete nonprincipal ultrafilter over then the characteristic function of is an example of such a probability measure
Proof: Fix a witnessing probability so is -additive, singletons are null, and is either nonatomic or bivalued.
is regular. Because, by -completeness, every bounded subset of in particular every ordinal below is null, and therefore cannot be the union of fewer than many ordinals below since any such union has measure zero.
is limit. For 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 -completeness of there must be some such that has positive measure.
By the pigeonhole principle, there is some fixed such that for many distinct values of Recall now that whenever By the pigeonhole principle again, since there is some nonzero such that for uncountably many ordinals This is clearly a contradiction.
Assume now that is measurable. Then is strong limit. Let be the ultrafilter corresponding to Let and suppose that and It is enough to see that for some since this shows that is -complete, so Notice that for each there is a (least) such that Let be the function that to each assigns the corresponding Then since is at least -complete.
Homework problem 11. Show that if is real-valued measurable then by showing that where is as above. For this, begin by showing that for any there is some such that
Of course, Theorem 2 implies that the existence of measurable cardinals is not provable in since if is strongly inaccessible then
Whether all inaccessible cardinals are measurable is a different story. In fact, if is measurable, then it is the -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 -complete probability space 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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