By Paul C. Eklof

ISBN-10: 0444885021

ISBN-13: 9780444885029

This can be a longer therapy of the set-theoretic options that have reworked the examine of abelian crew and module thought over the past 15 years. a part of the booklet is new paintings which doesn't seem somewhere else in any shape. furthermore, a wide physique of fabric which has seemed formerly (in scattered and occasionally inaccessible magazine articles) has been widely remodeled and in lots of circumstances given new and superior proofs. The set concept required is punctiliously built with algebraists in brain, and the independence effects are derived from explicitly acknowledged axioms. The booklet comprises workouts and a consultant to the literature and is appropriate to be used in graduate classes or seminars, in addition to being of curiosity to researchers in algebra and good judgment.

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**Additional resources for Almost Free Modules: Set-Theoretic Methods**

**Example text**

If there is a measurable cardinal, then there is an w-measurable cardinal K such that there are exactly K wl-complete ultrafilters on K . ] 12. Suppose K is a measurable cardinal. (i) For any subset X of K of cardinality K there is a K-complete ultrafilter on K which contains X . (ii) There is a subset S of P ( K )such that S has cardinality 2" and for every X # Y in S, IX n YI < K [cf. 51. (iii) There are 2" different 6-complete ultrafilters on K .

Then we turn to some notions which are, perhaps, less familiar. In section 2, we discuss filters and ultrafilters and use these to define some large cardinals, including measurable cardinals. In section 3, we define ultraproducts, and encounter for the first time the key notion of a %-freemodule, which will play a major role in the rest of the book. In section 4, we introduce a couple of other definitions which will be very important, namely that of a stationary set and of a %-filtration. Finally, we conclude, in section 5 , with a short discussion of game theory and of trees.

There is no filter on I which properly contains D. 2 Filters and large cardinals 23 PROOF. If D is an ultrafilter and D’is a filter on I containing D , then every X in D’ must belong to D,because otherwise, I \X would belong to D and then we would have 0 = X n ( I \ X ) E D’, which contradicts the definition of a filter. Conversely, suppose that D is a maximal filter on I . 2, contained in a filter D‘. But then the maximality of D implies that D’= D,so Y belongs to D. For any X I , it is easy to verify that either { X } U D or { I \ X} U D has FIP.

### Almost Free Modules: Set-Theoretic Methods by Paul C. Eklof

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