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02-convergence - Lecture 2 2 The Concept of Convergence...

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Lecture 2 2. The Concept of Convergence: Ultrafilters and Nets In this lecture we discuss two points of view on the notion of convergence. The first one employs a set theoretical concept, which turns out to be technically useful. Definition. Suppose X is a fixed (non-empty) set. A filter in X is a (non- empty) family F of non-empty subsets of X which has the property 2 : (f) Whenever F and G belong to F , it follows that F G also belongs to F . What is important here is that all the sets in the filter are assumed to be non- empty . The set of all filters in X can be ordered by inclusion. A simple application of Zorn’s Lemma yields: For each filter F there exists at least one maximal filter U with U F . Maximal filters will be called ultrafilters . An interesting feature of ultrafilters is given by the following: Lemma 2.1. Let X be a non-empty set, and let U be a filter on X . The following are equivalent: (i) U is an ultrafilter. (ii) For any subsets A X , it follows that either A or X A belongs to U , but not both! Proof. ( i ) ( ii ). Assume U is an ultrafilter. First remark that X always belongs to U . (Otherwise, if X does not belong to U , the family U ∪ { X } will be obviously a new filter which will contradict the maximality of U ). Let us assume that A is non-empty and it does not belong to U . This means that the family M = U ∪ { A U | U U } is no longer a filter (otherwise, the maximality of U will be contradicted). Note that if F and G belong to M , then automatically F G belongs to M . This means that the only thing that can prevent M from being a filter, must be the fact that one of the sets in M is empty. That is, there is some set V U such that A V = . In other words, V X A . But then, it follows that for any U U we have U ( X A ) U V = and then the set N = U ∪ { U ( X A ) | U U } will be a filter. By maximality, it follows that N = U , in particular, X A belongs to U . It is obvious that A and X A cannot simultaneously belong to U , because this will force = A ( X A ) to belong to U . 2 Some textbooks may use a slightly different definition. 9
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10 LECTURE 2 ( ii ) ( i ). Assume property (ii) holds, but U is not maximal, which means that there exists some ultrafilter V with V U . Pick then some set A V U . Since A U , by (ii) we must have X A U . This would force both A and X A to belong to V , which is impossible. Exercise 1 . Let U be an ultrafilter on X , and let A U . Prove that the collection U A = { U A : U U } is an ultrafilter on A . Remark 2.1. If U is an ultrafilter on X , and A U , then U contains all sets B with A B X . Indeed, if we start with such a B , then by the above result, either B U or X B U . Notice however that in the case X B U we would get U ( X B ) A = , which is impossible. Therefore B must belong to U .
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