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Topic: Ulimit of a sequence (Read 6706 times) 

0.999...
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Ulimit of a sequence
« on: Apr 29^{th}, 2012, 10:02am » 
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I am currently reading Set Theory by Thomas Jech (3rd Millennium ed.), and did not like my solution to exercise 7.6: Let U be an ultrafilter on and let <a_{n}> be a bounded sequence of countably infinitely many real numbers. Prove that there exists a unique Ulimit a such that for every >0, the set {n: a_{n}a < } U. Here's my proof. If anyone could either lead me toward something more elegant (e.g. a way to link both cases, perhaps a different line of reasoning in Case I would do that) or confirm that this is the best I can do, I would be grateful. hidden:  Case I: U contains only infinite sets (i.e. U is nonprincipal). Since the given sequence is bounded, it has one or more cluster points. Define a mapping from pairs (c, ) to subsets of : A(c, ) = {n: a_{n}c < }, where c is a cluster point and > 0. For the sake of contradiction, suppose there is > 0 such that A(c, ) U for all cluster points c. Since A(c, ) A(c, ) when < , there exists r such that the property holds for all < r and for no > r. Furthermore, A(c,r) U; otherwise, U would be a principal ultrafilter. Choose > r such that < 3r. The set {n: r <= a_{n}c < } is in U, and so is infinite, which implies that there is a cluster point c' and 0< <= (r)/2 < r such that A(c', ) U. This is a contradiction. As > 0 gets smaller, the cluster points c such that A(c, ) U are required to be closer to to each other (cc'/2 <= ). Thus, there is a unique cluster point which satisfies the Ulimit criterion. Of course, no other point x does, since for some the set {n: a_{n}x< } is finite. Case II: U contains a singleton set {m}. It is clear that {n: a_{n}a_{n} < } {m}, and thus is in U. 

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