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        riddles >> putnam exam (pure math) >> U-limit of a sequence
        
(Message started by: 0.999... on Apr 29^th, 2012, 10:02am)

Title: U-limit of a sequence
Post by 0.999... on Apr 29^th, 2012, 10:02am

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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbn.gif and let <a_n> be a bounded sequence of countably infinitely many real numbers. Prove that there exists a unique U-limit a such that for every http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif>0, the set {n: |a_n-a| < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif 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.

[hideb]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, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif) to subsets of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbn.gif: A(c, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif) = {n: |a_n-c| < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif}, where c is a cluster point and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif > 0. For the sake of contradiction, suppose there is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif > 0 such that A(c, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif U for all cluster points c. Since A(c, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/delta.gif) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subset.gif A(c, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif) when http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/delta.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif, there exists r such that the property holds for all http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif < r and for no http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif > r. Furthermore, A(c,r) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif U; otherwise, U would be a principal ultrafilter. Choose http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif > r such that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif < 3r. The set {n: r <= |a_n-c| < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif} is in U, and so is infinite, which implies that there is a cluster point c' and 0< http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/delta.gif <= (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif-r)/2 < r such that A(c', http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/delta.gif) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U. This is a contradiction.
As http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif > 0 gets smaller, the cluster points c such that A(c, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif U are required to be closer to to each other (|c-c'|/2 <= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif). Thus, there is a unique cluster point which satisfies the U-limit criterion. Of course, no other point x does, since for some http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gifthe set {n: |a_n-x|< http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif} is finite.
Case II: U contains a singleton set {m}. It is clear that {n: |a_n-a_n| < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varepsilon.gif} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supset.gif {m}, and thus is in U.[/hideb]