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Math Help - Revised: A sequence of non-empty, compact, nested sets converges to its intersection

  1. #1
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    Revised: A sequence of non-empty, compact, nested sets converges to its intersection

    Proposition 2.4.7 *  S is a metric space with metric  \rho .     {A_n} is a sequence of descending, non-empty, compact sets.

    Then for  \epsilon > 0,\     lim A_n = A = \bigcap A_n in the Hausdorff sense.

    In that sense, one must show that

    (1)  A \subseteq N_\epsilon (A_n) and (2)  A_n \subseteq N_\epsilon (A )

    Prof Edgar shows (1) clearly, but I have difficulty with his demonstration of (2). I shall try to restate it:
    By the  \epsilon neighborhood of  A , he means:

     N_\epsilon (A) = \{y: \rho (x,y) < \epsilon \ for \  some \ x\in A \} .  N_\epsilon (A) is an open set, he states, which is straightforward.

    He then states that the family  \{N_\epsilon (A) \}\bigcup \{S \backslash A_n : n \in N \}

    is an open cover of  A_1 . Herein  S\backslash A_n is the complement of  A_n . It seems to me that the family described covers not only  A_1 but the entire metric space  S , all points of the intersection and all points "outside" the intersection . So, yes, it does cover  A_1 . The author then continues that there is therefore a finite subcover of  A_1 , since A_1 is compact by hypothesis. He states that:

     N_\epsilon (A) \bigcup \{S\backslash A_n\} \supseteq A_1 for all  n \geq M , where M is some positive integer is a finite sub-cover. The author concludes with

    Therefore,  A \sqsubseteq N_\epsilon (A_n) , and (2), above, is satisfied.

    My difficulty: The expression for the finite sub-cover does not seem finite to me because it is a family for all n > M.

    Can anyone explain?

















    *"Measure, Topology, and Fractal Geometry", by Gerald A. Edgar (Springer Verlag, New York, 2000) p.68
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  2. #2
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    Re: Revised: A sequence of non-empty, compact, nested sets converges to its intersec

    You don't seem to have stated what it is you want to prove!
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    Re: Revised: A sequence of non-empty, compact, nested sets converges to its intersec

    Well, for starters, how about the last and next to last sentences of my post? Could you help with those?
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    Re: Revised: A sequence of non-empty, compact, nested sets converges to its intersec

    Now that I have restated more clearly my question with LaTex, where is Mr "Plato" ?
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    Re: Revised: A sequence of non-empty, compact, nested sets converges to its intersec

    My own solution is that the one offered by Prof Edgar does not make sense. Maybe his graduate assistant screwed up and got away with it. Maybe I should not have mentioned Edgar's name in my posting. That mention might have discouraged a critical review . In any case, I believe I have proved Proposition 2.4.7 by recalling that in taking the limit of the sequence, the index goes to "infinity". With that in mind, I skipped all the stuff about the Heine-Borel theorem and simply looked at the limit in the Hausdorff metric. So, this posting's question has been answered - by the poster! Thanks for all the help.
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  6. #6
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    Re: Revised: A sequence of non-empty, compact, nested sets converges to its intersec

    here is how i look at it:

    since A1 is compact, and we have exhibited an open cover of A1, there is some finite sub-cover. note that we can take Nε(A) to be one member of the subcover, so we only need the complements of finitely many of the Aj to "meet up with Nε(A)" (the rest of the infinite family lies inside the set Nε(A)).

    if the book really DOES say: n ≥ M, instead of n ≤ M, then that is a typo.
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  7. #7
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    Re: Revised: A sequence of non-empty, compact, nested sets converges to its intersec

    It really, really does read as I reported; also, the new edition of 2008 reads the same. Thus, it is not just a typo but carelessness, if in the author's mind he meant less than or equal. Furthermore, the sub-cover, if you are correct about the "typo", omits a "ring" of points in sets of index greater than M+1 and without the intersection of all - unless M is unbounded - because the indexed sets are complements of the given sequence's sets.

    My view is far simpler: lim A_n = Intersection of all A_n because the members are descending. The second part of the Hausdorff requirement is thus satisfied. The limit has to be taken in any case to show that the Hausdorff "distance" is zero.
    Last edited by goedelite; September 14th 2012 at 03:11 PM.
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