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Can anybody help me to understand this proof?

  1. Sep 30, 2012 #1
    Let [itex](F_n)[/itex] be a decreasing sequence [i.e., F_1 contains F2 which contains F_3...etc] of closed bounded nonempty sets in [itex]R^k[/itex]. Then [itex]F = \cap^{\infty}_{m=1} F_n[/itex] is also closed, bounded and nonempty.

    The proof (from our textbook) says:

    Clearly F is closed and bounded. It is the nonemptiness that needs proving! For each n, select an element [itex](x_n)[/itex] in [itex]F_n[/itex]. By the Bolzano-Weierstrass theorem 13.5, a subsequence [itex](x_n_m})^{\infty}_{m=1}[/itex] of [itex](x_n)[/itex] converges to some element [itex]x_0[/itex] in [itex]R^k[/itex]. To show [itex]x_0 \in F[/itex], it suffices to show [itex]x_0 \in F_n_0[/itex] with [itex]n_0[/itex] fixed.

    Why does it suffice to show that?

    If [itex]m \geq n_0[/itex], then [itex]n_m \geq n_0[/itex], so [itex]x_n_m \in F_n_m \subseteq F_n_0[/itex].

    But what if m<n_0?

    Hence the sequence [itex]{x_n_m}^{\infty}_{m=1}[/itex] consists of points in [itex]F_n_0[/itex] and converges to [itex]x_0[/itex]. Thus [itex]x_0[/itex] belongs to [itex]F_n_0[/itex] by (b) of proposition 13.9 (which says “The set E is closed if and only if it contains the limit ofevery convergent sequence of points in E.)

    Thanks in advance
     
    Last edited: Sep 30, 2012
  2. jcsd
  3. Sep 30, 2012 #2
    As you could see, your post is very hard to understand. This wouldn't have happened had you used the "Preview Post" option before posting it.

    There is no [Latex] option here but itex enclosed in square parentheses. Please do fix your post.

    DonAntonio
     
  4. Sep 30, 2012 #3
    But I don't know how to use itex...??
     
  5. Sep 30, 2012 #4
    Inside [itex] [/itex] you have to write in latex language.
     
  6. Sep 30, 2012 #5
    But I did do that...it still didn't change...
     
  7. Oct 1, 2012 #6

    micromass

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    Last edited: Oct 1, 2012
  8. Oct 2, 2012 #7

    Erland

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    I agree that the post needs to be fixed, but I think I understand it as it is:


    i agree that this is confusing. It is poorly expressed. The author must mean something like this:

    To show x_0 in F, it suffices to show that for each fixed n_0: x_0 in F_n_0.
    This doesn't matter. The convergence behaviour of a sequence does not depend upon its first finite number of elements. If x1,x2,x3,x4,x5.... converges to x, so does e.g. y1,y2,y3,x4,x5....
     
  9. Oct 2, 2012 #8
    Thanks for answering, but I'm not sure if I understood the first answer. What exactly does "for each fixed n_0" mean?
     
  10. Oct 2, 2012 #9

    Erland

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    That this should be proved for every number n_0 (which will be fixed in the succeeding part of the proof), not just that there is one n_0 for which this holds.
     
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