1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can't understand this Bolzano-Weierstrass proof

  1. Jul 28, 2012 #1
    I can't understand the Bolzano-Weierstrass theorem's proof from here on page 2:
    http://www.u.arizona.edu/~mwalker/econ519/Econ519LectureNotes/Bolzano-Weierstrass.pdf

    I'll type out the proof and cease typing it at the part that I don't understand.

    The Bolzano-Weierstrass Theorem: Every bounded sequence of real numbers has a
    convergent subsequence.

    Proof: Let [itex]\{x_n\}[/itex] be a bounded sequence and without loss of generality assume that every term of the sequence lies in the interval [itex][0,1][/itex]. Divide [itex][0,1][/itex] into two intervals, [itex][0,0.5][/itex] and [itex][0.5,1][/itex]. (Note: this is not a partition of [itex][0,1][/itex].) At least one of the halves
    contains infinitely many terms of [itex]\{x_n\}[/itex], denote that interval by [itex]I_1[/itex], which has length 0.5,.....


    My misunderstanding

    The part that I don't understand is "At least one of the halves contains infinitely many terms of [itex]\{x_n\}[/itex]" ... Why can't [itex]\{x_n\}[/itex] be [itex]\{0.1,0.2\}[/itex], in which case one of the intervals contains 2 terms and the other contains 0 terms (with neither containing infinitely many)? No where did they state that [itex]\{x_n\}[/itex] couldn't have finitely many terms..?
     
  2. jcsd
  3. Jul 28, 2012 #2
    Ummm, they are probably assuming that x_n is finite. In real analysis, a sequence is usually assumed to be infinite unless stated otherwise.
     
  4. Jul 28, 2012 #3
    Oh ok that makes sense then.

    I wonder why the same theorem's proof on Page 49 of Beyer's "Calculus and Analysis" starts out the proof with "In case [itex]S[/itex] is finite, there is a subsequence...". I guess that's the trivial case the the other document is not even bothering with.
     
  5. Jul 28, 2012 #4
    Haha, yes - the finite case is trivial and not interesting and so is usually not dealt with.

    Ahh, I remember learning all this stuff a while back. Great stuff and very illuminating, I have to say. It's definitely something someone interested in math should look into. Real analysis offers many answers to question relating to why numbers and concepts are defined in a certain way, etc. :)
     
  6. Jul 28, 2012 #5
    Thanks for your help. I'm learning maths for fun .. analysis is great but hard!
     
  7. Jul 28, 2012 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In fact, if the sequence were finite, then the Bolzano-Weierstrasse theorem would not be true.
     
  8. Jul 28, 2012 #7
    Hmm.. Can you elaborate?

    The way I'm seeing it: no matter the cardinality of the finite [itex]S := \{x_n\}_{n \in \mathbb{N}} [/itex], we can always extract a single element of S as the sub-sequence, and this constant is itself always convergent (hence B-W is true for a finite S). Or, if |S|≥2, we can extract any number of distinct elements from S and hence have convergence in this sub-sequence because it'll be monotone increasing and bounded? (... This is probably a misapplication of knowledge from infinite sequences to finite ones, I'm new to this.)
     
    Last edited: Jul 28, 2012
  9. Jul 28, 2012 #8
    Well, it depends if you consider a finite sequence to be a "convergent".
     
  10. Jul 28, 2012 #9
    [retracted, mistake]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can't understand this Bolzano-Weierstrass proof
Loading...