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Homework Help: 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:

    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


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    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]
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