Can't understand this Bolzano-Weierstrass proof

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In summary, the Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. The proof involves dividing the bounded sequence into two intervals and selecting the one with infinitely many terms. This assumption is based on the fact that sequences in real analysis are usually assumed to be infinite. The trivial case of a finite sequence is not considered, as it does not hold for the theorem.
  • #1
operationsres
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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..?
 
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  • #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.
 
  • #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.
 
  • #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. :)
 
  • #5
Thanks for your help. I'm learning maths for fun .. analysis is great but hard!
 
  • #6
In fact, if the sequence were finite, then the Bolzano-Weierstrasse theorem would not be true.
 
  • #7
HallsofIvy said:
In fact, if the sequence were finite, then the Bolzano-Weierstrasse theorem would not be true.

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.)
 
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  • #8
Well, it depends if you consider a finite sequence to be a "convergent".
 
  • #9
[retracted, mistake]
 

Related to Can't understand this Bolzano-Weierstrass proof

1. What is the Bolzano-Weierstrass theorem?

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. In other words, if a sequence of real numbers is bounded (meaning its values do not exceed a certain limit), then there exists a subsequence that converges to a finite limit.

2. What does the Bolzano-Weierstrass theorem prove?

The Bolzano-Weierstrass theorem is an important result in real analysis that is used to prove the convergence of sequences and series. It is also a key step in proving other theorems, such as the Heine-Borel theorem and the intermediate value theorem.

3. How is the Bolzano-Weierstrass theorem proved?

The Bolzano-Weierstrass theorem can be proven using a method known as the nested interval theorem, which involves dividing a bounded interval into smaller subintervals and finding a subinterval that contains infinitely many terms of the sequence. This process is repeated to create a nested sequence of intervals, and the limit of this nested sequence is the limit of the subsequence that converges.

4. What is the significance of the Bolzano-Weierstrass theorem?

The Bolzano-Weierstrass theorem is significant because it allows us to prove the convergence of a sequence without knowing its exact limit. This is useful in many areas of mathematics, such as calculus, where finding exact values of limits can be difficult. It also has important applications in engineering, physics, and other fields where sequences and series are used to model real-world phenomena.

5. Can the Bolzano-Weierstrass theorem be extended to higher dimensions?

Yes, the Bolzano-Weierstrass theorem can be extended to higher dimensions. In fact, it is a fundamental result in multivariable calculus and is known as the Bolzano-Weierstrass theorem for sequences in n-dimensional Euclidean space. It states that every bounded sequence of n-dimensional vectors has a convergent subsequence.

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