Can't understand this Bolzano-Weierstrass proof

[operationsres]

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 \{x_n\} be a bounded sequence and without loss of generality assume that every term of the sequence lies in the interval [0,1]. Divide [0,1] into two intervals, [0,0.5] and [0.5,1]. (Note: this is not a partition of [0,1].) At least one of the halves
contains infinitely many terms of \{x_n\}, denote that interval by I_1, 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 \{x_n\}" ... Why can't \{x_n\} be \{0.1,0.2\}, 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 \{x_n\} couldn't have finitely many terms..?

 

[who_]

Ummm, they are probably assuming that x_n is finite. In real analysis, a sequence is usually assumed to be infinite unless stated otherwise.

 

[operationsres]

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 S is finite, there is a subsequence...". I guess that's the trivial case the the other document is not even bothering with.

 

[who_]

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. :)

 

[operationsres]

Thanks for your help. I'm learning maths for fun .. analysis is great but hard!

 

[HallsofIvy]

In fact, if the sequence were finite, then the Bolzano-Weierstrasse theorem would not be true.

 

[operationsres]

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 S := \{x_n\}_{n \in \mathbb{N}}, 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.)

 

[who_]

Well, it depends if you consider a finite sequence to be a "convergent".

 

[operationsres]

[retracted, mistake]