# Homework Help: Can't understand this Bolzano-Weierstrass proof

1. Jul 28, 2012

### 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..?

2. Jul 28, 2012

### 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.

3. Jul 28, 2012

### 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.

4. Jul 28, 2012

### 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. :)

5. Jul 28, 2012

### operationsres

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

6. Jul 28, 2012

### HallsofIvy

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

7. Jul 28, 2012

### operationsres

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.)

Last edited: Jul 28, 2012
8. Jul 28, 2012

### who_

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

9. Jul 28, 2012

### operationsres

[retracted, mistake]