Intro to Analysis (Boundedness of Cauchy)

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Homework Statement



Prove the Boundedness Theorem for Cauchy Sequences by a contrapositive argument.

Homework Equations





The Attempt at a Solution



If {an} is not bounded then {an} is not a cauchy sequence.

We will prove that {an} is not a cauchy sequence by showing that there exists an ε>0 so that for every pos. int. N there exists an n,m>N so that abs(an-am)≥ε.

Set ε=1
Consider positive N arbitrary.
Since by hypothesis {an} is not bounded to mean for every B there exists an n so that abs(an)>B. Choose such an n.

This is where I am stuck, how do I find/choose an m? Also how do I get the inequalities to work out at the end?
 
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The same way you found n. You know that {a_n} is unbounded, so for any B, there exists something greater than it, right? well, what about a_n? Shouldn't there be something greater than a_n?

Also, look at some consequences of the triangle inequality.
 
Alright I am following what you are saying but get stuck at:

abs(an-am) ≥ abs(an) - abs(am)

Then I do not know how to make it bigger than one.
 
So take a peek at the reverse triangle inequality. There's another one, like you have, but a little different.

But if we know that we can some large a_n, well, since {a_n} is unbounded, there exists some |a_m| > |a_n|, right? Then certainly there's some |a_m| > |a_n| + 1?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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