# Intro to Real Analysis: Question on a completed proof.

• Nov 1st 2008, 08:57 AM
ilikedmath
Intro to Real Analysis: Question on a completed proof.
(The fact that http://www.cramster.com/Answer-Board...2187502633.gif is bounded is crucial.)

I found this:

http://i33.tinypic.com/2yxoxl2.jpg

My question is where the red arrow is pointing and the red boxed part:
How do I know how to choose 1/M for ε?
In this case we want to prove | $a_n$ $b_n$| < ε
so I need to choose any ε value right? How do I know to choose 1/M?
Is there algebra I can do that will lead me to choosing 1/M?
I think I understand the rest of the proof up to that point where the red arrow is pointing to.

Any help, suggestions are appreciated. Thanks for your time!
• Nov 1st 2008, 09:15 AM
Plato
Quote:

Originally Posted by ilikedmath
How do I know how to choose 1/M for ε?
In this case we want to prove | $a_n$ $b_n$| < ε
so I need to choose any ε value right? How do I know to choose 1/M?
Is there algebra I can do that will lead me to choosing 1/M?

You are to prove that given any $\varepsilon >0$ it is possible to insure that $
\left| {a_n b_n - 0} \right| < \varepsilon$
.
So from the fact that $\left( {a_n } \right) \to 0$ you use the definition of sequence convergence.
But in place of just $\varepsilon >0$ you are free to use $\frac{\varepsilon}{M} >0$.

$\left| {a_n b_n - 0} \right| = \left| {a_n b_n } \right| = \left| {a_n } \right|\left| {b_n } \right| \leqslant \left| {a_n } \right|M < \frac{\varepsilon }{M}M = \varepsilon$
• Nov 1st 2008, 10:21 AM
ilikedmath
Quote:

Originally Posted by Plato
You are to prove that given any $\varepsilon >0$ it is possible to insure that $
\left| {a_n b_n - 0} \right| < \varepsilon$
.
So from the fact that $\left( {a_n } \right) \to 0$ you use the definition of sequence convergence.
But in place of just $\varepsilon >0$ you are free to use $\frac{\varepsilon}{M} >0$.

$\left| {a_n b_n - 0} \right| = \left| {a_n b_n } \right| = \left| {a_n } \right|\left| {b_n } \right| \leqslant \left| {a_n } \right|M < \frac{\varepsilon }{M}M = \varepsilon$

So basically all they did was work with the inequality $\varepsilon >0$ and divided both sides by M to get the epsilon they wanted? :) It was that 'simple'? (Smirk)
• Nov 1st 2008, 10:27 AM
Plato
Quote:

Originally Posted by ilikedmath
So basically all they did was work with the inequality $\varepsilon >0$ and divided both sides by M to get the epsilon they wanted? :) It was that 'simple'?

NO that is not what happened.
$\frac{\varepsilon}{M} > 0$ was used in the definition of sequence convergence as the positive bound.
There are no "sides" here to be divided.
It is all a matter of properly using definitions.
• Nov 1st 2008, 12:43 PM
ilikedmath
Quote:

Originally Posted by Plato
It is all a matter of properly using definitions.

Ok, I see now. Thanks for clarifying.(Nod)