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

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

Let and be sequences of real numbers. If is bounded and = 0,
then = 0.
(The fact that is bounded is crucial.)

I found this:

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!

2. 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 $\frac{\varepsilon}{M} >0$.

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

3. 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'?

4. 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.

5. Originally Posted by Plato
It is all a matter of properly using definitions.
Ok, I see now. Thanks for clarifying.