• August 1st 2012, 09:00 AM
MagisterMan
Is this a valid proof?

Problem: Prove; If lim an < lim bn, then there exists N that belongs to the natural numbers such that n >= N implies an < bn

My solution:

Let lim an = L and lim bn = M. Assume L < M. Then for every ε there exists N1 such that n >= N1 implies L - ε < an < L + ε. For every ε there exists N2 such that n >= N2 implies M - ε < bn < M + ε. Let N = max{N1, N2}

=> L - ε< an < L + ε
M - ε < bn < M + ε

Assume an >= bn

Then L -
ε < M - ε < bn <= an.
an < L +
ε => L - ε < bn < L + ε => |bn - L| < ε => L=M which is a contradiction, thus an must be less than bn.
• August 1st 2012, 11:23 AM
HallsofIvy
What if $\epsilon< M-L$?
• August 1st 2012, 11:44 AM
emakarov
Quote:

Originally Posted by MagisterMan
|bn - L| < ε => L=M

L = M does not follow from the fact that |bn - L| < ε for some particular n and ε.

The main problem with your proof is that it silently goes from statements that quantify over n and ε to statements about some particular n and ε and then back. If you say "For every ε there exists N1 such that n >= N1 implies L - ε < an < L + ε," the variables ε and n exist only inside that phrase. Once you finish the sentence, you can't refer to ε or n. Therefore, you can't continue by writing "L - ε < an < L + ε."

Every variable in a proof must be properly introduced. If you are proving a statement of the form "For all n, ...," you can say "Fix n" and use that n for the rest of the proof of that statement. If you have an assumption or a proven fact of the form "There exists an n such that ...," you can say "Consider an n such that ..." and again use it until the end. A third way to introduce a variable is by an explicit definition: e.g., "Let ε = M - L." Then you can instantiate an assumption or a proven fact of the form "For every ε, ..." using this particular ε. In this particular problem, choosing a proper ε is an important step.
• August 1st 2012, 12:03 PM
MagisterMan