3
$\begingroup$

This is the problem:

Prove that if $a_n \le b_n$ for $n \ge 1, L = \lim_{n \to \infty} a_n$ and $M = \lim_{n \to \infty} b_n$, then $L \le M$

EDIT: Progress

Proof

Assume $L >M$ and $a_n \leq b_n$, then

(1) $|a_n -L| < \epsilon$ when n > $N_1$

(2) $|b_n-M| < \epsilon$ when n > $N_2$

Expanding (1) and (2) gives

$L - \epsilon < a_n < \epsilon + L$ and $M - \epsilon < b_n < \epsilon + M$

Since $a_n \leq b_n$, we have $L-\epsilon

OKay I am stuck now, but I feel I am getting close

EDIT: alternate proof from text

Proof

Let $\lim_{n\to\infty}b_n -a_n=M -L$. Therefore for any $\epsilon > 0$, $\exists N:$

$|b_n - a_n - (M - L)| <\epsilon$ whenever $n > N$

Take $\epsilon = L - M$ and we get $|b_n - a_n - (M - L)| whenever $n > N$ and since $a \leq |a|$, we have $b_n - a_n - (M - L) < L -M \iff a_n >b_n$, but this contradicts the assumption and therefore $L > M$ must be false

  • 2
    You haven’t proved anything: you’ve simply asserted the conclusion. You will need to use the definition of limit somewhere; I suggest that you assume that $L>M$ and use the definition of limit to derive a contradiction.2012-09-23
  • 0
    @BrianM.Scott, yeah I felt that the proof seemed circular. Is contradiction the only way?2012-09-23
  • 0
    It may not be the only way, but it’s by far the easiest.2012-09-23
  • 0
    OKay I will edit what I did. I thnk I got it2012-09-23
  • 0
    @jak: There's a gap between $M$ and $L$, so what should you choose $\epsilon$ to be in order to get a contradiction? Also, the flow of your argument could be improved slightly.2012-09-23
  • 0
    I want $\epsilon\to0$, but I am not sure if I can do that. How should the flow be improved? Is assuming $a_n \leq b_n$ no good here?2012-09-23
  • 0
    @jak: What I mean by "flow" is this: you should make your choice of $\epsilon$, and *then* invoke the definition of the limit. Otherwise, $\epsilon$ is unquantified and $N_1,N_2$ depend on it; this can sometimes lead to mistakes.2012-09-23
  • 0
    But don't you need to go through the algebra first to get the $\epsilon$? Why do mathematicans omit that (which I think is important)?2012-09-23
  • 0
    @jak: That's absolutely correct - you need to do some experimentation to get the right "magic" values. But you don't need to put that working in the finished product (the proof), because it's irrelevant to the logic and it can usually be inferred.2012-09-23

2 Answers 2

5

Suppose $L>M$. Let $\epsilon=\frac{L-M}{2}>0$. Then there are positive integers $A$ and $B$ such that $$L-\epsilon for all $n>A$ and $$M-\epsilon for all $n>B$. It follows that $$a_n>L-\epsilon=M+\epsilon>b_n$$ for all $n>A+B$, a contradiction.

  • 0
    How the heck did you come up with $\epsilon=\frac{L-M}{2}$?2012-09-23
  • 2
    @jak: http://imgur.com/g5X2A2012-09-23
  • 1
    @jak It's a classic.2012-09-23
  • 0
    @WillHunting, in your picture, does that mean I should omit the assumption that $a_n \leq b_n$?2012-09-23
  • 0
    @jak: No, the point is that by assuming $M and choosing $\epsilon$ appropriately, you can contradict the fact that $a_n \le b_n$ for all $n \ge 1$.2012-09-23
  • 0
    But by assuming $M < L$, I can't assume $a_n \le b_n$2012-09-23
  • 0
    @WillHunting, kinda like what I did before? Because those two inequalities are the only ones I know2012-09-23
  • 0
    I keep getting back to where I started (getting back to $L < M + 2\epsilon$). Trying to incorporate the hint about the choice of epislon you gave me and I don't know where/when I should introduce it.2012-09-23
  • 1
    @jak: You are given certain facts, and they tell you that you can pick any $\epsilon > 0$ you want and you will receive a $N$ such that $|a_n - L| < \epsilon$ whenever $n>N$. The key thing here is that the choice of $\epsilon$ is *up to you*; you must use this to your advantage. It happens that choosing $\epsilon = (L-M)/2$ contradicts $M, so $M cannot be true. (Have you taken a proof writing course? If not, you should!)2012-09-23
  • 0
    @wj32, I am going to try one more time2012-09-23
  • 0
    I am reading my other book and they used a different method (kind of) and I don't understand it either. I am going to add it here and maybe you could give me some comments2012-09-23
0

Use these facts. If $x_n\to L$ then for any open interval $I$ containing $L$, $x_n\in I$ for all but finitely many $n$. You can structure an argument from this fact. Give it a whirl.