Prove the limit theorem: a_n < b_n -> A-B

[shoescreen]

Homework Statement

Given the sequence {a_n} converges to A and {b_n} converges to B, and a_n <_ b_n for all n>_ n*, prove A <_ B

Homework Equations

x+ epsilon < y for every positive real epsilon, implies x <_ y
A - B = (A - a_n) + (b_n - B) + (a_n - b_n)

The Attempt at a Solution

I want to show A - B is not positive, I know a_n - b_n is not positive by hypothesis, and A - a_n and b_n - B are bounded by (-e/2, e/2) where e is any positive number, hence their sum is bounded by (-e,e). But since their sum is not bounded by zero, I can't figure out how to apply (x + e) - y negative implies (x - y) non positive.

 

[LCKurtz]

If you suppose A > B and take epsilon (1/3)(A-B) can you find a contradiction?

 

[shoescreen]

If you suppose A > B and take epsilon (1/3)(A-B) can you find a contradiction?

i should have mentioned earlier, the problem statement explicitly states do not prove by contradiction.

 

[LCKurtz]

If you see how to do the indirect argument I am hinting at, you can easily modify it to be a direct argument. (Shhh... I won't tell anybody if you look at the indirect argument first).

 

[shoescreen]

If you see how to do the indirect argument I am hinting at, you can easily modify it to be a direct argument. (Shhh... I won't tell anybody if you look at the indirect argument first).

Well I was asked to prove the this by contradiction before and i was able to do it with (a-b)/2 and i still don't see how to do it directly :(

EDIT:
hah i think i got it!
EDIT:
just kidding...

 

[LCKurtz]

Would a proof that showed for any ε > 0 that A ≤ B + ε be considered a direct proof?

 

[shoescreen]

so if I use
|a_n - A| < f(A,B)
|b_n - A| < g(A,B)

I can split up each inequality and ultimately end up with something like
a_n > -f(A,B) + A
b_n < G(A,B) + B

so a_n - b_n > -f(A,B) - g(A,B) + A - B

so if i want to show a_n - b_n > A - B, the function -f + -g must be positive. But this is impossible since both of these functions must be strictly positive, hence negative f plus negative g is also negative.
So I'm still stuck

 

[fmam3]

If you know liminf, limsup, then this is easy.

Since $$a_n \leq b_n$$, or $$0 \leq b_n - a_n$$. Then it follows that $$0 \leq \liminf (b_n - a_n) \leq \limsup (b_n - a_n)$$. But since $$\lim a_n = A$$ and $$\lim b_n = B$$, it follows that $$\lim (b_n - a_n) = B - A$$, which implies the limit inferior and limit superior must equal. That is, $$\limsup (b_n - a_n) = \liminf (b_n - a_n) = B - A$$. Thus, we have that $$0 \leq B - A$$, rearrange then you're done.

 

[Landau]

For every e>0 there is N such that n>N implies |a_n-A|<e and |b_n-B|<e.
In other words: A-e<a_n<A+e and B-e<b_n<B+e. Try to put these two together.

\\edit: as LCKurtz remarked, I gave away too much, so I removed the last part.

 

[LCKurtz]

And here I've been under the impression that we aren't supposed to actually work homework problems for them.