# Does the following argument work?

1. Feb 5, 2009

### Quantumpencil

1. The problem statement, all variables and given/known data

Suppose that a sequence {p}->p, and {q}->q. Show that if p_n < q_n for infinitely many n, then p<q.

2. Relevant equations

3. The attempt at a solution I'm sure I know why this is true intuitively. The convergence guarantees that all but a finite number of p_n are within some distance of p, and all but a finite number of q_n are within some distance of q, for large enough n this distance is arbitrarily small, therefore, if an infinite number of p_n < q_n, then an infinite number of p_n < q_n when in some neighborhood of p which can be made as small as we like; hence in the limit, p<q.

Now, for actual rigor... Does it suffice to say that the d(p, p_n) < r, d(q, q_n) < r for some n if we take n > max (N, N') (where N and N' are the natural numbers such that the above inequalities hold in each case). Therefore in any interval around p and q, there are an infinite number of p<q, ect.

I know this should be an epsilon argument, but I don't know how to make it less "qualitative"

This is my problem with math. Urgh. Help plz.

Edit: Yeah, I should've said < or =, but there is = on the keyboard, so I let that slide.

Last edited: Feb 5, 2009
2. Feb 5, 2009

### Kurret

Actually this does not hold, we could also have p=q. For example consider the sequences q_n=1/n and p_n=-1/n.

If I would prove this, i would make a proof by contradiction. Assume that p>q, and then use the epsilon-delta definition for the limits to arrive at a contradiction.

3. Feb 5, 2009

### Office_Shredder

Staff Emeritus
Rigor: If a sequence converges to a point, every subsequence converges to that point.

For infinitely many n, we have pn<qn. We can thus take the sequence of all such n, which we'll label n(k) (I'm not going to use subscripts for that as double subscripting doesn't work out on the forum)

pn(k)<qn(k) for all k

What can you do with that?

4. Feb 5, 2009

### Quantumpencil

You can then say that since the sub-sequence converges to p and q, p must be less than q, since every p_nk < q_nk and in the limit of large n they are approximately equal.

Still not rigorous enough though, I know.

Last edited: Feb 5, 2009
5. Feb 5, 2009

### Office_Shredder

Staff Emeritus
Wrong! You don't NEED to use epsilons and deltas. You should have proven: If an<bn for all n, and an converges to a, bn converges to b, then a is less than or equal to b (which you shouldn't forget the = part, as Kurret pointed out).

So pn(k)<qn(k) for all k, and hence p<=q. That's it No more work to be done

6. Feb 5, 2009

### Quantumpencil

That is in fact substantially easier. Hm. I had heard from class-mates it was an Epsilon proof and wasn't able to get that to work out. All my triangle inequalities yielded useless information.

If it doesn't take too much effort, how would one do this using the given distance relationships?

7. Feb 5, 2009

### Office_Shredder

Staff Emeritus
The fastest way would be to just take the proof that if one sequence is smaller than another, the limits share the inequality, and apply it directly to the subsequence you have (replacing the arbitrary sequence that you have in the general proof with the specific one here). All you're doing is changing the names of the sequences