Convergence implies maximum/minimum/both

1. Nov 4, 2012

peripatein

Hello,

Could anyone please assist in proving that given a sequence converges it has a maximum/minimum/both?
I have hitherto written that granted it converges it must be bounded and have a supremum and an infimum. Now, how may I proceed to prove that the latter are indeed within (the neighbourhood) of the limit itself?

2. Nov 4, 2012

Zondrina

So if your sequence has a maximum, it is bounded above. If your sequence has a min, it is bounded below.

What do you know about a bounded sequence of real numbers?

3. Nov 4, 2012

peripatein

What do you mean? I understand that any bounded sequence of real numbers must have a maximum and/or minimum, but how do I formally demonstrate that? Namely, how do I show that its supremum and infimum are within L, its limit?

4. Nov 4, 2012

SammyS

Staff Emeritus
Look at some examples of convergent sequences.

Does each have both a maximum and minimum?

Do all have only a maximum?

Do all have only a minimum?

Do they have to have at least one or the other?

Do any have both?

5. Nov 4, 2012

peripatein

I can think of examples for sequences having either one of the two or both, but that still does not help me in formally proving the statement.

6. Nov 4, 2012

SammyS

Staff Emeritus
Do they have to have at least one or the other (minimum, maximum)?

7. Nov 4, 2012

peripatein

Yes, they do.

8. Nov 4, 2012

SammyS

Staff Emeritus
How about proving the contra-positive, that a sequence which has neither a maximum nor a minimum does not converge?

9. Nov 5, 2012

peripatein

I am still not sure how to formulate that mathematically. May you please advise?

10. Nov 5, 2012

HallsofIvy

Staff Emeritus
if sequence an converges to A, then there exist N such that if n> N, |an- A|< 1 which is the same as saying A-1< an< A+1. Do you see how that gives upper and lower bounds for all an with n> N? And an for $n\le N$ is a finite set.

11. Nov 5, 2012

peripatein

Naturally, but how do I use that to demonstrate that the supremum and infimum are within the neighbourhood of L.

12. Nov 5, 2012

peripatein

This is very frustrating. I am still unable to prove that the supremum and infimum are within L. May someone please assist?

13. Nov 5, 2012

SammyS

Staff Emeritus
What do you mean by the supremum and infimum are within L ?

14. Nov 5, 2012

peripatein

Hence, within L+epsilon and L-epsilon.

15. Nov 5, 2012

peripatein

Where sequence converges to limit L.