# 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.