Convergence implies maximum/minimum/both

[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?

 

[STEMucator]

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?

 

[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?

 

[SammyS]

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?

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?

 

[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.

 

[SammyS]

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.

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

 

[peripatein]

Yes, they do.

 

[SammyS]

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

 

[peripatein]

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

 

[HallsofIvy]

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

 

[peripatein]

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

 

[peripatein]

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

 

[SammyS]

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

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

 

[peripatein]

Hence, within L+epsilon and L-epsilon.

 

[peripatein]

Where sequence converges to limit L.