# Homework Help: Can't do this supremum question

1. Apr 27, 2013

### kramer733

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

here's the picture and it's the second part of question 5:

http://imgur.com/ybSW4v4

2. Relevant equations

N/A

3. The attempt at a solution

so by intuition, I suspect that b = sup{a_n: n is in the natural numbers}

If we can show that, then it will follow from the first part of question 5 that what we are trying to prove does hold.

Ok well the problem is this. I'm given the definition of Least Upper Bound:

1) b is an upperbound
2) if y is an upperbound of the set, then (y > b) or (y = b).

I'm not sure how to prove 2) at all.. I need major help =/ Please..

2. Apr 27, 2013

### Simon Bridge

The question is: $$\text{Q5. }\text{Suppose that }a_n\leq b\text{ for all } n \text{ and that } a=\lim_{n\rightarrow \infty} a_n \text{ exists. }\\ \text{5.1 Show that }a\leq b\text{. }\\ \text{5.2 Conclude that } a\leq \sup_n a_n = \sup_n \{ a_n : n\in \mathbb{N} \}$$
... hmmm, how to describe it...

if A={1,2,3}, then sup[A] = what?,
if an ≤ b, what is b?
Is b supposed to be a single number or a set of them?
What does it mean to say "$a=\lim_{n\rightarrow \infty} a_n$ exists"?

3. Apr 28, 2013

### kramer733

Oh crap..... For whatever reason, i parsed "b" to be the number such that it is greater than AND equal to an. Thank you. So b is a set of numbers that satisfy an≤ b right?

4. Apr 28, 2013

### Zondrina

I'm working on this same question. I believe b is supposed to be a real number so that the sequence is bounded.

5. Apr 28, 2013

### christoff

I've posted a hint in Zondrina's post; you can check it out there: https://www.physicsforums.com/showthread.php?p=4364873#post4364873

Basic idea: consider two subsequences of your sequence $a_n$, one that is decreasing, and one that is increasing. Show that if you have an decreasing sequence, then your inequality is satisfied. Show that this also holds for increasing sequences. Conclude that this inequality is valid for your original sequence.