# Homework Help: Real analysis- Convergence/l.u.b

1. Sep 11, 2007

### Scousergirl

I'm having a little difficulty understanding Epsilon in the definition of convergence. From what the book says it is any small real number greater than zero (as small as you can imagine???). Also, since I don't quite grasp what this epsilon is and how it helps define convergence, I am having difficulty applying it to the following problem:

Let b=Least upper bound of a set S (S is a subset of the real numbers) that is bounded and non empty. Then Given epsilon greater than 0, there exists an s in S such that (b-Epsilon)<= s <= b.

I started by proving that there exists an s in S, but I cannot figure out how to relate this all to epsilon. What is confusing me I guess is the actual definition of S. Can the set {1*, b} satisfy the requirements of s (it is a subset of the real numbers, bounded above by b and non empty) but then how do we show that the statement is true for s=1*. Also, b doesn't have to part of S right?

Last edited: Sep 11, 2007
2. Sep 11, 2007

### Dick

A bounded set has many upper bounds. The least upper bound b is the one that is low enough that if you pick any number less than b (b-Epsilon) no matter how small the Epsilon, then that number is not an upper bound. And yes, b doesn't have to be in S. After that, you confuse me. What's 1*?

3. Sep 11, 2007

### Scousergirl

Ok, I think I understand the least upper bound part of the question, I just don't know how to go about proving that ANY subset of the real numbers (S) that is non empty and bounded above by b must contain an element s such that (b-epsilon)<= s <= b.

1* is a dedekind cut (the real number 1 basically). Doesn't a set say S={1 , b} satisfy these conditions yet if I choose a small epsilon, 1 is not neccesarily greater or equal to (b-epsilon).

4. Sep 11, 2007

### Dick

Mmm. That's the DEFINITION of least upper bound. I.e. it's the definition of 'b'. You don't have to prove there is such an s. If b is least upper bound of S there is such an s for every epsilon. Otherwise it's not a least upper bound. A dedekind cut is two subsets of the rationals with special properties. Neither subset is even bounded. The least upper bound of S={1,b} is 1 if b>1 and b if b<1. I think this concept is much less complicated than you think it is.

5. Sep 11, 2007

### Scousergirl

hmmm...I think maybe I am confusing multiple concepts here. The question that I am having difficulty with is how to prove that such an s exists for all subsets of the real numbers. (b-epsilon)<= s <= b.

I tried tackling it by contradiction:

Assume there does not exist an s in S such that (b-epsilon)<= s. Thus for all s in S, s< (b-epsilon). Is this a contradiction to the fact that b is the least upper bound of S?

6. Sep 11, 2007

### Dick

Such an s does not exist for all subsets of the real numbers. Some aren't bounded. So there is no b. On the other hand for this "Assume there does not exist an s in S such that (b-epsilon)<= s. Thus for all s in S, s< (b-epsilon). Is this a contradiction to the fact that b is the least upper bound of S?". Yes, that contradicts b being least upper bound. But this is getting really confusing for me as well as for you, it's getting existential. Try dealing with well defined subsets of the reals and figuring out what the least upper bound is and what it means. It sounds much more confusing in the abstract than what it really is. Trust me.