# Lower bound proof/definition

1. Aug 22, 2007

### ice109

in my book this is called the lower bound but it implies that it might be called the greatest lower bound elsewhere.

lower bound: some quantity m such that no member of a set is less than m but there is always one less than m + $\epsilon$

definition using Dedekind section

there are quantities a such that no member of a set is less than any a. there are quantities b such that some members of a set are less than any b. all a(s) are less than all b(s)[anyone have a problem with this???]. and all quantities of the same dimension are either a(s) or b(s). a(s) define the L class and b(s) define the R class. now make a partition, m, that has to be in L. for if it were in R then there would be some member of the set, k, less than it and and there would be no a(s) between m and k and hence it wouldn't partition correctly. m is then the greatest lower bound. m + $\epsilon$ is in R which by prior definition must have a member less than it.

the part in brackets i'm not sure about specifically besides not being sure about the whole thing, despite the fact that i basically followed the Dedekind proof/defition for least upper bound. this is the first mathematical proof by argument i have ever done and boy was it more abstract than any physics problem i've ever done.

Last edited: Aug 22, 2007
2. Aug 23, 2007

### HallsofIvy

You are right. If your book really does define "lower bound" that way, it is very peculiar. The definition of "lower bound" I have always seen is "a number that is less than or equal to any member of the set". If there is always a a value less than $m+\epsilon$ there cannot be any number larger than m which is also an lower bound and so m is the greatest lower bound. A set can have an infinite number of lower bounds but only one greatest lower bounmd.

What, exactly, are you trying to prove? That if a set has a lower bound, then it must have a greatest lower bound? I would not "copy" the proof of the least upper bound. I would start with a set having a lower bound, multiply everything by -1 and then USE the least upper bound property.

3. Aug 23, 2007

### ice109

i was trying to define the greatest lower bound using a dedekind cut. in writing this definition i was guided by the definition of the least upper bound in the book

4. Aug 23, 2007

### HallsofIvy

Actually, I find your use of "L" and "Q" a bit peculiar. I learned the definition of Dedekind cut (from Rudin) as a set of rational numbers satisfying:
1) It is non-empty
2) It does not contain all rational numbers
3) If a< b and b is in the cut, then so is a
4) It has no largest member

In other words, what I would call a Dedekind cut corresponds only to your set "L" which is guarenteed to be non-empty by 1. Your set "R" is the complement of L (in the rational numbers).
The proof that a set of real numbers with an upper bound has a least upper bound consists of taking the union of all dedekind cuts (your "L" sets) in the set and showing that that is itself a dedekind cut and is the least upper bound.
To find the greatest lower bound of a set having a lower bound, you probably want to look at the union of all the "R" sets. Take its complement (in the rational numbers) and call that "L". Does that make a cut in your sense? Is it the greatest lower bound>?

5. Aug 23, 2007

### tronter

I learned the definition of Dekekind cut as follows: A pair of subsets $$A,B$$ of $$\mathbb{Q}$$ such that (i) $$A \cup B = \mathbb{Q}, A \neq \emptyset, B \neq \emptyset, A \cap B = \emptyset$$, (ii) if $$a \in A$$ and $$b \in B$$ then $$a < b$$ and (iii) $$A$$ contains no largest element. Denote the cut by $$x = A|B$$.

6. Aug 24, 2007

### HallsofIvy

Basically, the same, then. Your "A" is, as I said, what I would consider "the cut" and your "B" is its complement in the rational numbers.