Prove the uniqueness of a greatest lower bound if it exists.

[daon]

:?: I didn't get too far into this proof, as I don't know where to start. First of all let me see if I have the right definition (we had to make our own...)

A set S has a greatest lower bound if there exists a Real number x, such that for all elements s in S, s >= x and, for all other Real lower bounds y, x>=y.

Makes sense intuitively to me, but I want to make sure...

Okay, so... I have a definition now for a greatest lower bound. I need to show that this bound is unique. To do this I believe this will accomplish it:

Assume that a set A has a greatest lower bound. Further, if A has two greatest lower bounds x and x' then x = x'.

I believe I am headed on the right path here, but I am not sure what to do next. Help me!

 

[pka]

This is what I call a ‘definitional’ proof. It really depends upon the definition.
The statement that b is the greatest lower bound of a non-empty set A means that b is a lower bound of A and if c is also a lower bound of A the c≤b.

First you need to understand the intricate working of GLB’s.
We can prove the following based on that definition: If A is a number set and b=GLB(A) then if for any x>b, x is not a lower bound of A; that means the there is a number c in A and b≤c<x.

Now suppose that A has two GLBs, d & e. Either d<e or e<d, they are not equal.
Lets say that d<e. Well from the above that means that e is not a lower bound.
But that is a contradiction. Likewise, if e<d then d is not a lower bound!

 

[daon]

This is what I call a ‘definitional’ proof. It really depends upon the definition.
The statement that b is the greatest lower bound of a non-empty set A means that b is a lower bound of A and if c is also a lower bound of A the c≤b.

First you need to understand the intricate working of GLB’s.
We can prove the following based on that definition: If A is a number set and b=GLB(A) then if for any x>b, x is not a lower bound of A; that means the there is a number c in A and b≤c<x.

Now suppose that A has two GLBs, d & e. Either d<e or e<d, they are not equal.
Lets say that d<e. Well from the above that means that e is not a lower bound.
But that is a contradiction. Likewise, if e<d then d is not a lower bound!

Sheesh... Thanks once again. I don't know why I didn't see it... Seems so obvious now.

I also have a separate question. You said that you've teached Discrete Mathematics, right? Well did you cover this material in your class? When I took it over a year ago at a community college, the proofs we did were far more simple then the ones I'm doing now and we barely needed any rigor in our proofs. I feel like I would have been better prepared for this class (An intro class for higher mathematics) if I was challenged in my Discrete class. If you don't mind telling me, what book do you teach out of and would you reccomend it? Or maybe you have another book you'd reccomend. Thanks again.

 

[pka]

a separate question. You said that you've teached Discrete Mathematics, right? Well did you cover this material in your class? When I took it over a year ago at a community college, the proofs we did were far more simple then the ones I'm doing now and we barely needed any rigor in our proofs. I feel like I would have been better prepared for this class (An intro class for higher mathematics) if I was challenged in my Discrete class. If you don't mind telling me, what book do you teach out of and would you reccomend it? Or maybe you have another book you'd reccomend. Thanks again.

No, not in discrete mathematics. I taught a year long course: first term was logic and sets, second term was combinatics and introductory graph theory and networks.
I would teach this material in a first course in analysis.