Greatest lower bound

  • Thread starter steven187
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  • #1
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hello all

I know this might be a simple question to ask, but i want to find other ways of proving it anyway here we go
propve that if A is a subset of R and is non empty and bounded below, then it has a greatest lower bound.

This is how i did it:

let b be a lower bound of A. then for every a an element of A b<=a so -a<=-b. now we notice that -b is an upper bound for -A, where -A={-x,x an element ofA}. since -A is non empty and it is bounded above then by the least upper bound axiom -A has a least upper bound , lets call this least upper bound -L , since -L is the least upper bound of -A, we must have -L<=-b or b<=L. and so this is true for any b which is a lower bound of A and hence L must be the greatest lower bound of A

is there another method to do this problem if not is there anyway of simplyfying this proof?

thanxs
 

Answers and Replies

  • #2
rachmaninoff
It's the shortest proof I can think of, if you're already given the greatest upper bound axiom.
 

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