so should I just show that -m = sup(-A), then we have m= - sup(-A) by the least upper bound axiom?
If A is a nonempty subset of R that is bounded below, show that A has a greatest lower bound. That is, show that there is a number satisfying: 1) m is a lower bound for A; and 2) if x is a lower bound for A, then x<=m. [Hint: Consider the set -A={ } and show that m = -sup(-A) works.]
The hint is confusing me, I'm really not good at proving, any help is appreaciated!
Hi wopashui,
This problem is incorrect. There are sets for which there exist lower bounds but not a greatest lower bound. For example let,
be the set of Rational numbers and Clearly there are lower bounds for the set A, such as, 1,0,-1 etc. But a greatest lower bound does not exist. Since if you take any lower bound (say ) there exist infinity many rational elements in between and .
Of course it is. As I have mentioned in my previous post examples of lower bounds for A include, 1,0,-1 etc.
The greatest lower bound should be inside the superset. In my example, . I think you have to carefully go though the definitions Lower and Upper bound of a set, Supremum and Infimum.
Did you denote the the set of Real numbers by R? I thought it was a general set not specifically .Oh because the problem I have is in R, the example you gave is in Q, yes R and Q appear quite different when we examine the existence of lub or glb fir biunded sets.