Math 381                                                                                             Name: Key

Quiz 3

  1. Let MATH Find $\inf S$ and $\sup S$ if they exist.

    2. MATH
     
  2. Let MATH be a bounded nonempty set and let $c>0.$$\ $Prove: If MATH, then MATH
Proof:    Premises: $S$ is bounded and nonempty, $c>0.$
Since $S$ is bounded, $\sup S$ exists by the Completeness Axiom.
By definition, MATH
Thus, since $c>0$, we have MATH
So, MATH
That is, $c\cdot \sup S$ is an upper bound for $M.$
Now we need to show it is the least upper bound.
 

Suppose, for the sake of contradiction, that there is a real number $n$ such that $n<c\cdot \sup S$ and $n$ is an upper bound for $M.$

Then MATH
Thus, (since $\sup S$ is the least upper bound for $S$), $\exists s\in S$ with $\dfrac{n}{c}<s.$
Hence $n<c\cdot s.$ But $c\cdot s\in M$ and $n$ is an upper bound for $M.$
This is a contradiction (no element of a set can be bigger than an upper bound for the set.)
Thus there is no upper bound for $m$ that is less than $c\cdot \sup S.$
Conclusion: MATH
 
 
This document created by Scientific WorkPlace 4.0.