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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Discussion 2 -- Completeness of ${\bf R}$, continued}
\centerline{September 8, 2000}
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\noindent
{\it Background}
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In class on Wednesday, we introduced the {\it L.U.B., or Completeness
Axiom} for the real number system:
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\proclaim Completeness (LUB) Axiom.  Every nonempty set of real numbers
that is bounded above has a least upper bound (lub) in the real numbers.\par

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\noindent
Today, we want to begin to explore some of the consequences of this
innocuous-seeming but {\it extremely important} property.

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\noindent
{\it Discussion Questions}
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\item{A)}  Identify a lub for each of the following nonempty sets of 
real numbers.  You may use facts you know from calculus if needed.
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\itemitem{1)}  $A = \{x\ |\ x^2 - 3x - 6 \le 0\}$
\itemitem{2)}  $B = \left\{{2n\over n+3}\ |\ n \in {\bf N}\right\}$
\itemitem{3)}  $C = \{\ \cos x\ |\ 0 < x \le \pi/2\}$
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\item{B)}  In this problem, you will show that if $A$ is a set of 
real numbers that is bounded above and nonempty, then ${\rm lub}(A)$
is unique -- we are actually justified in saying {\it ``the least
upper bound''} instead of {\it ``a least upper bound''}.  To 
show that something is unique, a standard approach is to assume 
you have any two of them, then show that the two must be equal.
So suppose that $\alpha$ and $\alpha'$ are both least upper bounds of 
$A$.  Then $\ldots$  (you complete the proof).  ({\it Hint}:  
Go back to our original definition of a lub and use what that says.)
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\item{C)}  For sets that are bounded {\it below} there is a corresponding
notion of {\it greatest lower bound}.
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\itemitem{1)}  Give a precise definition of what it should mean for
$\beta \in {\bf R}$ to be a greatest lower bound for a set $B$
that is bounded below.
\itemitem{2)}  Prove that the Completeness Axiom above implies
that every nonempty subset of the real numbers that is bounded
below has a greatest lower bound.  ({\it Hint:  The best way to 
think about this is that you can deduce what you want from 
the LUB axiom if you can
``turn lower bounds into upper bounds'' somehow -- for instance
by doing some operation to your set that reverses all inequalities.})
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\item{D)}  For each of the following statements that is true,
give a short proof; for those that are false, give a counterexample.
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\item{A)}  Suppose $\alpha = {\rm lub}(A)$.  Then
$|\alpha| = {\rm lub}(\{|x| | x \in A \})$.  
\item{B)}  Suppose $\alpha = {\rm lub}(A)$ and $\beta = {\rm lub}(B)$.
Then $\alpha + \beta ={\rm lub}(\{x+y | x \in A, y \in B\}$.
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\noindent
{\it Assignment}
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Group writeups due in class, Wednesday, September 13.
\bye
