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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Exam 1 -- February 21, 2014}
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{\it Directions:}  Do all work in the blue exam booklet (do not place any work 
that you want to have considered for credit on this sheet).  There are 100 regular and 
10 extra credit points.
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I.
\item{A)} (10)  {\it Define:}  The real number $a$ is a {\it least upper bound} 
of $A \subset {\bf R}$, and state the Least Upper Bound Axiom for ${\bf R}$. 
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\item{} {\it Solution:}  $a$ is a least upper bound of $A$ if (1) $a \ge x$ for
all $x\in A$, and (2) if $a' \ge x$ for all $x \in A$, then $a' \ge a$.  The LUB Axiom
states that if $A$ is a nonempty subset of ${\bf R}$ that is bounded above, then 
$A$ has a least upper bound in ${\bf R}$.  
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\item{B)} (10) Let
$$A = \bigcap_{n=1}^\infty \left(1 + {1\over n} , 4 - {1\over n}\right)$$
Explain why $A$ is bounded and determine the least upper and greatest lower bounds for $A$.
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\item{} {\it Solution:}  When $n = 1$, we have $\left(1 + {1\over n}, 4 - {1\over n}\right) = 
(2,3)$.  For $n > 1$, $1 + {1\over n} < 2$ and $4 - {1\over n} > 3$.  Hence the intersection
of all the intervals $\left(1 + {1\over n} , 4 - {1\over n}\right)$ is equal to the interval
for $n = 1$:  $A = (2,3)$.  Hence ${\rm glb}(A) = 2$ and ${\rm lub}(A) = 3$.  
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\item{C)} (15)  Let $A$ be a bounded subset of ${\bf R}$ with ${\rm lub}(A) = 2$,
and let $B = \{-4 x + 3 \mid x \in A\}$.
What can be said about ${\rm glb}(B)$?  Prove your assertion.
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\item{} {\it Solution:}  ${\rm glb})(B)$ must exist and equal $-5$.  Proof:  First, 
$2$ is an upper bound for $A$, so for all $x \in A$, $x \le 2$.  Hence $-4x \ge -8$,
and so $-4x + 3 \ge -5$.  Next if $b$ is any lower bound for $B$, we have $b \le -4x + 3$
for all $x \in A$, hence ${b - 3\over -4} = {3 - b\over 4} \ge x$ for all $x$ in 
$A$.  Since $2 = {\rm lub}(A)$, this implies ${3 - b\over 4} \ge 2$, so 
$3 - b \ge 8$, and hence $b \le -5$.  This shows ${\rm glb}(B) = -5$.   
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II.  (15) Let $x_n$ be the sequence defined by the rules $x_1 = 1$ and $x_{n+1} = -{2\over 5} x_n + 1$
for all $n \ge 1$.  Show by mathematical induction that
$$0 \le x_n \le 1 \ {\rm for\ all\ } n\ge 1.$$
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{\it Solution:}  The base case for the induction is $n = 1$ and $x_1 = 1 \ge 0$, so there
is nothing else to be proved there.  Now for the induction step, assume $0 \le x_k \le 1$.  Then 
the ${-2\over 5} \le {-2\over 5} x \le 0$, so ${-2\over 5} + 1 \le {-2\over 5} x + 1 \le 1$, 
which shows what we want since ${-2\over 5} + 1 \ge 0$.   
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III.   Let $x_n = {3 n^2 \over 5n^2 + 3n + 1}$ for all natural numbers $n \ge 1$.
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\item{A)}  (10) Determine $\lim_{n\to\infty} x_n$ intuitively.
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\item{} {\it Solution:}  Dividing numerator and denominator
by $n^2$ have 
$$x_n = {3 \over 5 + {3\over n} + {1\over n^2}}$$
so we expect $x_n \to {3\over 5}$ as $n \to \infty$.   
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\item{B)}  (20)  Use the $\varepsilon$, $n_0$ definition of convergence
to prove that $\{x_n\}$ converges to the number you identified in part A.
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\item{} {\it Solution:}  Let $\varepsilon > 0$.  Since ${\bf N}$ is not bounded in
the real numbers, there exists $n_0$ in ${\bf N}$ with $n_0 > {2\over 5\varepsilon}$, so
${2\over 5n_0} < \varepsilon$.  Then for any $n \ge n_0$, we have 
$$\eqalign{\left|{3n^2 \over 5n^2 + 3n + 1} - {3\over 5}\right| &= {9n + 3\over 25n^2 + 15n + 5}\cr
&< {10 n \over 25 n^2} \cr
&= {2 \over 5n}\cr
&\le {2\over 5n_0} \cr
&< \varepsilon.\cr}$$
This shows $x_n \to {3\over 5}$ as $n \to \infty$.   
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IV.  True-False.  For each true statement, give a short proof or reason.  For each false
statement, give an explicit counterexample.
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\item{A)} (10)  If $A$ and $B$ are two nonempty bounded sets of real numbers and 
${\rm lub}(B) > {\rm lub}(A)$, then $y > x$ for all $y \in B$ and all $x\in A$.
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\item{} {\it Solution:}  This is FALSE.  A counterexample would be 
something like $A = [0,1]$ and $B = [-1, 2]$
Then ${\rm lub}(A) = 1$ and ${\rm lub}(B) = 2$, but $B$ also contains
elements smaller than elements in $A$:  $-1 \in B$ but $-1 < x$ for some $x \in A$
(in fact all $x$ in $A$, but that would not be necessary to get a counterexample).  
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\item{B)} Let 
$$A = \{x \in {\bf R} \mid 0 < x < 1,\ {\rm and }\ x = r\sqrt{2}\ {\rm for\ some}\ 
r \in {\bf Q}\}.$$
Then ${\rm lub}(A) = 1$. 
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\item{} {\it Solution:}  This is TRUE.  First, $x < 1$ for all $x\in A$
by definition.  Now, let $0 < a < 1$ be any real number.  
We claim that there are elements of $A$ between $a$ and $1$, so that 
$a$ cannot be an upper bound for $A$.   To see this, note
that by a theorem we proved in class, there must be rational numbers $r$ in the interval 
$\left({a\over \sqrt{2}}, {1\over {\sqrt 2}}\right)$.  So for any such $r$, 
$a < r\sqrt{2} < 1$.  Hence $r \sqrt{2} \in A$.  This shows that $a$ cannot 
be an upper bound for $A$.  Hence ${\rm lub}(A) = 1$.   
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{\it Extra Credit} (10)  Is it possible to produce a sequence $x_n$ whose
terms include all the positive and negative integers? 
If so, give an indication how to construct such a sequence.
If not, give a reason why there cannot exist such a sequence.
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{\it Solution:}  There is such a sequence, and an example can be constructed like this:
Let 
$$x_n = \cases{{1 - n \over 2} & if $n$ is odd\cr
                {n \over 2} & if $n$ is even\cr}$$
Listing out the first few terms we see
$$0,1,-1,2,-2,3,-3,\ldots$$
so it is clear that every integer will be in the set of terms of the sequence.
The positive integers come from $n$ even and the non-positive integers come from $n$ odd. 
\bye
