\headline={\bf Math 4606, Summer 2003:\it \hskip .1in  Binary
Searches\hfill\rm\howmany}
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\def\dable{differentiable} 
\def\dably{differentiability}
\def\dve{derivative} 
\def\th{\theta} 
{\bf Introduction}
\gap
``Binary Search'' really refers to a \sla{method} for finding a number with a
desired property, or else demonstrating that such a number does not exist. For
example, how does we find a needle in a haystack? Maybe we'd have a strong magnet,
knowing that the needle is made of steel, and we'd need permission to dismantle the
haystack. But we'd have to reassemble it, I suppose. First, we'd divide the stack
in half, then use the magnet to find out which half the needle is in. If we could
not tell, we'd divide both halves in half. If we could, we'd divide the half with
the needle in half again, use the magnet, and keep on until we find it. That's the
whole idea.
\gap
We'll use the Binary Search Method in some examples.
\gap
{\bf Finding the supremum and infimum of a non-empty bounded set}
\gap
Let $S$ denote our nonempty bounded set. We are given that \te numbers $A$ and
$B$ that bound the set $S.$ That is, for every $x$ in $S,$ $B\le x\le A.$ Let's
assume that $B<A.$ Otherwise, $S$ has only one point, and we don't have to look for
our sup and inf! We know by the Completeness Axiom that the number
$\sup S$ exists. We need to find it.
\gap
Since $S$ is nonempty we know that at least one of the intervals $[B,\,(B+A)/2]$
and $[(B+A)/2,\,A]$ contains a point of $S$ (why?). We're looking for big elements
of $S$ so we choose the right-most of the two intervals. To avoid
complicated long expressions like $(B+(B+A)/2)/2$ we'll use a notational convention
in our search: we'll let our original interval be denoted $[L_1,\,R_1].$ Then, if
$[(B+A)/2,\,A]\cap S\ne \emptyset,$ we'll write $[L_2,\,R_2]$ for $[(B+A)/2,\,A].$
Otherwise, $[L_2,\,R_2]$ will stand for $[B,\,(B+A)/2].$ It's important to notice
that $L_1\le L_2$ and that $R_2\le R_1.$ It's not as important to notice that
exactly one of the statements $``L_1=L_2\quot$ and $``R_1=R_2\quot$ is true. 
\sla{It is very important} to notice that $R_2-L_2=(R_1-L_1)/2=(A-B)/2.$
\gap
We now repeat this process, dividing our chosen interval in half. Since the chosen
interval contained at least one point of $S$ one of the halves of the chosen
interval contains a point of $S$ (why?). {\bf We will always choose as our next
interval the half that is rightmost.} In this way, we get a \seq of intervals
$[L_n,\,R_n].$ According to our method of choosing intervals, there are no points
of $S$ larger than $R_n$ (why?). Thus each $R_n$ is an upper bound of $S.$ The \seq
$\{R_n\}$ is a decreasing \seq \bdd\ below by
$B.$ The \seq $\{L_n\}$ is an increasing \seq \bdd\ above by $A.$ Thus both \seqnd s
$\{L_n\}$ and $\{R_n\}$ converge. And
we know (by induction) that $0<R_n-L_n=2(A-B)/2^n\to 0.$ Thus both \seqnd s have
the same limit, say $P.$ 
\gap
{\bf Claim: } $P=\sup S.$ We need to show that $P$ satisfies the {\bf defining}
conditions for being the sup of $S.$ \sla{First: } $P$ is an upper bound for $S$
because if $y>P$ then for all $n$ sufficiently large, $0\le R_n<y,$ and $R_n$ is an
upper bound for $S.$ Thus no element of $S$ can be larger than $P$ (why?), so $P$
is an upper bound for $S.$ \sla{Second: } if $x<P,$ then there must exist $L_n>x$
because $L_n\to P$ (the argument is the same one we used when $y$ was bigger than
$P,$ but we used the numbers $R-n$ there). But then, since there must be a point of
$S$ in $[L_n,\,R_n],$ \tE a point of $S$ that exceeds $x.$ These two conditions are
the conditions that need to be true for a number to be the sup of a set.
\gap
{\bf Exercise: } Decide what has to be changed in the proof so far in order to find
the infimum of $S,$ then make the changes and check that you have a proof.
\gap
{\bf Showing that every real number is the limit of a \seq of rational numbers}
\gap
We suppose that $x_0$ is a given rational number. We want to find a \seq of
rational numbers that converges to $x_0.$ We may as well assume that $x_0$ is an
irrational number. Otherwise, we could let our \seq
$\{x_0,\,x_0,\,x_0,\,\dots,\,x_0,\,\dots\}.$ We will actually construct a \seq of
\sla{dyadic} rational numbers $r_n$ that converge to $x_0$ (the dyadic rational n
umbers are those whose denominators are powers of $2.$ We may also assume that
$x_0>0$ (why?). We know \tE a natural number $n_0$ larger than $x_0$ (why?).
This means that $x_0\in I_0:=[0,\,n_0].$ We divide $I_0$ in half. One of the halves
contains $x_0.$ Both halves cannot, since then $x_0$ would be the average of two
rational numbers, which is a rational number, and we assumed that $x_0$ is
irrational. We choose as $I_1$ the half of $I_0$ that contains $x_0.$ Both
endpoints of $I_1$ are dyadic rationals, since each endpoint is either one of the
original endpoints or is $n_0/2.$ 
\gap
We now repeat this process, dividing our chosen intervals in half. Since the chosen
intervals contain $x_0,$ one of the halves of each chosen
interval contains $x_0$ (why?). {\bf We will always choose as our next
interval the half that contains $x_0.$} In this way, we get a \seq of intervals
$[L_n,\,R_n].$ As before, the \seq $\{R_n\}$ is a decreasing \seq
\bdd\ below by $0$ and $\{L_n\}$ is an increasing \seq \bdd\ above by $n_0.$
Moreover, $R_n-L_n=2n_0/2^n\to 0,$ so (again) both \seqnd s converge, and they
converge to the same limit. By the Squeeze Principle, since $L_n<x_0<R_n$ we have 
$\lim_{n\to\infty}L_n=x_0=\lim_{n\to\infty}R_n.$ We prove by induction that each
$R_n$ and each $L_n$ is a dyadic rational. This completes the proof. It's worth
noticing that each of our \seqnd s was monotone! This allows us to show that each
real number is the limit of a \seq of \sla{irrational} numbers: we just have to
show that between any two rational numbers there is a rational multiple of (for
example) $\sqrt{2}.$
\gap
{\bf The Bolzano-Weierstrass Theorem: } \sla{Every bounded infinite set of real
numbers has a limit point.}
\gap
\Pf We have to show this: for every set $S\sub \real$ that is a bounded infinite
set, there exists a point $y\in\real$ that is a limit point ``for'' $S.$
\gap
Since $S$ is bounded, \te numbers $A$ and
$B$ \st for every $x$ in $S,$ $B\le x\le A.$
We define $I_0:=[L_0,\,R_0]:=[A,\,B].$ We will divide $I_0$ into its two halves
(they overlap at the midpoint), and select as the interval
$I_1$ the left-most half that has infinitely many element of $S$ in it.
One of the halves \sla{must} contain infinitely many element of $S$ (why?).
We name the left and right endpoints of $I_1$ $L_1$ and $R_1,$ \rspy.
\gap
We then continue the process, always selecting as $I_{n+1}$ the left-most half
of $I_n$ that contain infinitely many element of $S.$
\gap
Now $\{L_n\}$ is an increasing \seq \bdd\ above by $A$ and $\{R_n\}$ is a
decreasing \seq \bdd\ below by $B.$ Moreover, $R_n-L_n=(A-B)/2^n,$ so both \seqnd s
converge and they converge to the same limit, say $y.$ Our task now is to verify
that $y$ is a limit point ``for'' $S.$
\gap
Let $\del>0$ be given. We need to show that $(y-\del,\,y+\del)$ contains a point of
$S$ that is different than $y.$ 
\gap
Since $\{L_n\}$ is an increasing \seq that
converges to $y,$ we know that $L_n\le y$ \fA $n.$ Similarly, $y\le R_n$ \fA $n.$
\gap
We now let $\del$ play the r\^ole of $\ep$ in the definition of convergence
of $\{L_n\}$ to $y.$ We know \tE $N_1\in\nat$ \st $n\ge N_1\impl
|L_n-y|=y-L_n<\del.$ Similarly, \tE $N_2\in\nat$ \st $n\ge N_2\impl
|R_n-y|=R_n-y<\del.$ We now select $N:=\max\{N_1,\,N_2\}.$ 
\gap
By our ``construction'' of the intervals $I_n,$ we know that $I_N=[L_N,\,R_N]$
contains infinitely many points of $S.$ We will show that 
$[L_N,\,R_N]\sub (y-\del,\,y+\del).$ Once we do so, we will know that 
$(y-\del,\,y+\del)$ contains infinitely many points of $S,$ hence must contain 
two points of $S$ that are different from each other. At least one of them must be
unequal to $y.$ This will complete the proof.
\gap
To show that $[L_N,\,R_N]\sub (y-\del,\,y+\del)$ we will show that 
$L_N>y-\del$ and show that $R_N<y+\del.$ 
\gap
Now, $N\ge N_1,$ so $|L_N-y|=y-L_N<\del,$ so $L_N>y-\del.$
\gap
{\bf Exercise: } Complete the proof by showing that $R_N<y+\del.$
\bye  
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