diff git a/ibl.cls b/ibl.cls
index b9c50a5a43b172c0717e21a74299ef22d6388df5..001f38ffc787835f860c0b01919f5f6d926f8e4b 100644
 a/ibl.cls
+++ b/ibl.cls
@@ 3,7 +3,7 @@
% Jim Hefferon jhefferon (at) smcvt.edu
%  Class structure: identification part
% 
\ProvidesClass{test}[2013/04/13 version 1.1 of Inquirybased Learning book class (Jim Hefferon)]
+\ProvidesClass{ibl}[2013/04/13 version 1.1 of Inquirybased Learning book class (Jim Hefferon)]
\NeedsTeXFormat{LaTeX2e}
% ========================================
@@ 342,6 +342,24 @@ qed=\qedsymbol
% \newcommand{\remark}{\textit{Remark:} }
+% For alternate complement notation
+% credit: Danie Els, https://tex.stackexchange.com/a/22101
+% \newcommand*\xoverline[2][0.75]{%
+% \sbox{\myboxA}{$\m@th#2$}%
+% \setbox\myboxB\null% Phantom box
+% \ht\myboxB=\ht\myboxA%
+% \dp\myboxB=\dp\myboxA%
+% \wd\myboxB=#1\wd\myboxA% Scale phantom
+% \sbox\myboxB{$\m@th\overline{\copy\myboxB}$}% Overlined phantom
+% \setlength\mylenA{\the\wd\myboxA}% calc width diff
+% \addtolength\mylenA{\the\wd\myboxB}%
+% \ifdim\wd\myboxB<\wd\myboxA%
+% \rlap{\hskip 0.5\mylenA\usebox\myboxB}{\usebox\myboxA}%
+% \else
+% \hskip 0.5\mylenA\rlap{\usebox\myboxA}{\hskip 0.5\mylenA\usebox\myboxB}%
+% \fi}
+
+
% ====================================================
% VERSION number
\newcommand{\version}{1.1}
diff git a/ibl.sty b/ibl.sty
index f5b067daaa931bf56a64e629a80dfb728e7fe06a..ca67834d4b3d39b9e0d6ecd039c3f872d5c5683e 100644
 a/ibl.sty
+++ b/ibl.sty
@@ 29,7 +29,7 @@
\RequirePackage{euscript}
\newcommand{\partition}[1]{\EuScript{#1}}
% mathdots fot dots from lowerleft to upper right
+% mathdots for dots from lowerleft to upper right
\RequirePackage{mathdots}
% Misc commands
diff git a/ibl.tex b/ibl.tex
index 0bbb14d8669a1709c61457461bc4996bb68a826b..73f0a2c377e1264657d7b941c53170fb35bcbe91 100644
 a/ibl.tex
+++ b/ibl.tex
@@ 35,7 +35,7 @@
\pagestyle{bodypage}
\chapter{Numbers}
We begin with results about the integers $\Z=\set{\ldots,2,1,0,1,2,\ldots}$.
+We begin with results about the integers $\Z=\set{\ldots\,2,1,0,1,2,\ldots}$.
In this chapter, ``number'' means integer.
Some statements refer to the natural numbers $\N=\set{0,1,2,\ldots}$
or the positive integers $\Z^+=\set{1,2,\ldots}$.
@@ 1884,13 +1884,13 @@ Let $p$ be prime. Prove each.\label{ex:EuclidsOtherLemma}
% and
% http://gowers.wordpress.com/2011/11/18/provingthefundamentaltheoremofarithmetic/
\begin{problem} \notetext{Fundamental Theorem of Arithmetic}
 Any number $n>1$ can be expressed as a product of primes
+ Any number $n>1$ can be expressed as a factorization into primes
$n=p_0^{e_0}p_1^{e_1}\cdots p_s^{e_s}$.
What's more, under the condition that the primes are distinct,
this expression is
 unique:~any two prime factorizations of~$n$ contain the same
 primes, to the same powers.
 We do this in two parts.
+ unique:~ for each prime $p_i$, any two prime factorizations of~$n$
+ have the same power $e_i$ of $p_i$.
+ We prove this in two parts.
\begin{exes}
\begin{exercise}
Prove that any $n>1$ can be written as a product of primes.
@@ 2066,29 +2066,29 @@ Together these show that $\gcd(a,b)\cdot\lcm(a,b)=ab$.)
%===================================================
\chapter{Sets}
\begin{df}
A \definend{set} is a collection that is definite\Dash every
object either
definitely is contained in the collection or definitely is not.
+%\begin{df}
+A \definend{set} is a collection that is definite, that is, welldetermined,
+so that every thing either definitely is contained in the collection
+or definitely is not.
An object~$x$ that belongs to a set~$A$ is an \definend{element}
or \definend{member}
of the set, written $x\in A$
(to denote that $x$ is not an element of $A$ write~$x\notin A$).
+of that set, denoted $x\in A$.
+We denote that $x$ is not an element with~$x\notin A$.
Two sets are equal if and only if they have the same elements.
\end{df}
\noindent Read `$\in$' as ``is an element of'' rather than ``in'' to avoid
+% \end{df}
+%\noindent
+(Read `$\in$' as ``is an element of'' rather than ``in'' to avoid
confusion between this and
the subset relation defined below.
As a synonym for set we sometimes say ``collection.''

We usually specify a set either by listing it or by
describing its elements.
Thus we may write
the set of the primes less than ten either
as $P=\set{2, 3, 5, 7}$
or as~$P=\setbuilder{p\in\N}{\text{$p$ is prime and $p<10$}}$
(read the vertical bar as ``such that''; some authors use a colon~`$:$' instead
of a vertical bar).
+the subset relation defined below.)
+% As a synonym for set we sometimes say ``collection.''
+
+We usually describe a set either by listing its elements or by
+stating a criteria for membership.
+Thus we can write the set of primes less than ten as
+$\set{2, 3, 5, 7}$
+or as $\setbuilder{p\in\N}{\text{$p$ is prime and $p<10$}}$
+(read the vertical bar as ``such that'';
+some authors instead use a colon,~`$:$').
\begin{problem}[\midlength]
Decide if each is true and justify your decision.
@@ 2103,9 +2103,12 @@ Decide if each is true and justify your decision.
\begin{items}
\item These sets are equal because they contain the same elements,
$1$, $3$, and~$5$.
+ With sets, order does not matter.
\item These sets are equal because they contain the same elements,
$2$, $4$, and~$6$.
\item These sets are not equal; for one thing, the set on the right contains~$0$
+ With sets, repeats collapse.
+\item These are not equal sets.
+ For one thing, the set on the right contains~$0$
while the set on the left does not.
\item This is false.
The set on the right has three elements,
@@ 2124,11 +2127,12 @@ We write $B\subseteq A$.
\end{df}
\begin{df}
The set without any elements is the \definend{empty set} $\emptyset$.
+The set without any elements is the \definend{empty set}, denoted $\emptyset$.
\end{df}
\begin{problem} Decide each, with justification.
\begin{items}
+\item $\set{1, 3, 5}\subseteq\set{1, 3, 5, 7, 9}$
\item $\set{1, 3, 5}\in\set{1, 3, 5, 7, 9}$
\item $\set{1,3,5}\subseteq\setbuilder{n\in\N}{\text{$n$ is prime}}$
\item $\emptyset\subseteq\set{1, 2, 3, 4}$
@@ 2138,11 +2142,14 @@ The set without any elements is the \definend{empty set} $\emptyset$.
\end{items}
\begin{answer}
\begin{items}
\item This is true because, by inspection, for each element of the set
 on the left, $1$, $3$, and~$5$, it is an element of the set on the right.
\item The set $\set{1,3,5}$ is not an element of the set on the right.
 All elements of the set on the right are numbers.
\item This is not true since~$1$ is an element of the set on the left
+\item This is true.
+ By inspection, each element of the set
+ on the left, $1$, $3$, and~$5$, is an element of the set on the right.
+\item This is false.
+ The set $\set{1,3,5}$ is not an element of the set on the right.
+ The set $\set{1,3,5}$ is not equal to the number $1$ from the set on the
+ right, it is not equal to the number $3$, \ldots\@
+\item This is false because~$1$ is an element of the set on the left
but not of the set on the right.
\item This is true since $x\in\emptyset$ implies that $x\in\set{1, 2, 3, 4}$.
(The implication is vacuous.)
@@ 2164,10 +2171,15 @@ Prove.
\begin{answer}
For the first, let $x$ be an element of the set on the left, $A$.
Then $x$ is an element of the set on the right, $A$, obviously.
+
For the second, observe that the implication
`if $x\in\emptyset$ then $x\in A$' is true
 because any implication with a false first clause is true
 (that is, the implication is vacuously true).
+ because any implication with a false antecedent clause is true.
+ (That is, the implication is vacuously true.
+ Another way to express this is to note that this implication has
+ no counterexamples\Dash you cannot find a number that is an element of
+ $\emptyset$ but that is not an element of~$A$, simply because there
+ are no numbers that are element of~$\emptyset$.)
\end{answer}
\begin{exercise}
The empty set is unique: if the set~$A$ is empty and
@@ 2197,13 +2209,16 @@ Prove, for any sets $A$, $B$, and~$C$.
\end{exercise}
\begin{answer}
Suppose that $x\in A$.
 Because $A\subseteq B$ we have that~$x\in B$ and with that, because
 $B\subseteq C$ we have that $x\in C$.
 Thus $x\in A$ implies that $x\in C$, and therefore~$A\subseteq C$.
+ Because $A\subseteq B$ we have that~$x\in B$.
+ With that, because
+ $B\subseteq C$, we have that $x\in C$.
+ Thus $x\in A$ implies that $x\in C$ and therefore~$A\subseteq C$.
\end{answer}
\end{exes}
\end{problem}
+Mutual inclusion is the most common way to show that two sets are equal.
+
\begin{problem}[\maxlength]
For each, give an example of three sets satisfying the conditions,
or show that no example is possible.
@@ 2221,14 +2236,16 @@ or show that no example is possible.
\end{ans}
\end{problem}
In this book statements talk about things inside of
some \definend{universal set},
+Usually mathematical statements are in a context of
+some \definend{universe},
denoted $\universalset$.
For instance, in the first chapter on number theory
the universal set is the integers~$\universalset=\Z$.
+For instance, in the first chapter
+the universe is the integers,~$\universalset=\Z$.
There, if we say that we are considering
the set of things less than~$100$ then
we are considering the set of integers less than~$100$.
+Another example is that in first semester calculus
+the universe is the set of real numbers,~$\universalset=\R$.
%% \begin{df}
%% The \definend{characteristic function} of a set~$A$ is a map
@@ 2250,19 +2267,21 @@ themselves~$D=\setbuilder{S}{S\notin S}$.
Show that assuming $D$ is an element of itself leads to a contradiction.
\end{exercise}
\begin{answer}
 If~$D$ were to have itself as an element
 then it wouldn't satisfy the condition
 $S\notin S$, so $D$ would not be in the collection
 $\setbuilder{\text{set~$S$}}{S\notin S}$.
 That is, in this case $D\notin D$, a contradiction.
+ Suppose that $D$ were to have itself as an element, $D\in D$.
+ Then $D$ would not be an element of the collection
+ $\setbuilder{\text{set~$S$}}{S\notin S}$ because it wouldn't satisfy the
+ condition $S\notin S$.
+ Thus $D\notin D$, a contradiction.
\end{answer}
\begin{exercise}
Show that assuming $D$ is not an element of itself also leads to a
contradiction.
\end{exercise}
\begin{answer}
 If $D$ does not have itself for an element then it satisfies the
 condition $S\notin S$, so $D\in D$, a contradiction.
+ Suppose that $D$ were to not have itself as an element, $D\notin D$.
+ Then $D$ would be an element of the collection
+ $\setbuilder{\text{set~$S$}}{S\notin S}$.
+ Thus $D\in D$, which is a contradiction.
\end{answer}
\end{exes}
\end{problem}
@@ 2277,17 +2296,16 @@ themselves~$D=\setbuilder{S}{S\notin S}$.
\begin{df}
The \definend{complement} of a set~$A$, denoted $\setcomp{A}$, is the
set of objects that are not elements of~$A$.
+set of things that are not elements of~$A$.
+Some authors denote the complement with a bar, $\xoverline{A}$.
\end{df}
\noindent\remark
working inside of a universal set makes the complement
operation sensible.
+working inside of a universe makes the complement sensible.
For instance, in a number theory discussion where $\universalset=\Z$,
if we consider the set of things less than~$100$ then we can
take the complement and the result is another subset of
$\universalset$, so
we are still in number theory.
+$\universalset$, so we are still in number theory.
\begin{df}
Let $A$ and $B$ be sets.
@@ 2307,100 +2325,103 @@ Picture set operations with \definend{Venn diagrams}.
\hspace*{3em}
\grf{asy/venn_comp.pdf}{$\setcomp{A}$}
\end{center}
In each diagram
the region inside the rectangle depicts the universal set and the
region inside each circle depicts the set.
+In each diagram the region inside the rectangle depicts the universe and the
+region inside a circle depicts a set.
On the left the darker color shows
the union as containing all of the two sets joined,
+the union as containing all of the two sets,
the middle shows the intersection
containing only the region common to both,
and on the right the dark region, the complement, is all but the set~$A$.
+and on the right the complement is all but the set~$A$.
\begin{problem}
Another tool illustrating set relationships is this table describing
two sets.
+Another tool illustrating set relationships is this table.
+It describes
+two sets in the universe $\universalset=\set{0,1,2,3}$.
+(It uses $0$ and~$1$ in place of $F$ and~$T$ so that the right side of each
+row is the binary representation of its number.)
\begin{center} \small
 \begin{tabular}{rcc}
 \multicolumn{1}{r}{\textit{row number}} &$x\in A$ &$x\in B$ \\ \hline
+ \begin{tabular}{ccc}
+ \multicolumn{1}{r}{\textit{number}} &$x\in A$ &$x\in B$ \\ \hline
$0$ &$0$ &$0$ \\
$1$ &$0$ &$1$ \\
$2$ &$1$ &$0$ \\
$3$ &$1$ &$1$
\end{tabular}
+ \hspace*{3em}
+ \vcenteredhbox{\includegraphics{asy/venn_two.pdf}}
\end{center}
It uses $0$ and~$1$ instead of $F$ and~$T$ so that the right side of each
row is the binary representation of its row number.
The top row says that $0\notin A$ and $0\notin B$,
+The table's top row says that $0\notin A$ and $0\notin B$,
while the second row says that $1\notin A$ but $1\in B$.
In total, $A=\set{2,3}$, $B=\set{1,3}$,
and~$\universalset=\set{0,1,2,3}$.
For each of these simple results about set operations,
apply the statement to the sets $A$ and~$B$.
Also pick one and prove it.
+apply the statement to these example sets.
+Also, pick one statement and prove it.
\begin{items}
\item the complement of the complement is the original set
 $\setcomp{(\setcomp{A})}=A$
+ $\smash{\setcomp{(\setcomp{A})}}=A$
\item $A\intersection\emptyset=\emptyset$ and $A\union\emptyset=A$
\item \notetext{Idempotence} $A\intersection A=A$ and $A\union A=A$
\item $A\intersection B\subseteq A\subseteq A\union B$
\item \notetext{Commutativity}
$A\intersection B=B\intersection A$ and
$A\union B=B\union A$
\item \notetext{Associativity}
 $(A\intersection B)\intersection C=A\intersection (B\intersection C)$
 and
 $(A\union B)\union C=A\union (B\union C)$
+ \item \notetext{Associativity}
+ $(A\intersection B)\intersection C=A\intersection (B\intersection C)$
+ and
+ $(A\union B)\union C=A\union (B\union C)$
+ (extend the table to add a third set,~$C$).
\end{items}
\begin{answer}
For convenience here is the defintion of the sets again, as well
as a Venn diagram.
+For convenience here is the table again, along with the Venn diagram.
\begin{center} \small
\begin{tabular}{ccc}
 $x\in A$ &$x\in B$ &row number \\ \hline
+ $x\in A$ &$x\in B$ &\multicolumn{1}{l}{\textit{number}} \\ \hline
$0$ &$0$ &$0$ \\
$0$ &$1$ &$1$ \\
$1$ &$0$ &$2$ \\
$1$ &$1$ &$3$
\end{tabular}
 \hspace*{3em}
 \vcenteredhbox{\includegraphics{asy/venn_two.pdf}}
+ \hspace*{3em}
+ \vcenteredhbox{\includegraphics{asy/venn_two.pdf}}
\end{center}
\begin{items}
\item Applied to the set~$A$, the complement is
 $\setcomp{A}=\set{0,1}$ while
 the complement of the complement is~$\setcomp{\setcomp{A}}=\set{2,3}$.
+\item Applying the statement to the example, with this
+ universe $\universalset=\set{0,1,2,3}$, the complement is
+ $\setcomp{A}=\set{0,1}$, and
+ complement of the complement is~$\setcomp{\setcomp{A}}=\set{2,3}$.
 The proof
 is by mutual inclusion, that both~$A\subseteq\setcomp{(\setcomp{A})}$
 and~$\setcomp{(\setcomp{A})}\subseteq A$.
+ We prove this
+ by mutual inclusion, that both~$A\subseteq\smash{\setcomp{(\setcomp{A})}}$
+ and~$\smash{\setcomp{(\setcomp{A})}}\subseteq A$.
For the first inclusion suppose that $a\in A$.
By the definition of complement $a\notin\setcomp{A}$ and
 so $a\in\setcomp{(\setcomp{A})}$.
 Thus $A\subseteq\setcomp{(\setcomp{A})}$.
+ thus $a\in\smash{\setcomp{(\setcomp{A})}}$.
+ That shows $A\subseteq\smash{\setcomp{(\setcomp{A})}}$.
Inclusion in the other direction is similar.
 If $x\in\setcomp{(\setcomp{A})}$
+ If $x\in\smash{\setcomp{(\setcomp{A})}}$
then $x\notin\setcomp{A}$ and
 by definition of the complement $x\in\setcomp{A}$.
 Thus $\setcomp{(\setcomp{A})}\subseteq A$.
+ by definition of the complement, $x\in\setcomp{A}$.
+ Thus $\smash{\setcomp{(\setcomp{A})}}\subseteq A$.
\item The set $A\intersection\emptyset$
is the collection of
 elements of~$\universalset$ that are members of both~$A$ and~$\emptyset$,
 so applied to $A=\set{2,3}$ and $\universalset=\set{0,1,2,3}$ it
+ things that are members of both~$A$ and~$\emptyset$.
+ Applied to $A=\set{2,3}$, it
is~$\set{}$.
 The second one is just as trivial: the collection of all of the members of
 $\universalset=\set{0,1,2,3}$ that are members of either~$A=\set{2,3}$
+ The second one is just as trivial:~the collection of all of things
+ that are members of either~$A=\set{2,3}$
or~$\emptyset=\set{}$ is~$\set{2,3}$.
The proof of the first equality is: $x\in A\intersection\emptyset$
 iff $x\in A$ and~$x\in\emptyset$,
 which, because an `and' statement holds if and only if both halves hold
+ iff $x\in A$ and~$x\in\emptyset$.
+ Because an `and' statement holds if and only if both halves hold
and here the second clause never holds,
 is equivalent to $x\in\emptyset$.
+ this is true if and only if $x\in\emptyset$.
For the second equality,
 $x\in A\union\emptyset$ iff either $x\in A$ or~$x\in\emptyset$,
 which is equivalent to $x\in A$.
+ $x\in A\union\emptyset$ iff either $x\in A$ or~$x\in\emptyset$.
+ An `or' statement holds if either of its two clauses holds,
+ and here the second clause never holds,
+ so this is equivalent to $x\in A$.
\item The first statement applied to the set $A=\set{2,3}$
says that the collection of members of both
$\set{2,3}$ and~$\set{2,3}$ is the set~$A=\set{2,3}$.
@@ 2410,23 +2431,24 @@ as a Venn diagram.
$x\in A\intersection A$ if and only if
both $x\in A$ and~$x\in A$,
which is true if and only if $x\in A$.
 The proof of the second is similar.
+ The second's proof is similar.
\item The statement has two containments.
 The application of the statement on the left is that
+ The application of $A\intersection B\subseteq A$ to the example sets is that
$A\intersection B=\set{3}\subseteq A=\set{2,3}$.
 On the right the application is~$A=\set{2,3}\subseteq A\union B=\set{1,2,3}$.
+ The application of $A\subseteq A\union B$
+ is~$A=\set{2,3}\subseteq A\union B=\set{1,2,3}$.
 To prove the lefthand containment $A\intersection B\subseteq A$ observe that
+ To prove that $A\intersection B\subseteq A$, observe that
$x\in A\intersection B$ implies that
$x\in A$ and $x\in B$, and so in particular $x\in A$.
 The righthand containment's argument is that
 $x\in A$ implies that $x\in A$ or~$x\in B$, and so $A\subseteq A\union B$.
+ The argument for the righthand containment $A\subseteq A\union B$ is that
+ $x\in A$ implies that $x\in A$ or~$x\in B$.
\item For the intersections the application gives that
$A\intersection B$ is the collection of elements common to
$\set{2,3}$ and $\set{1,3}$, which is $\set{3}$,
and $B\intersection A=\set{3}$ also.
 The application to the statement about unions is similar:
+ The application of the statement about unions is similar:
$A\union B=\set{1,2,3}=B\union A$.
For the proof that
@@ 2436,16 +2458,16 @@ as a Venn diagram.
=\setbuilder{x}{\text{$x\in B$ and $x\in A$}}=B\intersection A$.
(The middle equality holds by the commutativity of `and'.)
 Equality for union is similar since, like `and', the connective
+ Equality for union is similar since, like `and', the logical connective
`or' is also commutative.
\item This is a three set relationship so we use this table.
+\item Here is the extended table and the associated Venn diagram.
\begin{center} \small
\begin{tabular}{cccc}
 $x\in A$ &$x\in B$ &$x\in C$ &row number \\ \hline
+ $x\in A$ &$x\in B$ &$x\in C$ &\multicolumn{1}{l}{\textit{number}} \\ \hline
$0$ &$0$ &$0$ &$0$ \\
$0$ &$0$ &$1$ &$1$ \\
$0$ &$1$ &$0$ &$2$ \\
 $0$ &$1$ &$1$ &$3$ \\[.5ex] \hline
+ $0$ &$1$ &$1$ &$3$ \\[.5ex] % \hline
$1$ &$0$ &$0$ &$4$ \\
$1$ &$0$ &$1$ &$5$ \\
$1$ &$1$ &$0$ &$6$ \\
@@ 2454,9 +2476,11 @@ as a Venn diagram.
\hspace*{3em}
\vcenteredhbox{\includegraphics{asy/venn_three.pdf}}
\end{center}
 To apply associativity of intersection, $A\intersection B=\set{6,7}$ and so
+ To apply associativity of intersection to these example sets,
+ $A\intersection B=\set{6,7}$ and so
$(A\intersection B)\intersection C=\set{7}$, while
 $B\intersection C=\set{3,7}$ so $A\intersection(B\intersection C)=\set{7}$.
+ $B\intersection C=\set{3,7}$
+ and so $A\intersection(B\intersection C)=\set{7}$.
Associativity of union is similar.
The proof that intersection is associative goes
@@ 2510,13 +2534,46 @@ as a Venn diagram.
% \end{problem}
\begin{problem}[\maxlength]
Prove that the following are equivalent:
+Prove that the following statements are equivalent:
(i)~$A\subseteq B$,
(ii)~$A\union B=B$,
and (iii)~$A\intersection B=A$.
+(\hint~one approach is to
+show that (i) implies~(ii),
+that (ii) implies~(iii),
+and that (iii) implies~(i).)
\begin{answer}
One way to prove this is to
show that (i) holds iff (ii)~holds, and also show that (i) holds iff~(iii)
+We follow the hint.
+For (i) implies~(ii), assume that $A\subseteq B$.
+To show that $A\union B=B$ we do mutual inclusion.
+One of those inclusions, that $B\subseteq A\union B$, holds by the definition
+of union.
+So, we only need to show that $A\union B\subseteq B$.
+For that, let $x$~be an element of $A\union B$ and then
+there are two cases, that $x\in A$ or that $x\in B$, and
+we must verify that $x\in B$ in both cases.
+The $x\in A$ case is
+immediate from the $A\subseteq B$ assumption of~(i).
+The second case, the $x\in B$ case, is trivial.
+
+For (ii) implies~(iii), assume that $A\union B=B$.
+We show that $A\intersection B=A$ by mutual inclusion.
+Half, that $A\intersection B\subseteq A$, is one of the parts of the
+prior exercise.
+For the other half, that $A\subseteq A\intersection B$, assume that $x\in A$.
+Then $x\in A\union B$ and since we've assumed that $A\union B=B$,
+we have $x\in B$.
+Thus $x\in A\intersection B$.
+
+The third implication is that (iii) implies~(i).
+Assume that $A\intersection B=A$.
+Let $x\in A$.
+Then $x\in A\intersection B$ and so $x\in B$.
+
+\smallskip
+\noindent\textit{Alternate proof.}
+Another way to prove this is to
+show that (i) holds iff (ii)~holds, and also to show that (i) holds iff~(iii)
holds.
The second logical equivalence
is very like the first so here we only show the first.
@@ 2548,8 +2605,8 @@ The \definend{symmetric difference} is
$A\symdiff B=(AB)\union(BA)$.
\end{df}
\noindent If $A\subseteq X$ then $XA$ is the same as $\setcomp{A}$
where $X$ is the universal set $\universalset=X$.
+\noindent\remark if $A\subseteq X$ then $XA$ is the same as $\setcomp{A}$
+where $X$ is the universe, $\universalset=X$.
% \begin{problem}
% Draw the Venn diagram for difference and symmetric difference.
@@ 2580,17 +2637,17 @@ where $X$ is the universal set $\universalset=X$.
(and thus $AB\subseteq A$).
\end{exercise}
\begin{answer}
 We will show the sets are equal by mutual inclusion.
 For the lefttoright inclusion $AB\subseteq A\intersection \setcomp{B}$
+ We will show these are equal by mutual inclusion.
+ For the $AB\subseteq A\intersection \setcomp{B}$ inclusion,
suppose that $x\in AB$.
By the definition of set difference both $x\in A$ and $x\notin B$ hold.
Therefore $x$ is an element of the intersection
$x\in A\intersection\setcomp{B}$.
 For the $A\intersection\setcomp{B}\subseteq AB$ inclusion suppose that
+ For the $A\intersection\setcomp{B}\subseteq AB$ inclusion, suppose that
$x\in A\intersection\setcomp{B}$.
Then $x\in A$ and $x\in\setcomp{B}$, so $x\notin B$.
 By the definition of set difference then, $x\in AB$.
+ Thus, by the definition of set difference, $x\in AB$.
\end{answer}
\begin{exercise}[\midlength]
For all pairs of sets, $AB=BA$.
@@ 2611,37 +2668,39 @@ where $X$ is the universal set $\universalset=X$.
\end{problem}
\begin{problem}
\notetext{De Morgan's Laws} Prove, for all sets $A$, $B$, and~$C$.
+Prove, for all sets $A$, $B$, and~$C$.
\begin{exes}
\begin{exercise} \notetext{De Morgan's Laws}
$\setcomp{(A\intersection B)}=\setcomp{A}\union\setcomp{B}$
and
$\setcomp{(A\union B)}=\setcomp{A}\intersection\setcomp{B}$
\end{exercise}
+\Writetofile{answers}{\protect\item[\protect\remark] these two are
+ consequences of the fact that logical negation
+ satisfies that the logic statement `not($P$ and $Q$)' has the same truth value
+ as `not($P$) or not($Q$)'.
+}
\begin{answer}
 \remark these two are a consequence of the fact that logical negation
 satisfies that `not($P$ and $Q$)' has the same truth value, for any
 values of $P$ and~$Q$, as `not($P$) or not($Q$)'.

We show $\setcomp{(A\intersection B)}=\setcomp{A}\union\setcomp{B}$
by mutual containment.
The other identity is similar.
Suppose first that $x\in \setcomp{(A\intersection B)}$.
 Then $x\notin A\intersection B$, so
 either $x\notin A$ or~$x\notin B$,
+ Then $x\notin A\intersection B$.
+ So either $x\notin A$ or~$x\notin B$,
and so $x\in \setcomp{A}\union\setcomp{B}$.
Thus $\setcomp{(A\intersection B)}\subseteq\setcomp{A}\union\setcomp{B}$.
Suppose now that $x\in \setcomp{A}\union\setcomp{B}$.
Then $x\in\setcomp{A}$ or~$x\in\setcomp{B}$.
 Restated, that is $x\notin A$ or $x\notin B$,
 which gives that $x\notin (A\intersection B)$, and
+ Restated, either $x\notin A$ or $x\notin B$.
+ Hence $x\notin (A\intersection B)$, and
so $x\in\setcomp{(A\intersection B)}$.
Thus $\setcomp{A}\union\setcomp{B}\subseteq\setcomp{(A\intersection B)}$.
 \remark we could combine those paragraphs into a single chain of
 double implications, that
+ \smallskip
+ \noindent\textit{Alternative proof.}
+ The statement
$x\in \setcomp{(A\intersection B)}$ holds if and only if
$x\notin A\intersection B$, which is true if and only if
either $x\notin A$ or~$x\notin B$,
@@ 2653,8 +2712,11 @@ where $X$ is the universal set $\universalset=X$.
$A\intersection (B\union C)=(A\intersection B)\union (A\intersection C)$
\end{exercise}
\begin{answer}
 For $A\union (B\intersection C)=(A\union B)\intersection (A\union C)$
 first suppose that $x\in A\union (B\intersection C)$.
+ We will do $A\union (B\intersection C)=(A\union B)\intersection (A\union C)$
+ by mutual containment;
+ the other is similar.
+
+ First suppose that $x\in A\union (B\intersection C)$.
Then $x\in A$ or $x\in B\intersection C$.
In the case that $x\in A$ we have both that $x\in A\union B$ and that
$x\in A\union C$.
@@ 2673,9 +2735,6 @@ where $X$ is the universal set $\universalset=X$.
So in this case also $x\in A\union (B\intersection C)$.
Thus
$(A\union B)\intersection (A\union C)\subseteq A\union (B\intersection C)$.

 Therefore, by mutual containment, the two sets are equal.
 The other identity is similar.
\end{answer}
\end{exes}
\end{problem}
@@ 2698,23 +2757,18 @@ Two sets are \definend{disjoint} if their intersection is empty.
\begin{problem}
Find three sets $A$, $B$, and $C$, such that
$A\intersection B\intersection C$ is empty but the sets are
 not pairwise disjoint, that is, none of $A\intersection B$,
 $A\intersection C$, or $B\intersection C$ is empty.
+ not pairwise disjoint, that is, $A\intersection B$,
+ $A\intersection C$, and $B\intersection C$ are all nonempty.
\begin{answer}
One example is $A=\set{5,6}$, $B=\set{3,6}$, and
$C=\set{3,5}$.
 \remark
+ \noindent\remark
As with many set examples,
 to get this one we used the threeset table repeated below with
 the diagram.
 Think of it as listing eight regions that could either be occupied or not.
 We want to have the overlap of all three, area~$7$, empty.
 But we want the pairwise overlaps, areas $3$, $5$, and~$6$, to
 be nonempty.
+ to get this one we started with the threeset table repeated below.
\begin{center} \small
\begin{tabular}{cccc}
 $x\in A$ &$x\in B$ &$x\in C$ &row number \\ \hline
+ $x\in A$ &$x\in B$ &$x\in C$ &\textit{number} \\ \hline
$0$ &$0$ &$0$ &$0$ \\
$0$ &$0$ &$1$ &$1$ \\
$0$ &$1$ &$0$ &$2$ \\
@@ 2727,6 +2781,12 @@ Two sets are \definend{disjoint} if their intersection is empty.
\hspace*{3em}
\vcenteredhbox{\includegraphics{asy/venn_three.pdf}}
\end{center}
+ To modify it to make the required example,
+ think of it as showing eight regions that could either be occupied or not.
+ We want that the overlap of all three, area~$7$, is empty.
+ We also want that the pairwise overlaps, areas $3$, $5$, and~$6$, are
+ nonempty.
+ That gives the three sets.
\end{answer}
\end{problem}
@@ 2734,92 +2794,113 @@ Two sets are \definend{disjoint} if their intersection is empty.
For a finite set~$A$, the \definend{order} $A$ is the number of elements.
\end{df}
% \begin{problem}
% For finite sets, if $A\subseteq B$ then $A\leq B$.
% \begin{ans}
% Since $A$ finite we can list its elements
% $A=\set{a_0,\ldots,a_{n1}}$.
% Similarly, $B$ is finite so we can list its elements, and because
% $A\subseteq B$, the elements of $A$ are elements of~$B$.
% Thus $B=\set{a_0,\ldots,a_{n1},b_n,\ldots,b_k}$.
% Clearly then $A\leq B$.
% \end{ans}
% \end{problem}
+\begin{problem}
+ For finite sets, if $A\subseteq B$ then $A\leq B$.
+\begin{ans}
+Since $A$ finite we can list its elements
+$A=\set{a_0,\ldots\,a_{n}}$.
+Similarly, $B$ is finite so we can list its elements, and because
+$A\subseteq B$, the elements of $A$ are elements of~$B$.
+Thus $B=\set{a_0,\ldots\,a_{n},b_{n+1},\ldots\,b_{n+k}}$,
+and $A\leq B$.
+\end{ans}
+\end{problem}
\begin{df}
For a set~$A$ the \definend{power set} $\powerset(A)$ is the set of all
+For a set~$A$, the \definend{power set} $\powerset(A)$ is the set of all
subsets of~$A$.
\end{df}
\begin{problem}[\midlength]
 List the elements of each power set
 $\powerset(\set{0,1})$,
 $\powerset(\set{0,1,2})$,
 $\powerset(\set{0})$, and
 $\powerset(\emptyset)$.
 Find the order of each.
\begin{answer}
The power set of the twoelement set
+ List the elements of each power set:
+ \begin{items}
+ \item $\powerset(\set{0,1})$,
+ \item $\powerset(\set{0,1,2})$,
+ \item $\powerset(\set{0})$, and
+ \item $\powerset(\emptyset)$.
+ \end{items}
+ State the order of each.
+\begin{answer}
+\begin{items}
+\item
+The power set of the twoelement set
$\powerset(\set{0,1})=\set{\emptyset, \set{0}, \set{1}, \set{0,1}}$
has four elements, that is, is of order four.
+has four elements.
+That is, the power set is of order four.
+\item
The power set of the threeelement set
\begin{equation*}
\powerset(\set{0,1,2})=\set{\emptyset,\set{0},\set{1},\set{2},
\set{0,1},\set{1,2},\set{0,2},\set{0,1,2}}
\end{equation*}
has eight elements.
+\item
The power set of the singleton $\set{0}$ has two elements
$\powerset(\set{0})=\set{\emptyset,\set{0}}$.
+\item
The power set of the empty set has one element
$\powerset(\emptyset)=\set{\emptyset}$.
+$\powerset(\emptyset)=\set{\emptyset}$.
+\end{items}
\end{answer}
\end{problem}
\begin{problem}
Let $A=\set{\emptyset,\set{\emptyset}}$.
Decide, and justify, whether each is true or false.
+Decide, and justify, whether each is true or false:
\begin{items}
\item
 $\emptyset\in\powerset(A)$,
 $\emptyset\subseteq\powerset(A)$
\item $\set{\emptyset}\in\powerset(A)$,
 $\set{\emptyset}\subseteq\powerset(A)$
+ $\emptyset\in\powerset(A)$,
\item
 $\set{\set{\emptyset}}\in\powerset(A)$,
 $\set{\set{\emptyset}}\subseteq\powerset(A)$
+ $\emptyset\subseteq\powerset(A)$,
+\item
+ $\set{\emptyset}\in\powerset(A)$,
+\item
+ $\set{\emptyset}\subseteq\powerset(A)$,
+\item
+ $\set{\set{\emptyset}}\in\powerset(A)$,
+\item
+ $\set{\set{\emptyset}}\subseteq\powerset(A)$.
\end{items}
+\Writetofile{answers}{\protect\item[] For any twoelement set
+ $\protect\set{a,b}$, the power set is
+ $\protect\powerset(\protect\set{a,b})
+ =\protect\set{\protect\emptyset,\protect\set{a},\protect\set{b},\protect\set{a,b}}$.
+ Taking $\protect\emptyset$ for~$a$ and $\protect\set{\protect\emptyset}$
+ for~$b$ we have
+ $\protect\powerset(A)=\protect\set{\protect\emptyset, \protect\set{\emptyset}, \protect\set{\protect\set{\protect\emptyset}},
+ \protect\set{\protect\emptyset, \protect\set{\protect\emptyset}}}$.
+}
\begin{answer}
\begin{items}
\item
The power set of a twoelement set $\set{a,b}$ is
$\powerset(\set{a,b})=\set{\emptyset,\set{a},\set{b},\set{a,b}}$.
Taking $\emptyset$ for~$a$ and $\set{\emptyset}$ for~$b$ we have
$\powerset(S)=\set{\emptyset, \set{\emptyset}, \set{\set{\emptyset}},
 \set{\emptyset, \set{\emptyset}}}$.

 So, the first clause is true by inspection;
 the empty set is the first element
 of $\powerset(A)$ listed.
 The second is true by the earlier result~\ref{ex:EmptySetUnique}
+ So $\emptyset\in\powerset(A)$ is true by inspection;
+ the empty set is the first element listed in $\powerset(A)$.
+
+\item
+ This is true by the earlier result, Exercise~\ref{ex:EmptySetUnique},
that the empty set is a subset of all sets.
\item
 Again the first is true by inspection,
+ This is true by inspection of the above,
since one of the elements of $\powerset(A)$ is
$\set{\emptyset}$. (It is listed second.)
 The second is true.
+\item
+ This true.
To verify that one set is a subset of another we can check that for all
elements~$x$ of the first, $x$ is also an element of the second.
 In this case the subset~$\set{\emptyset}$ has only one element to check,
 it is indeed an element of~$\powerset(A)$.
+ In this case the subset~$\set{\emptyset}$ has only one element to check
+ and it is indeed an element of~$\powerset(A)$.
+
+\item
+ This is true by inspection, since it is the element that is listed
+ third in $\powerset(A)$.
+
\item
 The first is true by inspection, since it is the element that is listed
 third.
 To verify the second,
+ To verify
that $\set{\set{\emptyset}}$ is a subset of~$\powerset(A)$
we can check that each of its elements are elements of the power set.
 It has only one element, namely~$\set{\emptyset}$.
+ It has only one element, namely the set~$\set{\emptyset}$.
That is indeed one of the elements of~$\powerset(A)$.
\end{items}
\end{answer}
@@ 2833,7 +2914,7 @@ This is true.
Let $A\subseteq B$ and consider $x\in\powerset(A)$.
Then~$x$ is a subset of~$A$.
Any subset of~$A$ is a subset of~$B$ by
the Transitivity property of~\ref{ex:PropertiesOfSubset},
+the Transitivity property of Exercise~\ref{ex:PropertiesOfSubset},
so $x$ is a subset of~$B$.
Therefore $\powerset(A)\subseteq\powerset(B)$.
\end{answer}
@@ 2845,8 +2926,8 @@ Where~$A$ is a finite set, prove that $\powerset(A)=2^{A}$.
We use induction on the number of elements in~$A$.
The base step is that~$A$ is empty and so has zeromany elements.
The power set of the empty set has
one element $\powerset(\emptyset)=\set{\emptyset}$ and
since $2^0=1$, the statement is true in this case.
+one element $\powerset(\emptyset)=\set{\emptyset}$,
+and $2^0=1$.
For the inductive step suppose that the statement is true for the
$n=0$, \ldots, $n=k$ cases and consider a set
@@ 2856,23 +2937,21 @@ Elements $S\in\powerset(A)$\Dash that is, subsets~$S\subseteq A$\Dash
fall into one of two categories: either $a_k\notin S$ or $a_k\in S$.
The inductive hypothesis gives that
the number of subsets in the first category is $2^{k}$.
Showing that
the second category also contains~$2^{k}$many elements will finish the proof
because then
the total of the two categories is
$2^{k}+2^{k}=2\cdot(2^{k})=2^{k+1}$.
We will show that the collection of subsets containing~$a_k$
+We will argue that the collection of subsets containing~$a_k$
has the same number of elements as the collection of subsets that do not
contain~$a_k$.
(As noted in the prior paragraph, this collection has $2^{k}$many elements.)
To prove that, with every subset~$S$ of $A$ that contains~$a_k$ we associate
+contain~$a_k$, namely $2^k$many elements.
+That will finish the proof
+because then the total of the two categories is
+$2^{k}+2^{k}=2\cdot(2^{k})=2^{k+1}$.
+
+For that argument, with every subset~$S$ of $A$ that contains~$a_k$ we associate
a subset $\hat{S}$ that does not contain~$a_k$,
namely $\hat{S}=S\set{a_k}$.
Clearly for each such~$S$ there is one and only one such~$\hat{S}$.
So, the two collections\Dash the set of subsets of $A$ containing~$a_k$
and the set of subsets of~$A$ that do not contain~$a_k$\Dash match
up elementbyelement, and therefore have the same number of elements.
+Clearly for each~$S$ there is one and only one such~$\hat{S}$.
+Therefore the two collections, the set of subsets of $A$ that contain~$a_k$
+and the set of subsets of~$A$ that do not, match
+up elementbyelement and thus have the same number of elements.
\end{answer}
\end{problem}
@@ 3067,7 +3146,7 @@ Thus $A^0=B^0=\set{\sequence{}}$.
$\setcomp{A}\times\setcomp{B}$.
\end{exercise}
\begin{answer}
 Fix the universal set~$\universalset=\Z$.
+ Fix the universe~$\universalset=\Z$.
If
$A=\set{0}$ and $B=\set{1}$ then $A\times B=\set{\sequence{0,1}}$,
so $\setcomp{(A\times B)}=\Z^2\set{\sequence{0,1}}$,
@@ 3247,7 +3326,7 @@ Thus the number of possible function is $C^{D}$.
\begin{df}
The \definend{characteristic function} of a set~$A$ is a map
$\charfcn{A}$ (or $\mathbb{1}_A$), whose domain is the universal set, such that
+$\charfcn{A}$ (or $\mathbb{1}_A$), whose domain is the universe, such that
$\charfcn{A}(x)=1$ if~$x\in A$, and $\charfcn{A}(x)=0$ if~$x\notin A$.
\end{df}
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