\documentclass[a4paper,twoside]{article}
\title{Revisions for \emph{Sets and Classes,} 2007.03.02 ed.}
\date{\today}
\author{David Pierce}

%\usepackage{parskip}   % paragraphs not indented, but separated by spaces
\setlength{\parindent}{0pt}

\usepackage[polutonikogreek,english]{babel}
\usepackage{bm}     % bold math
\usepackage{stmaryrd}   % for \shortrightarrow and \binampersand
\usepackage{upgreek}     % for \upomega
\usepackage[mathscr]{euscript}
\usepackage{amssymb,amsmath,amsthm}
\usepackage{ifthen}
\usepackage{url}
\usepackage{verbatim}  % allows a comment environment:
\usepackage{eco}  % gives old-style numerals
\usepackage{calc}  %  used with the following counters
\usepackage{showlabels}

\newcounter{mynum}
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\swapnumbers
\newtheorem{theorem}{Theorem}
\newtheorem{thmdefn}[theorem]{Theorem and Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}

\theoremstyle{definition}
\newtheorem{parag}[theorem]{}

\newcommand{\parref}[1]{\textnormal{\P~\ref{#1}}}

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%\newenvironment{change}{}{\hfill$\dashv$\vspace{1\baselineskip}}
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\newcommand{\defn}[1]{\textbf{#1}}   % for words being defined
%\newcommand{\defin}[1]{\textbf{#1}}   % for words being defined
\newcommand{\tech}[1]{\textsl{#1}}     % for technical terms

\newcommand{\Gk}[1]{\selectlanguage{polutonikogreek}{#1}\selectlanguage{english}}
\newcommand{\Eng}[1]{\emph{#1}}
\newcommand{\Tur}[1]{\textsf{#1}}
\newcommand{\Lat}[1]{\textsc{#1}}
\newcommand{\lett}[1]{\textsf{#1}}
\renewcommand{\phi}{\varphi}
\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}

\renewcommand{\theenumi}{\roman{enumi}}
\renewcommand{\labelenumi}{(\theenumi)}

\newcommand{\qsep}{\;}                 % follows a quantified variable
\newcommand{\Forall}[1]{\forall{#1}\qsep }
\newcommand{\Exists}[1]{\exists{#1}\qsep }
\newcommand{\existsunique}{\exists!}
\newcommand{\Existsunique}[1]{\exists!\,{#1}\qsep }

\newcommand{\setim}[1][]{''{#1}} % image of set under preceding function 
\newcommand{\setimb}[1]{[\,#1\,]}    % image of set under preceding function 
\newcommand{\conv}[1]{\breve{#1}}       % converse of a relation
\newcommand{\To}{\longrightarrow}
\newcommand{\inv}{^{-1}}                % mult. inverse
\newcommand{\rest}[1]{\restriction{#1}}% restriction of function to #1
\newcommand{\id}[1]{\operatorname{id}_{#1}}
\newcommand{\mapset}[2]{{}^{#1}#2}     % set of functions from #1 to #2

\newcommand{\Iff}{}
\let\Iff\iff
\renewcommand{\iff}{\Leftrightarrow}
\newcommand{\lto}{\Rightarrow}
\renewcommand{\land}{\mathrel{\binampersand}}

%%%%%%%%  sets:

\newcommand{\set}[2]{\{#1\colon #2\}}
\newcommand{\cls}[2]{\{#1\colon #2\}}
\newcommand{\class}[1]{\bm{#1}}
\newcommand{\cA}{\class A}
\newcommand{\cB}{\class B}
\newcommand{\cC}{\class C}
\newcommand{\cD}{\class D}
\newcommand{\cE}{\class E}
\newcommand{\cF}{\class F}
\newcommand{\cG}{\class G}
\newcommand{\cH}{\class H}
\newcommand{\cK}{\class K}
\newcommand{\cM}{\class M}
\newcommand{\cR}{\class R}
\newcommand{\cS}{\class S}
\newcommand{\cT}{\class T}

\newcommand{\universe}{\mathrm{\mathbf V}}
\newcommand{\cOm}{\bm{\mathit{\Omega}}}
\newcommand{\on}{\mathbf{ON}}           % the class of ordinals
\newcommand{\cn}{\mathbf{CN}}           % the class of cardinals
\newcommand{\wf}{\mathbf{WF}}           % class of well-founded sets
\newcommand{\rwf}{\mathrm{\mathbf R}}   % R used to define WF
\newcommand{\defr}[2]{\mathbf{D}_{#1}(#2)} % definable n-ary relations
\newcommand{\cu}{\mathbf{L}}            % constructible universe L
\newcommand{\const}{\mathrm{\mathbf L}} % C used to define L


\newcommand{\str}[1]{\mathfrak{#1}}     % structure
%\newcommand{\lang}{\mathcal{L}}        % a language or signature
\newcommand{\End}[1]{\operatorname{End}(#1)} % endomorphisms

% set-theoretic operators:

% nullary:

\renewcommand{\emptyset}{\varnothing}  % [I like the other symbol better]

% unary:

\newcommand{\comp}{^{\mathrm{c}}}      % set-theoretic complement
\newcommand{\pow}[1]{\mathcal{P}(#1)}  % power set

% binary:
\renewcommand{\setminus}{\smallsetminus} % I don't like the standard symbol
\newcommand{\symdiff}{\vartriangle}    % symmetric difference

%%%%%%%%%%%%  Cardinality

\newcommand{\injects}{\preccurlyeq}    % injects in
\newcommand{\pinjects}{\prec}
\newcommand{\equip}{\approx}           % equipollent (is in bijection with)
\newcommand{\nequip}{\not\approx}      % not equipollent
\newcommand{\vscr}[1]{#1'}        % successor ordinal (v for von Neumann)
\newcommand{\ord}[1]{\operatorname{ord}(#1)} % order-type (ordinality)
\newcommand{\supr}[1]{\sup(#1)}        % supremum
%\newcommand{\size}[1]{\lvert#1\rvert}  % cardinality
\newcommand{\card}[1]{\operatorname{card}(#1)} % cardinality
\newcommand{\pl}{^{+}}                 % successor cardinal
\newcommand{\cntm}{\mathfrak c}        % cardinality of the continuum
\newcommand{\pred}[1]{\operatorname{pred}(#1)}
\newcommand{\rank}[1]{\operatorname{rank}(#1)}

%%%%%%%%%%%%%  Functions

\newcommand{\dom}[1]{\operatorname{dom}(#1)} % domain
\newcommand{\ran}[1]{\operatorname{rng}(#1)} % range
\newcommand{\fld}[1]{\operatorname{f{}ld}(#1)} % field
\newcommand{\injection}{\rightarrowtail}
\newcommand{\surjection}{\twoheadrightarrow}
%\newcommand{\bijection}{\rightarrowtail\mkern-17mu\shortrightarrow}
\newcommand{\bijection}{\rightarrowtail\mkern-18mu\twoheadrightarrow}
\newcommand{\supp}[1]{\operatorname{supp}(#1)} % support

%%%%%%%%%%%%%

\newcommand{\included}{\subseteq}      % [the name suggests the meaning here]
\newcommand{\nincluded}{\not\subseteq} % not included
\newcommand{\pincluded}{\subset}       % proper inclusion    
\newcommand{\includes}{\supseteq}

%%%%%%%%%%%%%%%

\newcommand{\axioms}[1]{\mathrm{#1}}   % set-theoretic axiom or set of axioms
\newcommand{\zf}{\axioms{ZF}}          % Zermelo-Frankel axioms
%\newcommand{\ac}{\axioms{AC}}	       % Axiom of Choice
\newcommand{\zfc}{\axioms{ZFC}}        % ZF + AC
%\newcommand{\ch}{\axioms{CH}}          % Continuum Hypothesis
%\newcommand{\pa}{\axioms{PA}}          % first-order Peano arithmetic

%%%%%%%%%%%%%%%%

% Standard structures

\newcommand{\vnn}{\upomega}       % von Neumann natural numbers

\newcommand{\stnd}[1]{\mathbb{#1}}
\newcommand{\N}{\stnd{N}}         % natural numbers
\newcommand{\scr}[1]{#1^{+}}           % successor
\newcommand{\Zp}{\Z_{+}}          % positive integers
\newcommand{\Z}{\stnd{Z}}         % integers
\newcommand{\Q}{\stnd{Q}}         % rationals
\newcommand{\R}{\stnd{R}}         % reals
\newcommand{\C}{\stnd{C}}         % complex numbers

\begin{document}
\maketitle
Some corrections and changes. 

\section{General}

\begin{change}
A list of symbols should be provided.
\end{change}

\begin{change}
  The relation symbolized by $\in$ and called
\textbf{containment} on p.~34 would be better called
\textbf{membership} (as on p.~19).
\end{change}


\section{Significant changes}


\begin{change}
p.~10, \P~1.1.2.  The latter part of this paragraph needs to be
rethought.  \emph{Set} and \emph{class} are not the most
`generally applicable' collective nouns; they are the most abstract.
For us, \emph{set} will be the name of something whose members are
other {sets.}
\end{change}

\begin{change}
P.~11: expand \P~1.2.2: say more about this `correspondence' between
$\cup$ and~$\lor$. 
\end{change}

\begin{change}
P.~21, \P~2.2.4, after the list: the comment `depending on the axioms'
makes
the truth\-{}-value of $a\in a$ sound arbitrary.  The Foundation Axiom
will say that the sentence is false; but \P~8.3.5 shows that this
axiom can be understood merely as a \emph{definition} of the sets that
we choose to study.
\end{change}

\begin{change}
\P~2.2.5:  The sets that we study can be called \emph{pure} sets
(Moschovakis) or \emph{hereditary} sets (Kunen).
\end{change}

\begin{change}
P.~22, \P~2.2.9:  The alternative formulation of $\universe$ should
also use the `official' language: so make it $\cls x{x\in x\lto x\in
  x\lto x\in x}$. 
\end{change}

\begin{change}
P.~25:  Not all of our theorems will have formal deductions even in
principle:  In \P~8.3.5, it will be noted that $\wf$ is a model of
$\zf$.  This conclusion can be formulated as an infinite list of
sentences in the official language, each with a formal proof.
We conclude that $\zf$ is consistent: this can be formulated (as
G\"odel showed) as a single sentence of the official language; but it
has no formal proof.
\end{change}

\begin{change}
Exercises might be added to Ch.~2.
\end{change}


\begin{change}
P.~32, bottom: the definition of \defn{field} is not really needed.
\end{change}

\begin{change}
P.~34, \P~3.5.6 (ii): the reference should be to \P~3.5.5.
\end{change}

\begin{change}
P.~35, \P~3.5.7: a \defn{least} element is also a \defn{minimum}
element.
\end{change}

\begin{change}
  P.~36, \P~3.6.2:  Instead of `virtual class' here, we might speak of
  a \textbf{family} of classes.  Then a class $\cC$ and a binary
  relation $\cR$ determine the families $\cls{x\cR}{x\in\cC}$ and
  $\cls{\cR x}{x\in\cC}$.  In particular, if
  $\cE\included\cC\times\cC$ and is an equivalence-relation on $\cC$,
  then $\cC/\cE$ is the family $\cls{x\cE}{x\in\cC}$.
\end{change}

\begin{change}
P.~37, \P~3.6.4:  item (v) should be item (ii); also, in (vi), the
notation $\pi_0$ and $\pi_1$ that will be used in \P~5.4.2 (Recursion
with Parameter) can be introduced.
\end{change}

\begin{change}
P.~46: replace \P~4.2.4 with:

\begin{parag}
Suppose $(\cC,\cF,i)$ is a recursive structure.  Then $\cC$ can be
denoted suggestively by
\begin{equation*}
  \{i,\cF(i),\cF(\cF(i)),\dots\}.
\end{equation*}
Some possibilities are depicted in Fig.~4.1.  Note well that possibly
$\cF$ is not injective, and possibly $i\in\cF\setimb{\cC}$.  (However,
these possibilities seem to be mutually exclusive.)
\end{parag}

\begin{parag}\label{par:rec}
Suppose again  $(\cC,\cF,i)$ is a recursive structure, and
$(\cD,\cG,j)$ is another iterative structure (not necessarily
recursive).  There \emph{may be} a function $\cH$ from $\cC$ to $\cD$
such that
\begin{enumerate}
  \item
$\cH(i)=j$,
\item
$a\in\cC\lto \cH(\cF(a))=\cG(\cH(a))$, that is,
  $\cH\circ\cF=\cG\circ\cH$ on $\cC$.
\end{enumerate}
The first rule says what $\cH(i)$ is; the second says how to
obtain $\cH(\cF(a))$ from $\cH(a)$.  By induction, $\cH$ is uniquely
determined by these rules: see Corollary~\ref{cor:rec-uni} below.
In this case, we say that $\cH$ is \defn{recursively defined} by the
given rules.
\end{parag}

\begin{parag}\label{par:rec2}
  Note first another possible kind of recursive definition:
  $(\cC,\cF,i)$ is recursive, 
$\cE\included\cD$, and
  $\cG:\cD\to\cD$, then perhaps there is a sub-class $\cR$
  of $\cC\times\cD$ such that (in the notation of \P~3.6.2)
  \begin{enumerate}
    \item
$i\cR=\cE$,
\item
$a\in\cC\lto\cF(a)\cR=\cG\setimb{a\cR}$.
  \end{enumerate}
Then $\cR$ too is uniquely determined by these rules, so it too is
\defn{recursively defined:}
\end{parag}

\begin{theorem}\label{thm:rec-uni}
  Suppose   $(\cC,\cF,i)$ is recursive,
  $\cG:\cD\to\cD$, and $\cE\included\cD$.  Then there is at most one
  relation $\cR$ as in \parref{par:rec2}.
\end{theorem}

\begin{proof}
  Suppose $\cR_0$ and $\cR_1$ are two such relations.  Let
  \begin{equation*}
    \cC_1=\cls x{x\in\cC\land x\cR_0=x\cR_1}.
  \end{equation*}
Since $i\cR_0=\cE=i\cR_1$, we have $i\in\cC_1$.  Suppose $a\in\cC_1$,
so $a\cR_0=a\cR_1$.  Then
\begin{equation*}
  \cF(a)\cR_0=\cG\setimb{a\cR_0}=\cG\setimb{a\cR_1}=\cF(a)\cR_1,
\end{equation*}
so $\cF(a)\in\cC_1$.  By induction (and Lemma 4.2.3), $\cC_1=\cC$.
Since $\dom{\cR_0}\included\cC$ and $\dom{\cR_1}\included\cC$, we
conclude that $\cR_0=\cR_1$.
\end{proof}

\begin{corollary}\label{cor:rec-uni}
  Suppose $(\cC,\cF,i)$ is recursive, and $(\cD,\cG,j)$ is iterative.
  Then there is at most one function $\cH$ as in \parref{par:rec2}.
\end{corollary}

\begin{proof}
  The function $\cH$ (if it exists) is a relation, namely a sub-class
  $\cR$ of $\cC\times\cD$.  Let $\cE=\{\cH(i)\}$.  Then
    \begin{enumerate}
      \item
$i\cR=\{\cH(i)\}=\cE$;
\item
$a\in\cC\lto\cF(a)\cR=
  \{\cH(\cF(a))\}=\{\cG(\cH(a))\}=\cG\setimb{\{\cH(a)\}}=\cG\setimb{a\cR}$.  
    \end{enumerate}
By the theorem, $\cR$ is unique, so $\cH$ is unique.
\end{proof}
\end{change}

\begin{change}
P.~48, \P~4.3.5:  The Recursion Theorem should be given more
generally:

\begin{theorem}[Recursion]\label{thm:rec}
  Suppose $(\cC,\cF,i)$ is an arithmetic structure,
  $\cG\colon\cD\to\cD$, and $\cE\included\cD$.
  Then there is (uniquely, by Theorem~\ref{thm:rec-uni}) a sub-class $\cR$ of
  $\cC\times\cD$ such that 
  \begin{enumerate}
\item
    $i\cR=\cE$,
\item
$a\in\cC\lto\cF(a)\cR=\cG\setimb{a\cR}$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Let $\cB$ be the sub-class
  \begin{multline*}
    \Bigl\{x\colon\Forall y\Bigl(y\in x\lto\Exists
    z\bigl(z\in\cE\land y=(i,z)\bigr)\lor{}\\ 
{}\lor\Exists u\Exists v\bigl((u,v)\in \cC\times\cD\cap x\land
    y=(\cF(u),\cG(v))\bigr)\Bigr)\Bigr\}  
  \end{multline*}
of $\pow{\cC\times\cD}$, and let $\cR=\bigcup\cB$, so
$\cR\included\cC\times\cD$.  If $a\in\cE$, then $\{(i,a)\}$ is a set
(by the Pairing Axiom), so it belongs to $\cB$, and hence
$i\mathrel{\cR}a$.  Suppose $b\mathrel{\cR}c$.  Then $(b,c)\in d$ for
some $d$ in $\cB$, so $d\cup\{(\cF(b),\cG(c))\}$ is a set (by the Weak
Union Axiom), and this set belongs to $\cB$, so
$\cF(b)\mathrel{\cR}\cG(c)$.  We now have the following
characterization of $\cR$:
\begin{equation*}
  a\mathrel{\cR}b\iff((a=i\land b\in\cE)\lor\Exists u\Exists
  v(u\mathrel{\cR}v\land a=\cF(u)\land b=\cG(v))).
\end{equation*}
Since $i\notin\cF\setimb{\cC}$, we have $i\mathrel{\cR}a\iff a\in\cE$,
so $i\cR=\cE$.  Since $\cF$ is injective, if $a\in\cC$, we have
$\cF(a)\mathrel{\cR}b\iff b\in\cG\setimb{a\cR}$, so
$\cF(a)\cR=\cG\setimb{a\cR}$. 
\end{proof}

\begin{corollary}\label{cor:rec}
  Suppose $(\cC,\cF,i)$ is an arithmetic structure and $(\cD,\cG,j)$
  is an iterative structure.  Then there is (uniquely, by
  Corollary~\ref{cor:rec-uni}) a function $\cH$ from $\cC$ to $\cD$ such
  that
  \begin{enumerate}
    \item
$\cH(i)=j$,
\item
$a\in\cC\lto\cH(\cF(a))=\cG(\cH(a))$, that is, $\cH\circ\cF=\cG\circ\cH$.
  \end{enumerate}
\end{corollary}

\begin{proof}
  Exercise.
\end{proof}
\end{change}

\begin{change}
P.~51, end of \S~4.4:  `We seem to have this if $i=\emptyset$\textbf{, and
$\cF$} and $\cG$ \textbf{are both} $x\mapsto x\cup\{x\}$.'
\end{change}

\begin{change}
P.~52, proof of 4.5.4, part (i): `Then $b\neq\alpha$, since $\alpha$
is well-ordered by containment, and such orderings are by definition
strict'. 
\end{change}

\begin{change}
P.~53, \P~4.5.8 should begin:  `The structure $(\on,x\mapsto\vscr
x,\emptyset)$ satisfies the hypothesis of Theorem 4.4.3 with respect
to the ordering $\in$; hence there is a class
$\{\emptyset,\vscr{\emptyset}, \vscr{\vscr{\emptyset}}\}\dots$'.
\end{change}

\begin{change}
\P\P~5.2.1--2 can be replaced with the following:

  \begin{parag}
We shall define the binary operation of \tech{addition} on $\N$ so
that
\begin{enumerate}
  \item
$m+0=m$,
\item
$m+\scr n=\scr{(m+n)}$.
\end{enumerate}
These rules tell how to add $0$, and they tell how to add $\scr n$,
provided one can add $n$.  The rules \emph{will} determine a unique
operation, by a variant of the Recursion Theorem (\parref{thm:rec}).
Moreover, suppose $(\cA,\cS,i)$ is a recursive structure.  (This will
be so throughout this section.)  Then we shall be able to define
addition on $\cA$ by the rules
\begin{enumerate}
  \item
$a+i=a$,
\item
$a+\cS(b)=\cS(a+b)$,
\end{enumerate}
even though the Recursion Theorem does not apply generally to all
recursive structures.
  \end{parag}

  \begin{lemma}\label{lem:+}
    Suppose $\cF\colon\cB\to\cC$ and $\cG\colon\cC\to\cC$.  Then there
    is a unique function $\cH$ from $\cB\times\N$ to $\cC$ such that
    \begin{enumerate}
      \item
$a\mathop{\cH}0=\cF(a)$,
\item
$a\mathop{\cH}\scr n=\cG(a\mathop{\cH}n)$.
    \end{enumerate}
  \end{lemma}

  \begin{proof}
    By the Recursion Theorem, there is a unique sub-class $\cR$ of
    $\N\times(\cB\times\cC)$ such that
    \begin{enumerate}
      \item
$0\cR=\cF$,
\item
$n\in\N\lto\scr n\cR=\bigl((x,y)\mapsto(x,\cG(y)\bigr)\setimb{n\cR}=
  \cls{(x,\cG(y))}{n\mathrel{\cR}(x,y)}$.  
    \end{enumerate}
By induction, if $n\in\N$, then $n\cR$ is a function from $\cB$ to
$\cC$.  Indeed, $0\cR$ is such a function (namely $\cF$), and if
$n\cR$ is such a function, then $\scr n\cR$ is its composition with
$\cG$.  Let the function $n\cR$ be denoted by $\cK_n$; then $\cK_{\scr
  n}=\cG\circ\cK_n$.  We can now define the binary function $\cH$ as
$(x,y)\mapsto\cK_y(x)$ on $\cA\times\N$.  Then
\begin{enumerate}
  \item
$a\mathop{\cH}0=\cK_0(a)=\cF(a)$,
\item
$a\mathop{\cH}\scr n=\cK_{\scr n}(a)=\cG(\cK_n(a))=\cG(a\mathop{\cH}n)$.
\end{enumerate}
So $\cH$ is as desired.  To see that $\cH$ is unique, note that $\cR$
determines $\cH$, and conversely.  Indeed,
\begin{align*}
\cH&=\cls{((x,y),z)}{y\mathrel{\cR}(x,z)},\\
  \cR&=\cls{(y,(x,z))}{x\mathop{\cH}y=z}.  
\end{align*}
Since $\cR$ uniquely satisfies the given conditions, so does~$\cH$. 
  \end{proof}

  \begin{thmdefn}\label{thm:+}
Suppose $(\cA,\cS,i)$ is recursive.  Then there is a unique binary
operation of \defn{addition} on $\cA$ given by
\begin{enumerate}
  \item
$a+i=a$,
\item
$a+\cS(b)=\cS(a+b)$.
\end{enumerate}
  \end{thmdefn}

  \begin{proof}
    By the lemma, there is a unique function $\cH$ from $\cA\times\N$
    to $\cA$ such that
    \begin{enumerate}
      \item
$a\mathop{\cH}0=a$,
\item
$a\mathop{\cH}\scr n=\cS(a\mathop{\cH}n)$.
    \end{enumerate}
So $\cH$ is recursively defined in its second argument.  We shall show
that it is also recursively definable in its first argument.  First, let
$\cF$ be the function $x\mapsto i\mathop{\cH}x$ from $\N$ into $\cA$.
Then 
\begin{equation}\label{eqn:+i}
  \begin{split}
    \cF(0)&=i,\\
    \cF(\scr n)&=\cS(\cF(n)).
  \end{split}
\end{equation}
(So $\cF$ is the unique homomorphism from $(\N,{}\scr{},0)$ into
$(\cA,\cS,i)$ guaranteed by Corollary~\ref{cor:rec}.)
By induction, $\ran{\cF}=\cA$; indeed, $i\in\ran{\cF}$, and
$a\in\ran{\cF}\lto\cS(a)\in\ran{\cF}$.  The equation
\begin{equation}\label{eqn:+n}
  \cS(a)\mathop{\cH}n=\cS(a\mathop{\cH}n)
\end{equation}
holds when $n=0$, since
$\cS(a)\mathop{\cH}0=\cS(a)=\cS(a\mathop{\cH}0)$.
Suppose~\eqref{eqn:+n} holds for some $n$ in $\N$.  Then
\begin{align*}
  \cS(a)\mathop{\cH}\scr n
&=\cS(\cS(a)\mathop{\cH}n)&&\text{[by definition of $\cH$]}\\
&=\cS(\cS(a\mathop{\cH}n))&&\text{[by inductive hypothesis]}\\
&=\cS(a\mathop{\cH}\scr n).&&\text{[by definition of $\cH$]}
\end{align*}
So~\eqref{eqn:+n} holds for all $n$ in $\N$.  Therefore each of the
operations $x\mapsto x\mathop{\cH}n$ is the operation $\cG_n$
recursively defined by
\begin{enumerate}
  \item
$\cG_n(i)=\cF(n)$,
\item
$\cG_n(\cS(a))=\cS(\cG_n(a))$.
\end{enumerate}
In particular,
\begin{equation*}
\cG_m=\cG_n\iff \cF(m)=\cF(n).
\end{equation*}
Now we can define addition on $\cA$ by
\begin{equation*}
  a+b=c\iff\Exists x(\cF(x)=b\land \cG_x(a)=c).
\end{equation*}
Then $a+i=\cG_0(a)=a\mathop{\cH}0=a$.  Also, if $b=\cF(n)$, so that
$\cS(b)=\cF(\scr n)$, then
\begin{equation*}
  a+\cS(b)
=\cG_{\scr n}(a)
=a\mathop{\cH}\scr n
=\cS(a\mathop{\cH}n)
=\cS(\cG_n(a))
=\cS(a+b).
\end{equation*}
Thus $+$ is as desired; it is unique by Theorem~\ref{thm:rec-uni}.
  \end{proof}
\end{change}

\begin{change}
  \P~5.2.3 should have a reference to Landau.  \P~5.2.4 can be
  slightly rewritten:

  \begin{parag}
    Suppose $(a,s,i)$ is a recursive \emph{set,} so that all
    operations on $a$ are sets.  Then we can
    establish addition on $a$ as follows.  By
    Corollary~\ref{cor:rec-uni}, for each $b$ in $a$, there is at most
    one singulary operation $f_b$ on $a$ such that
    \begin{enumerate}
      \item
$f_b(i)=b$,
\item
$f_b\circ s=s\circ f_b$.
    \end{enumerate}
Let $a_0$ be the subset of $a$ comprising those $b$ such that $f_b$
does exist; note well how this definition of $a_0$ requires $f_b$ to
be a set.  Then $f_i$ exists and is $\id a$, so $i\in a_0$.  If
$b\in a_0$, then $f_{s(b)}$ exists and is $s\circ f_b$.  By induction,
$a_0=a$.  Now we can define
  $b+c=f_b(c)$.
  \end{parag}
\end{change}

\begin{change}
  At the beginning of \S~5.2 should be inserted the following:

  \begin{lemma}\label{lem:.}
    Suppose $\cF\colon\cB\to\cC$ and $\cG\colon\cC\times\cB\to\cC$.  Then there
    is a unique function $\cH$ from $\cB\times\N$ to $\cC$ such that
    \begin{enumerate}
      \item
$a\mathop{\cH}0=\cF(a)$,
\item
$a\mathop{\cH}\scr n=(a\mathop{\cH}n)\mathop{\cG}a$.
    \end{enumerate}
  \end{lemma}

  \begin{proof}
    By the Recursion Theorem, there is a unique sub-class $\cR$ of
    $\N\times(\cB\times\cC)$ such that
    \begin{enumerate}
      \item
$0\cR=\cF$,
\item
$n\in\N\lto\scr n\cR=\bigl((x,y)\mapsto(x,y\mathop{\cG}x)\bigr)\setimb{n\cR}=
  \cls{(x,y\mathop{\cG}x))}{n\mathrel{\cR}(x,y)}$.  
    \end{enumerate}
By induction, if $n\in\N$, then $n\cR$ is a function $\cK_n$ from $\cB$ to
$\cC$.  Indeed, $0\cR$ is such a function, namely $\cF$; this then is
$\cK_0$.  If
$n\cR$ is such a function, as $\cK_n$, then $\scr n\cR$ is
$x\mapsto\cH_n(x)\mathop{\cG}x$; this then is $\cK_{\scr n}$.  We can
now define the binary function $\cH$ as
$(x,y)\mapsto\cK_y(x)$ on $\cA\times\N$.  Then
\begin{enumerate}
  \item
$a\mathop{\cH}0=\cK_0(a)=\cF(a)$,
\item
$a\mathop{\cH}\scr n=\cK_{\scr
  n}(a)=\cK_n(a)\mathop{\cG}a=(a\mathop{\cH}n)\mathop{\cG}a$. 
\end{enumerate}
So $\cH$ is as desired; its uniqueness is as in Lemma~\ref{lem:+}.
  \end{proof}
\end{change}

\begin{change}
  The proof of Theorem~5.3.2 can be supplied as follows:

  \begin{proof}
    We follow the pattern of the proof of Theorem~\ref{thm:+}.  By the
    lemma, there is a unique function
    $\cH$ from $\cA\times\N$ into $\cA$ such
    that
    \begin{enumerate}
      \item
$a\mathop{\cH}0=i$,
\item
$a\mathop{\cH}\scr n=a\mathop{\cH}n+a$.
    \end{enumerate}
By induction, $i\mathop{\cH}n=i$ for all $n$ in $\N$; indeed, this is
given when $n=0$, and if it holds when $n=m$, then $i\mathop{\cH}\scr
m=i\mathop{\cH}m+i=i\mathop{\cH}m=i$.  Let $\cF$ be the unique
homomorphism from $(\N,{}\scr{},0)$ into $(\cA,\cS,i)$.  The equation
\begin{equation}\label{eqn:.n}
  \cS(a)\mathop{\cH}n=a\mathop{\cH}n+\cF(n)
\end{equation}
holds when $n=0$, since
$\cS(a)\mathop{\cH}0=i=i+i=a\mathop{\cH}0+\cF(0)$.
Suppose~\eqref{eqn:.n} holds for some $n$ in $\N$.  Then
\begin{align*}
  \cS(a)\mathop{\cH}\scr n
&=\cS(a)\mathop{\cH}n+\cS(a)&&\text{[by definition of $\cH$]}\\
&=(a\mathop{\cH}n+\cF(n))+\cS(a)&&\text{[by inductive hypothesis]}\\
&=a\mathop{\cH}n+(\cF(n)+\cS(a))&&\text{[by associativity of $+$]}\\
&=a\mathop{\cH}n+\cS(\cF(n)+a)&&\text{[by definition of $+$]}\\
&=a\mathop{\cH}n+(\cS(\cF(n))+a)&&\text{[by Lemma~5.2.5]}\\
&=a\mathop{\cH}n+(\cF(\scr n)+a)&&\text{[because $\cF$ is a homomorphism]}\\
&=a\mathop{\cH}n+(a+\cF(\scr n))&&\text{[by commutativity of $+$]}\\
&=(a\mathop{\cH}n+a)+\cF(\scr n)&&\text{[by associativity of $+$]}\\
&=a\mathop{\cH}\scr n+\cF(\scr n).&&\text{[by definition of $\cH$]}
\end{align*}
So~\eqref{eqn:.n} holds for all $n$ in $\N$.  Therefore each of the
operations $x\mapsto x\mathop{\cH}n$ is the operation $\cG_n$
recursively defined by
\begin{enumerate}
  \item
$\cG_n(i)=i$,
\item
$\cG_n(\cS(a))=\cG_n(a)+\cF(n)$.
\end{enumerate}
In particular,
\begin{equation*}
\cG_m=\cG_n\iff \cF(m)=\cF(n).
\end{equation*}
Now we can define multiplication on $\cA$ by
\begin{equation*}
  a\cdot b=c\iff\Exists x(\cF(x)=b\land \cG_x(a)=c).
\end{equation*}
Then $a\cdot i=\cG_0(a)=a\mathop{\cH}0=i$.  Also, if $b=\cF(n)$, so that
$\cS(b)=\cF(\scr n)$, then
\begin{equation*}
  a\cdot\cS(b)
=\cG_{\scr n}(a)
=a\mathop{\cH}\scr n
=a\mathop{\cH}n+a
=\cG_n(a)+a
=a\cdot b+a.
\end{equation*}
Thus $\cdot$ is as desired; it is unique by Theorem~\ref{thm:rec-uni}.
  \end{proof}
\end{change}

\begin{change}
  The proof of Theorem~5.3.5 can be replaced with a reference to
  Lemma~\ref{lem:.}. 
\end{change}








\section*{Trivial changes}

\begin{change}
\P~4.4.3: it can be noted that the last part of the proof is by
contradiction. 
\end{change} 

\begin{change} 
P.~4, item (i), after `$\mathbf V$ in these notes': insert reference
to \P~2.2.7.
\end{change}

\begin{change}
P.~8: include Table 2.1 on p.~18 (best done by changing the Table to a
Figure). 
\end{change} 

\begin{change} 
P.~9: capitalize the letters after the hyphens in `Replacement-scheme'
and `Power-set'.
\end{change}

\begin{change}
P.~10: transpose \P~1.1.1 to read:
\begin{quote}
  A \textbf{set} is a thing that \textbf{contains} other things.  Those other
  things are called \textbf{members} or
  \textbf{elements} of the set.  
The set cannot be separated 
  from its elements the way a box can be emptied of its contents: 
the set \textbf{comprises} its 
  members, and the
  members \textbf{compose} the set.
  A set \emph{is} its elements, considered as one thing.  It is a
  multitude that is also a unity.
\end{quote}
\end{change}

\begin{change}
P.~15, caption to Table 1.1:  Replace sentence `A terminal
\Gk{w}\dots' with
\begin{quote}
  The vowels \Gk a, \Gk h, and \Gk w may have an iota subscript
  (\Gk{a|, h|, w|}). 
\end{quote}
\end{change}

\begin{change}
P.~16, after the first list of 3 items:  Delete repeated `recursively'
(and add to index). 
\end{change} 

\begin{change} 
After the second list of 3 items: change `is' to `of'; don't
capitalize `Parts'.
\end{change}

\begin{change}
P.~20, n.~4:  `The latter sequence that gives\dots': delete `that'.
\end{change}

\begin{change}
P.~21, \P~2.2.4, item (iii):  `Then $\Exists x\phi$ \emph{is} true\dots'
\end{change}

\begin{change}
P.~23, \P~2.3.5, item (iii): `(where $a$ is allowed to appear in
$\sigma$)': change $\sigma$ to $\phi$.
\end{change}

\begin{change}
P.~23, \P~2.3.5:  `\emph{this} rule allows \emph{us} to obtain the
sentence $\tau$\dots'
\end{change}

\begin{change}
P.~24: allow Fig.~2.2 to float to the top of a page?
\end{change}

\begin{change}
P.~30, \P~3.2.3: replace `However, $\bigcup a$ is a set' with
`However, the union of a \emph{set} is a set'.
\end{change}

\begin{change}
\P~3.5.9: the meaning of \emph{greater than} should perhaps have been
given explicitly in \P~3.5.5.
\end{change}

\begin{change}
P.~36, \P~3.6.3: In the formula displayed over two lines, the terminal
$\land$ on the first line should be repeated on the second (as this is
the convention I use elsewhere).
\end{change}

\begin{change}
In the following line, replace \defn{to} with \defn{(in)to}.
\end{change}

\begin{change}
Pp.~38 f., \P~3.7.1: change $\cE$ to $\cC$.
\end{change}

\begin{change}
P.~40: Exercise (3) should follow (5).
\end{change}

\begin{change}
P.~45, \P~4.1.8: slant \tech{chain} as a technical term.
\end{change}

\begin{change}
Last line of text, but two: `\dots will \emph{be (in \P~5.1.3)}
another example\dots' 
\end{change} 

\begin{change} 
\P~4.2.2 can be broken into 3 paragraphs.
\end{change}

\begin{change}
\P~4.3.1: `This means \emph{by \P~4.2.1}\dots'; in item (ii): delete
\emph{from $\cC$}; afterwards: `The five numbered conditions here
\emph{for being an arithmetic structure} are sometimes\dots'
\end{change}

\begin{change}
\P~4.4.1, last line but one: $\cC$ should be $\cD$.
\end{change}

\begin{change}
p.~55, \P~5.1.1, just before (5.1):  `Meanwhile \emph{we} have'.  In
(5.2) and (5.3), the functions $\cF$ are really \emph{sets} and should
be written that way.  (Actually they are variables\dots)
\end{change}

\begin{change}
\P~5.1.3 (iv): change $\cC$ to $a$ (both times).
\end{change}

\begin{change}
\P~5.1.5: Give the numerical reference (4.3.5) for the Recursion Theorem.
\end{change}

\begin{change}
\P~5.3.7:  Add: `For all $a$ and $b$ in $\cA$, and all $m$ and $n$ in
$\N$'; in (ii), replace $x\mapsto x^a$ with $x\mapsto x^m$.
\end{change}







\end{document}
