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\begin{document}% \tiny \scriptsize \footnotesize \small \normalsize \large \Large \LARGE \huge \Huge 
\begin{center}
{\large A few comments about ``Principles of Mathematical Analysis'' by Rudin}\bigskip 

Pierre-Yves Gaillard\footnote{ORCID https://orcid.org/0000-0002-7960-1698} 
\end{center} 

\noi I wanted to study the famous book\smallskip 

[R] \textbf{Principles of Mathematical Analysis} by Walter Rudin,\smallskip 

\noi and to write a series of reading notes about it; but, after a few days of work, I abandoned this project. I still decided to post what I had written so far. This short text is the result of this decision. \bigskip 

\hrule\bigskip 

\noi\textbf{1.1.} Recall the definition of $A$ and $B$: 
$$
A:=\{a\in\Q\ |\ a>0,\ a^2<2\},\quad B:=\{b\in\Q\ |\ b>0,\ b^2>2\}.
$$ 
We want to show that $A$ contains no largest number, and $B$ no smallest. We give two arguments which differ slightly from Rudin's. \smallskip 

\noi\textbf{First argument.} Given $c\in A\cup B$, we want to find a positive $\eps\in\Q$ such that $c+\eps\in A$ if $c\in A$, and $c-\eps\in B$ if $c\in B$. 

If $c$ is in $A$ we have $c^2<2$ and we want $(c+\eps)^2<2$, that is, $2c\,\eps+\eps^2<2-c^2$. If $c\in B$ we have $c^2>2$ and we want $(c-\eps)^2>2$, that is, $2c\,\eps-\eps^2<c^2-2$. Hence it suffices to have $\eps\,(2c+\eps)<|c^2-2|$ in both cases. If $\eps<1$ it is enough to have $\eps\,(2c+1)<|c^2-2|$. So, we see that we can set $\eps:=\min\left(\frac12\,,\frac{|c^2-2|}{4c+2}\right)$. \smallskip 

\noi\textbf{Second argument.} Rudin needs a rational fraction\footnote{In this section ``rational fraction'' means ``univariate rational fraction with rational coefficients''} $f(x)$ such that the three rational fractions 
$$
f(x),\quad g(x):=\frac{f(x)-x}{2-x^2}\ ,\quad h(x):=\frac{f(x)^2-2}{x^2-2}
$$ 
are defined on the interval $I:=\{x\in\mathbb Q\ |\ 1\le x\le\frac32\}$, and $g(x)$ and $h(x)$ are positive on $I$. Rudin's rational fractions are 
$$
f(x):=\frac{2x+2}{x+2}\ ,\quad g(x)=\frac1{x+2}\ ,\quad h(x)=\frac2{(x+2)^2}\ ,
$$ 
and his notation is $q:=f(p)$. George Bergman explains on p. 3 of the text \smallskip 

\centerline{\url{https://math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.pdf}} \smallskip 

\noi how one might guess this rational fraction. I don't fully understand Bergman's explanation. Here is another way to find such an $f(x)$ (which will be different from Rudin's rational fraction). 

Since $f(x)$ and $h(x)$ can be recovered from $g(x)$, it suffices to give conditions on $g(x)$ which ensure that $h(x)$ is defined and positive on $I$. So, we start by assuming that $g(x)$ is a rational fraction defined and positive on $I$. The equality 
$$
h(x)=1-2\,x\,g(x)+(x^2-2)\,g(x)^2
$$ 
shows that $h(x)$ is automatically defined on $I$, and that it is positive on $I$ if $g(x)$ is small enough on $I$. This suggests to assume that $g(x)$ is bounded on $I$. Let $u\in\Q,u>0$ be an upper bound for $g(x)$ on $I$, and let us try to find a condition on $u$ which ensures that $h(x)$ is defined and positive on $I$. Since 
$$
h(1)=1-2\,g(1)-g(1)^2,
$$ 
a necessary condition for having $h(1)>0$ is $g(1)<\frac12$. This implies that $u$ must be less than $\frac12\,$. We have 
$$
\frac{h(x)}{g(x)^2}=\left(\frac1{g(x)}-x\right)^2-2\ge\left(\frac1u-\frac32\right)^2-2
$$ 
for $x$ in $I$. Since $\frac1u>2$, a sufficient condition for having $h(x)>0$ on $I$ is $\frac1u-\frac32\ge\frac32\,$, so 
$$
u:=\frac13
$$ 
works, and we can take for instance $g(x):=\frac13$ for all $x$ in $I$, which gives 
$$
f(x)=\frac{-x^2+3\,x+2}3\ ,\quad g(x)=\frac13\ ,\quad h(x)=\frac{x^2-6\,x+7}9\ .
$$ 
Note that Rudin's $g(x)=\frac1{x+2}$ also satisfies $g(x)\le\frac13$ on $I$. \bigskip 

\hrule\bigskip 

\noi\textbf{1.8.} I prefer the following wording of the definition of $\sup$ and $\inf$ to that of Rudin: If $S$ is a poset and $E$ is a subset of $S$ such that the set $U$ of all upper bounds is nonempty and has a minimum $m$, then we set $m=\sup E$, and similarly for $\inf$. The definitions coincide for ordered sets, but not for posets. Example: take the poset $S=\{e_1,e_2,\ldots,a,b\}$ with $e_1<e_2<\cdots$, $e_i<a,e_i<b$, and set $E:=\{e_1,e_2,\ldots\}$. We get $U=\{a,b\}$, and $a$ and $b$ satisfy Rudin's condition, but they are not suprema. \bigskip 

\hrule\bigskip 

\noi\textbf{1.11.}  
\begin{thm}[LUB implies GLB]\label{TLG}
If $S$ is a poset having the least upper bound property, then $S$ has the greatest lower bound property. 
\end{thm} 

\begin{proof} 
Let $B$ be a nonempty subset of $S$ such that the set $L$ of all lower bounds of $B$ in $S$ is also nonempty. 

It suffices to show that $L$ has a maximum. 

Let $\overline B$ be the set of all upper bounds of $L$ in $S$. 

By definition of $L$ and $\overline B$ we have for all $s$ in $S$: 

\noi(a) $s\in L\iff s\le b$ for all $b\in B$, 

\noi(b) $s\in\overline B\iff s\ge\ell$ for all $\ell\in L$. 

Statements (a) and (b) imply 

\noi(c) $B\subset\overline B$. 

In particular $\overline B$ is nonempty, so $a:=\sup L:=\min\overline B$ exists, and we must verify that $a=\max L$. 

The definition of $a$, coupled with (a) and (c), entails that $a$ is in $L$. 

Finally, (b) implies that $a$ is the maximum of $L$.  
\end{proof} \bigskip 

\hrule\bigskip 

\noi\textbf{Appendix to Chapter 1.} Rudin shows the existence of ordered fields with the least-upper-bound property. It seems natural to ask if two such fields are isomorphic. The answer is yes. In fact a much stronger result holds. To state this stronger result we make a few comments. Given an ordered field $K$ with the least-upper-bound property, we assume (as we can) that $K$ contains $\Q$, and we freely use Theorem~1.20b in [R] which says that, for any $x$ and $y$ in $K$ with $x<y$, there is a $p$ in $\Q$ such that $x<p<y$. Let $K$ and $L$ be two ordered fields with the least-upper-bound property, and let $f:K\to L$ be a set theoretical map. We denote by 

$(M)$ the condition that $f$ is a morphism of fields, by 

$(Q)$ the condition that $f$ induces the identity of $\Q$, i.e. that $f(p)=p$ for all $p$ in $\Q$, by 

$(I)$ the condition that $f$ is increasing, and by 

$(S)$ the condition that $f(x)=\sup_L\Q_{<x}$ for all $x$ in $K$, where $\sup_L$ means ``sup in $L$'' and $\Q_{<x}$ is the set of all $p$ in $\Q$ which are less than $x$. 

\begin{thm}\label{Tun} 
Let $K$ and $L$ be ordered fields with the least-upper-bound property and $f:K\to L$ a set theoretical map. If $f$ is a morphism of fields, then $f$ satisfies the four above conditions. In particular, $f$ is the unique morphism of fields from $K$ to $L$. In fact we have 
$$
(M)\iff[(Q)\text{ and }(I)]\iff(S).
$$ 
\end{thm} 

It is clear that $(M)$ implies $(Q)$. The implication $(M)\implies(I)$ follows from the fact that, by Theorem~1.21 in [R], the set of positive elements of $K$ coincides with the squares, and similarly for $L$. It is also plain that $(S)$ implies $(Q)$. % and $(I)$. 
It remains to show 
%
\begin{equation}\label{Eqsi}
(S)\implies(I),
\end{equation} 
%
\begin{equation}\label{Eqis}
[(Q)\text{ and }(I)]\implies(S),
\end{equation} 
%
\begin{equation}\label{Eqim}
[(Q)\text{ and }(I)]\implies(M).
\end{equation} 
%
\begin{proof}[Proof of \eqref{Eqsi}] 
If $f$ satisfies (S), then it satisfies (Q) and is weakly increasing. Let $x$ and $y$ be in $K$ with $x<y$. For some $p$ and $q$ in $\Q$ we have $x<p<q<y$, and thus $f(x)\le p<q\le f(y)$. 
\end{proof} 

\begin{proof}[Proof of \eqref{Eqis}] 
Assume that $f$ satisfies $(Q)$ and $(I)$, and that $g:K\to L$ satisfies (S). It suffices to show $f=g$. As just observed, $g$ also satisfies $(Q)$ and $(I)$. Let $x$ be in $K$. We have, for all $p$ in $\Q$, 
$$
p<f(x)\iff p<x\iff p<g(x).
$$ 
This implies $f(x)=g(x)$. 
\end{proof} 

It only remains to prove $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)f(y)$ for all $x$ and $y$ in $K$ assuming $(Q)$ and $(I)$ (indeed, this will imply \eqref{Eqim}). We will only prove $f(xy)=f(x)f(y)$ (the proof of $f(x+y)=f(x)+f(y)$ is similar and easier). 

Let $x,y\in K$ and $\eps\in\Q$ with $0<\eps<1$. Let $r\in\Q$ with $r-1>|x|,|y|$. There are $p,q\in\Q$ with $|x-p|,|y-q|<\eps$ and $|p|,|q|<r$. Setting $a:=x-p,b:=y-q$, we get 
$$
|xy-pq|=|pb+qa+ab|\le|p||b|+|q||a|+|a||b|<3r\eps.
$$ 
Summarizing: 
\begin{equation}\label{Er-1}
r-1>|x|,|y|\text{ and }|x-p|,|y-q|<\eps\text{ and }|p|,|q|<r
\end{equation} 
implies 
\begin{equation}\label{Exy}
|xy-pq|<3r\eps. 
\end{equation} 
Note that \eqref{Er-1} also implies 
$$
r-1>|f(x)|,|f(y)|\text{ and }|f(x)-p|,|f(y)-q|<\eps, 
$$ 
and thus $|f(x)f(y)-pq|<3r\eps$. Note finally that \eqref{Exy} implies $|f(xy)-pq|<3r\eps$. We conclude that \eqref{Er-1} implies $|f(xy)-f(x)f(y)|<6r\eps$. Since $\eps$ can be taken arbitrarily small, this forces $f(xy)=f(x)f(y)$. \bigskip 

\hrule\bigskip 

\noi\textbf{2.32.} \textbf{Irrationality of} $e$. \smallskip 

Assume $e$ is rational. Let $p$ be a positive integer such that $\frac{p!}e$ is an integer. For some integer $q$ we have 
$$
\frac{p!}e=\sum_{n\ge0}(-1)^n\,\frac{p!}{n!}=q+\sum_{n\ge p}(-1)^n\,\frac{p!}{n!}=q+(-1)^p+\frac{(-1)^{p+1}}{p+1}+\frac{(-1)^{p+2}}{(p+1)(p+2)}+\cdots
$$ 
$$
=q+(-1)^p\left(1-\frac1{p+1}+\frac1{(p+1)(p+2)}-\cdots\right).
$$ 
Let $r$ be the last parenthesis. This is an integer, and, the series defining $r$ being alternating and decreasing in absolute value, we get $0<r<1$, contradiction.

\end{document}