UPDATE: clean up.

Author: gitcordier

FA_DM.pdf

@@ -44,7 +44,7 @@ Feel free to get copies of this great book by yourself.
 ## Files
 
 - FA_DM.pdfOutput from [Xelatex](https://www.tug.org) compilation.You can also get an html output from
- [Hevea](http://hevea.inria.fr) compilation; see HOWTO.
+ [Hevea](http://hevea.inria.fr) compilation, see HOWTO.
 - FA_DM.tex
 - FA_mainmatter.tex
 - FA_chapter_1.tex

README.md

@@ -6,7 +6,7 @@
 \begin{enumerate}
 \item{If $x,\,y\in X$ there is a unique $z\in X$ such that $x+z=y$.}
 \item{ $0\cdot x=0=\alpha\cdot 0 \quad (\alpha\in\field, x\in X)$.}
-\item{ $2A\subseteq A+A$.}
+\item{ $2A\subset A+A$.}
 \item{ $A$ is convex if and only if $(s+t)A=sA+tA$ %
  for all positive scalars $s$ and $t$.}
 \item{ Every union (and intersection) of balanced sets is balanced.}
@@ -25,7 +25,7 @@
 \begin{enumerate}
 %: (a)
 \item Such property only depends on the group structure of $X$: Each $x$ in
-$X$ has an opposite $\minus x$. Let $x'$ be any opposite of $x$, so that
+$X$ has an opposite $\minus x$. Let $x'$ be any opposite of $x$so that
 ${x-x=0}=x+x'$. %
 Thus, $\minus x +x -x = \minus x + x + x' $, %
 which is equivalent to $\minus x = x'$. So is established the uniqueness of %
@@ -70,7 +70,7 @@
 \begin{align}\label{double lies in sum}
  2A = \{2x: x\in A \} 
  = \{x + x: x \in A \} 
- \subseteq \{ x + y : (x,\,y) \in A^2 \} 
+ \subset \{ x + y : (x,\,y) \in A^2 \} 
  = A+A
 \end{align}
 %
@@ -79,7 +79,7 @@
 \item If $A$ is convex, then %
 %
 \begin{align}
- A \subseteq \frac{s}{s+t} A + \frac{t}{s+t} A \subseteq A;
+ A \subset \frac{s}{s+t} A + \frac{t}{s+t} A \subset A;
 \end{align}
 %
 which is %
@@ -104,33 +104,33 @@
 (or simply use the fact that $+$ is commutative!).
 So ends the proof. %
 %: (e)
-\item Let $A$ range over $B$ a collection of balanced subsets, so that %
+\item Let $A$ range over $B$ a collection of balanced subsetsso that %
 %
 \begin{align}
- \alpha \bigcap B \subseteq \alpha A \subseteq A \subseteq \bigcup B 
+ \alpha \bigcap B \subset \alpha A \subset A \subset \bigcup B 
 \end{align}
 %
 for all scalars $\alpha$ of magnitude $\leq 1$. %
-The inclusion $\alpha \bigcap B \subseteq A$ establishes the first part. %
+The inclusion $\alpha \bigcap B \subset A$ establishes the first part. %
 Now remark that %
 %
 \begin{align}
- \alpha A \subseteq \bigcup {B} 
+ \alpha A \subset \bigcup {B} 
 \end{align}
 %
 implies %
 %
 \begin{align}
- \alpha \bigcup {B} \subseteq \bigcup {B};
+ \alpha \bigcup {B} \subset \bigcup {B};
 \end{align}
 which achieves the proof. %
 %
 %: (f)
-\item Let $A$ range over $C$ a collection of convex subsets, so that %
+\item Let $A$ range over $C$ a collection of convex subsetsso that %
 %
 \begin{align}
- (s+t) \bigcap C \subseteq s\bigcap C + t\bigcap C \subseteq sA + tA 
- \overset{(d)}{\subseteq} (s+t)A
+ (s+t) \bigcap C \subset s\bigcap C + t\bigcap C \subset sA + tA 
+ \overset{(d)}{\subset} (s+t)A
 \end{align}
 %
 for all positives scalars $\mathit{s}$, $\mathit{t}$. %
@@ -159,7 +159,7 @@
 %
 So, $x_1, x_2$ are now elements of the convex set $C_2$. %
 Every convex combination of our $x_i$'s is then in %
-$C_2 \subseteq \bigcup \Gamma$. Hence (g). %
+$C_2 \subset \bigcup \Gamma$. Hence (g). %
 %
 %: (h)
 \item Simply remark that 
@@ -174,7 +174,7 @@
 \item Given any $\alpha$ from the closed unit disc, %
 %
 \begin{align}
-\alpha(A+B)=\alpha A+ \alpha B \subseteq A+B. 
+\alpha(A+B)=\alpha A+ \alpha B \subset A+B. 
 \end{align}
 %
 There is no more to prove: $A+B$ is balanced. %
@@ -197,7 +197,7 @@
 To prove the lemma, let $\mathit{S}$ %
 run through all nonempty subsets of $X$. %
 First, assume that (i) holds: Clearly, every $S$ is convex balanced. %
-Moreover, $S+S \subseteq S$. Conversely, $S = S + \{0\} \subseteq S + S $; %
+Moreover, $S+S \subset S$. Conversely, $S = S + \{0\} \subset S + S $; %
 which establishes (ii). %
 %
 Next, assume (only) (ii): A proof by induction shows that %
@@ -211,9 +211,9 @@
 Thus, %
 %
 \begin{align}
- nS \overset{(\ref{induction nS})}{\subseteq} S 
- \subseteq n\,\lambda S 
- \subseteq n^2 S, 
+ nS \overset{(\ref{induction nS})}{\subset} S 
+ \subset n\,\lambda S 
+ \subset n^2 S, 
 \end{align}
 %
 since $S$ is balanced. %
@@ -224,7 +224,7 @@
 Dividing the latter inclusions by $n$ shows that %
 %
 \begin{align} 
- S \subseteq \lambda S \subseteq nS \overset{(\ref{induction nS})}{\subseteq} S,
+ S \subset \lambda S \subset nS \overset{(\ref{induction nS})}{\subset} S,
 \end{align}
 %
 which is (iii). Finally, dropping (ii) in favor of (iii) leads to %
@@ -245,7 +245,7 @@
 %
 \begin{align} 
  \alpha S + \beta S = 0S + 0S\overset{(b)}{=} \{0\} 
- \overset{(b)}{=} 0S \subseteq S.
+ \overset{(b)}{=} 0S \subset S.
 \end{align}
 %
 Hence (i), which achieves the lemma's proof. %
@@ -257,10 +257,10 @@
 First, remark that every member of $V$ is convex balanced: %
 So is $I$ (combine (e) with (f)). %
 %
-Next, let $\mathit{Y}$ range over $V$, so that %
+Next, let $\mathit{Y}$ range over $V$so that %
 %
 \begin{align}
- I + I \subseteq Y + Y \subseteq Y; 
+ I + I \subset Y + Y \subset Y; 
 \end{align}
 %
 which yields
@@ -269,7 +269,7 @@
  I + I = I 
 \end{align}
 %
-(the fact that $I = I + \{0\} \subseteq I + I$ was tacitely used). %
+(the fact that $I = I + \{0\} \subset I + I$ was tacitely used). %
 %
 It now follows from the lemma's (ii) $\Rightarrow$ (i) that %
 $I$ is a vector subspace of $X$. %
@@ -279,18 +279,18 @@
 %
 To show that $U$ is more specifically a vector subspace, %
 we first remark that such total order implies that either %
-$Z \subseteq Y$ or $Y \subseteq Z$, as $\mathit{Z}$ ranges over $V$. %
+$Z \subset Y$ or $Y \subset Z$, as $\mathit{Z}$ ranges over $V$. %
 A straightforward consequence is that 
 %
 \begin{align}
- Y \subseteq Y + Z \subseteq Y\cup Z.
+ Y \subset Y + Z \subset Y\cup Z.
 \end{align}
 %
 Another one is that $Y \cup Z$ ranges over $V$ as well. %
 Combined with the latter inclusions, this leads to %
 %
 \begin{align}
- U \subseteq U + U \subseteq U.
+ U \subset U + U \subset U.
 \end{align}
 %
 It then follows from the lemma's (ii) $\Rightarrow$ (i) that %

chapter_1/1_1.tex

@@ -24,7 +24,7 @@
 %
  \begin{align}
  \label{definition of v}
- \sum_{e} V \subseteq W, 
+ \sum_{e} V \subset W, 
  \end{align}
 %
 since addition is continuous. Moreover, for each $e$, there exists $x_e$ %
@@ -56,7 +56,7 @@
 We now come back to (\ref{definition of v}) and so conclude that %
 %
  \begin{align}
- y \in \sum_e \Lambda (V) \subseteq \Lambda(W) 
+ y \in \sum_e \Lambda (V) \subset \Lambda(W) 
  \end{align}
 %
 for if $|y_e| < 1/s$; which proves (a) whether %
@@ -77,7 +77,7 @@
 theorem (see Exercise 1.9) asserts that $f$ is an isomorphism of $X/N$ %
 onto $Y$. %
 We now conclude with the help of [\ref{notations: vector spaces: finite-dimensional vector spaces}] %
-that $f$ is an homeomorphism of $X/N$ onto $Y$. %
+that $f$ is a homeomorphism of $X/N$ onto $Y$. %
 We have thus established that $f$ is continuous: So is $\Lambda = f\circ \pi$.
 \end{proof}
 % END

chapter_1/1_10.tex

@@ -34,21 +34,21 @@
 In terms of balls $B_i$, this reads as follows, %
 %
 \begin{align}
- B_1(a,\eta) \subseteq B_2(a,\epsilon).
+ B_1(a,\eta) \subset B_2(a,\epsilon).
 \end{align}
 %
 The second consequence is that the $\tau_1$-continuousness of %
 $\varphi^{\, \minus 1}$ yields similar inclusions %
 %
 \begin{align} \label{1_12_6}
-\quad B_2(a, \epsilon') \subseteq B_1 (a, \eta')
+\quad B_2(a, \epsilon') \subset B_1 (a, \eta')
 \end{align}
 %
 provided $\eta'>0$. At arbitrary $\epsilon$, the special case $\eta' = \eta$ %
 is the concatenation %
 %
 \begin{align}
- B_2(a, \epsilon') \subseteq B_1(a,\eta) \subseteq B_2(a,\epsilon); 
+ B_2(a, \epsilon') \subset B_1(a,\eta) \subset B_2(a,\epsilon); 
 \end{align}
 %
 which proves that $\tau_1 =\tau_2$. %

chapter_1/1_12.tex

@@ -22,7 +22,7 @@
 \begin{proof}%
 Let us equipp $\D_K$ with the inner product %
 %
- $\bra{f}\ket{g} = \int_0^1 f \, \bar{g}$, so that %
+ $\bra{f}\ket{g} = \int_0^1 f \, \bar{g}$so that %
  $\bra{f}\ket{f} = \| f \|_2^2$. %
 %
 The following %
@@ -31,7 +31,7 @@
  \int_0^1 1\,| D^n f| \leq \| 1 \|_2 \| D^n f \|_2 
 \end{align}
 %
-is then a Cauchy-Schwarz inequality; see Theorem 12.2 of \cite{FA}. %
+is then a Cauchy-Schwarz inequality, see Theorem 12.2 of \cite{FA}. %
 We so obtain %
 %
 \begin{align}\label{inequalities 1}
@@ -99,7 +99,7 @@
 The whole chain (\ref{inclusions}) then forces %
 %
 \begin{align}
- \tau_1 \subseteq \tau_2 \subseteq \tau_\infty \subseteq \tau_1;
+ \tau_1 \subset \tau_2 \subset \tau_\infty \subset \tau_1;
 \end{align}
 %
 which achieves the proof.