Added pseudonorms for TVSs.

src/cat/index.tex

@@ -3,3 +3,4 @@
 
 \input{./src/cat/cat/index.tex}
 \input{./src/cat/gluing/index.tex}
+\input{./src/cat/tricks/index.tex}

src/cat/tricks/dyadic.tex

@@ -0,0 +1,62 @@
+\section{Dyadic Rational Numbers}
+\label{section:dyadic}
+
+\begin{definition}[Dyadic Rational Number]
+\label{definition:dyadic}
+    Let $x \in \rational$, then $x$ is a \textbf{dyadic rational number} if there exists $n \in \natp$ and $k \in \nat$ such that $x = k/2^{-n}$.
+
+    For each $n \in \natp$, denote $\mathbb{D}_n = \bracs{k/2^{-n}|k \in \integer}$. The set $\mathbb{D} = \bigcup_{n \in \natp}\mathbb{D}_n$ is the collection of all dyadic rational numbers.
+\end{definition}
+
+\begin{definition}[Rank]
+\label{definition:dyadic-rank}
+    Let $q \in \mathbb{D}$, then
+    \[
+    \text{rk}(q) = \min\bracs{n \in \natp|x \in \mathbb{D}_n}
+    \]
+    is the \textbf{dyadic rank} of $q$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:dyadic-decompose}
+    Let $x \in \mathbb{D}$ with $n = \text{rk}(x) > 1$, then there exists a unique $y \in \mathbb{D}_{n - 1}$ such that $x = y + 2^{-n}$.
+\end{lemma}
+\begin{proof}
+    Since $x \in \mathbb{D}_n$, there exists a unique $k \in \integer$ such that $x = k/2^{-n}$. Given that $x \not\in \mathbb{D}_{n-1}$, $k$ is odd. Therefore $y = (k - 1)/2^{-n}$ is the unique element of $\mathbb{D}_{n - 1}$ with $x = y + 2^{-n}$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:dyadic-subset}
+    Let $x \in \mathbb{D} \cap [0, 1)$, then there exists a unique $M(x) \subset \natp \cap [1, \text{rk}(x)]$ such that $x = \sum_{n \in M(x)}2^{-n}$.
+\end{proposition}
+\begin{proof}
+    First suppose that $\text{rk}(x) = 1$. In which case, $x \in \bracs{0, 1/2}$, and either $M(x) = \emptyset$ or $M(y) = \bracs{1}$.
+
+    Assume inductively that the proposition holds for all dyadic rationals of rank $n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$. By \ref{lemma:dyadic-decompose}, there exists a unique $y \in \mathbb{D}_n$ such that $x = y + 2^{-n-1}$. Since $x > 0$, $x \ge 1/2^{-n-1}$, so $y \in [0, 1)$. By the inductive assumption, there exists a unique $M(y) \subset \natp \cap [1, n]$ such that $y = \sum_{k \in M(y)}2^{-k}$. In which case, $M(x) = M(y) \cup \bracs{n}$ is the desired set.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:dyadic-semigroup-order}
+    Let $G$ be a commutative ordered semigroup, and $\seq{g_n} \subset G$ such that for each $n \in \natp$, $g_{n+1} + g_{n+1} \le g_n$. For each $x \in \mathbb{D} \cap [0, 1)$, let $\phi(x) = \sum_{n \in M(x)}g_n$, then
+    \begin{enumerate}
+        \item For any $x, y \in \mathbb{D} \cap [0, 1)$ such that $x + y \in [0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$.
+        \item For any $x, y \in \mathbb{D} \cap [0, 1)$ with $x \le y$, $\phi(x) \le \phi(y)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Assume without loss of generality that $x, y \in (0, 1)$. First suppose that $\text{rk}(x) = 2$ and $\text{rk}(y) \le 2$. In which case, since $G$ is commutative, it is sufficient to consider two cases: $(x, y) = (1/4, 1/4)$ and $(x, y) \in (1/2, 1/4)$. In the first case,
+    \[
+    \phi(x) + \phi(y) = g_2 + g_2 \le g_1 = \phi(x + y)
+    \]
+    In the second, $\phi(x) + \phi(y) = \phi(x + y)$.
+
+    Now assume inductively that the proposition holds for all $x, y \in \mathbb{D} \cap (0, 1)$ with $\text{rk}(x) = n$ and $\text{rk}(y) \le n$. Let $x \in \mathbb{D}_{n+1} \setminus \mathbb{D}_n$ and $y \in \mathbb{D}_{n+1}$. By \ref{lemma:dyadic-decompose}, there exists $x_0 \in \mathbb{D}_n$ such that $x = x_0 + 2^{-n-1}$. If $y \in \mathbb{D}_n$, then
+    \[
+    \phi(x) + \phi(y) = \phi(x_0) + \phi(y) + g_{n+1} \le \phi(x_0 + y) + g_{n+1} = \phi(x + y)
+    \]
+    by the inductive assumption. Otherwise, there exists $y_0 \in \mathbb{D}_n$ such that $y = y_0 + 2^{-n-1}$, so
+    \[
+    \phi(x) + \phi(y) \le \phi(x_0) + \phi(y_0) + g_n = \phi(x_0) + \phi(y_0) + \phi(2^{-n}) \le \phi(x + y)
+    \]
+    by the inductive assumption.
+\end{proof}

src/cat/tricks/index.tex

@@ -0,0 +1,4 @@
+\chapter{Inequalities and Computations}
+\label{chap:tricks}
+
+\input{./src/cat/tricks/dyadic.tex}