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}

src/fa/tvs/index.tex

@@ -2,6 +2,7 @@
 \label{chap:tvs}
 
 \input{./src/fa/tvs/definition.tex}
+\input{./src/fa/tvs/metric.tex}
 \input{./src/fa/tvs/bounded.tex}
 \input{./src/fa/tvs/dual.tex}
 \input{./src/fa/tvs/continuous.tex}

src/fa/tvs/metric.tex

@@ -0,0 +1,116 @@
+\section{Pseudonorms}
+\label{section:tvs-pseudonorm}
+
+\begin{definition}[Pseudonorm]
+\label{definition:pseudonorm}
+    Let $E$ be a vector space over $K \in \RC$, then a \textbf{pseudonorm} is a function $\rho: E \to [0, \infty)$ such that
+    \begin{enumerate}
+        \item[(PN1)] $\rho(0) = 0$.
+        \item[(PN2)] For any $x \in X$ and $\lambda \in K$ with $\abs{\lambda} \le 1$, $\rho(\lambda x) \le \rho(x)$.
+        \item[(PN3)] For any $x, y \in X$, $\rho(x + y) \le \rho(x) + \rho(y)$.
+        \item[(PN4)] For any $x \in X$, $\lim_{\lambda \to 0}\rho(\lambda x) = 0$.
+        \item[(PN5)] For any $\lambda \in K$, and $\seq{x_n} \subset X$ with $\rho(x_n) \to 0$ as $n \to \infty$, $\limv{n}\rho(\lambda x_n) = 0$.
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Topology Induced by Pseudonorm]
+\label{definition:tvs-pseudonorm-topology}
+    Let $E$ be a vector space over $K \in \RC$ and $\seqi{\rho}$ be pseudonorms on $E$. For each $i \in I$, let
+    \[
+    d_i: E \times E \quad (x, y) \mapsto \rho_i(x - y)
+    \]
+    then $d_i$ is a pseudometric. The uniform topology on $E$ induced by $\seqi{d}$ is the \textbf{topology induced by $\seqi{\rho}$}, and
+    \begin{enumerate}
+        \item The topology induced by $\seqi{\rho}$ is a vector space topology.
+        \item The uniformity induced by $\seqi{\rho}$ is the translation-invariant uniformity for its topology.
+        \item For each $i \in I$, $x \in E$, and $r > 0$, let $B_i(x, r) = \bracs{y \in E|d_i(x, y) < r}$, then
+        \[
+        \bracs{\bigcap_{j \in J}B_j(0, r)|J \subset I \text{ finite}, r > 0}
+        \]
+        is a fundamental system of neighbourhoods at $0$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (3): By \ref{definition:pseudometric-uniformity}.
+
+    (2): Each $d_i$ is translation-invariant.
+
+    (1):
+    \begin{enumerate}
+        \item[(TVS1)] Let $x, x', y, y' \in E$, $J \subset I$ be finite, and $r > 0$. If for each $j \in J$, $d_j(x, x'), d_j(y, y') < r/2$, then $d_j(x + x', y + y') < r$ by (PN3).
+        \item[(TVS2)] Let $x, x' \in E$ and $\lambda, \lambda' \in K$, then
+        \[
+        \lambda x - \lambda' x' = \lambda(x - x') + (\lambda - \lambda')x'
+        \]
+        Let $i \in I$ and $\eps > 0$. By (PN5) then there exists $\delta > 0$ such that if $\rho_i(x - x') < \delta$, then $\rho_i(\lambda (x - x')) < \eps$.
+
+        On the other hand,
+        \[
+        (\lambda - \lambda')x' = (\lambda - \lambda')x + (\lambda - \lambda')(x' - x)
+        \]
+        By (PN4), there exists $\delta' \in (0, 1]$ such that if $\abs{\lambda - \lambda'} < \delta'$, then $\rho_i((\lambda - \lambda')x) < \eps$. In which case, since $\delta' \le 1$, (PN2) implies that
+        \[
+        \rho_i((\lambda - \lambda')x') < \eps + \rho_i(x' - x) < 2\eps
+        \]
+        Therefore $\rho_i(\lambda x - \lambda' x') < 3\eps$.
+    \end{enumerate}
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:tvs-pseudonorm-compatible}
+    Let $E$ be a TVS over $K \in \RC$, and $\rho: E \to [0, \infty)$ be a pseudonorm, then the following are equivalent:
+    \begin{enumerate}
+        \item $\rho \in UC(E; [0, \infty))$.
+        \item $\rho \in C(E; [0, \infty))$.
+        \item $\rho$ is continuous at $0$.
+        \item The topology on $E$ contains the topology induced by $\rho$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    $(4) \Rightarrow (1)$: By \ref{definition:tvs-pseudonorm-topology}, for each $r > 0$, $\rho^{-1}([0, r)) \in \cn_E(0)$. Thus for any $x, y \in E$, if $x - y \in \rho^{-1}([0, r))$, then $\abs{\rho(x) - \rho(y)} \le r$. Therefore $\rho \in UC(E; [0, \infty))$.
+\end{proof}
+
+\begin{lemma}[{{\cite[Theorem 1.6.1]{SchaeferWolff}}}]
+\label{lemma:tvs-sequence-pseudonorm}
+    Let $E$ be a vector space over $K \in \RC$, $\seq{U_n} \subset 2^E$ such that
+    \begin{enumerate}
+        \item[(a)] For each $n \in \natp$, $U_n$ is circled, radial, and contains $0$.
+        \item[(b)] For each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$.
+    \end{enumerate}
+    then there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $n \in \natp$,
+    \[
+    U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}
+    \]
+\end{lemma}
+\begin{proof}
+    For each $H \subset \natp$ finite, let
+    \[
+    U_H = \sum_{n \in H}V_n \quad \rho_H = \sum_{n \in H}2^{-n}
+    \]
+    Define
+    \[
+    \rho: E \to [0, 1] \quad x \mapsto \inf\bracs{\rho_H|x \in U_H}
+    \]
+    then
+    \begin{enumerate}
+        \item[(PN1)] Since $0 \in \bigcap_{H \subset  \natp \text{ finite}}U_H$, $\rho(0) = 0$.
+        \item[(PN2)] Let $x \in X$ and $\lambda \in K$ with $\abs{\lambda} \le 1$. By assumption (a), $\lambda U_n \subset U_n$ for each $n \in \natp$. Thus for any $H \subset \natp$ finite with $x \in U_H$,
+        \[
+        \lambda x \in \sum_{n \in H}\lambda U_n \subset \sum_{n \in H}U_n
+        \]
+        so $\rho(\lambda x) \le \rho(x)$.
+        \item[(PN3)] Let $x, y \in X$ and $M, N \subset \natp$ finite such that $x \in U_M$ and $y \in U_N$. Assume without loss of generality that $\rho_M + \rho_N < 1$, then there exists a unique $P \subset \nat$ finite such that $\rho_P = \rho_M + \rho_N$. In which case, $U_P \supset U_M + U_N$ by assumption (b). Therefore $\rho(x + y) \le \rho(x) + \rho(y)$.
+    \end{enumerate}
+    For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1} < 2^n$, so $U_{n+1} \subset \rho^{-1}([0, 2^{-n}))$ by \ref{proposition:dyadic-semigroup-order}. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}} = U_n$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
+    \begin{enumerate}
+        \item[(PN4)] Let $x \in X$ and $n \in \natp$. By assumption (a), there exists $\alpha > 0$ such that for any $\lambda \in K$ with $\abs{\lambda} \ge \alpha$, $x \in \lambda U_n$. Therefore for any $\lambda \in K$ with $\abs{\lambda} \le \alpha^{-1}$, $\lambda x \in U_n$, and $\rho(x) \le 2^{-n}$.
+        \item[(PN5)] Let $\lambda \in K$ and $n \in \natp$. By assumption (b), there exists $m \in \nat$ such that $\lambda U_{n-m} \subset \sum_{j = 1}^m U_{n-m}^j \subset U_n$.
+    \end{enumerate}
+\end{proof}
+
+\begin{remark}
+	\label{remark:tvs-sequence-pseudonorm}
+
+	As discussed in \ref{remark:uniform-sequence-pseudometric} on the proof of \ref{lemma:uniform-sequence-pseudometric}, constructing a pseudometric from entourages by building its level sets is difficult because composing symmetric entourages does not necessarily lead to symmetric entourages. The topological vector space does not have this shortcoming, and as such allows this construction.
+\end{remark}

src/topology/uniform/metric.tex

@@ -146,6 +146,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
 \end{remark}
 
 \begin{remark}
+\label{remark:uniform-sequence-pseudometric}
 	It may be tempting to construct the level sets of the pseudometric on the dyadic rational numbers by composing these sets, then proceed to construct the pseudometric as in Urysohn's lemma. However, this approach has a major shortcoming in that the composition of symmetric entourages are not necessarily symmetric. As such, it is difficult to construct symmetric level sets for the desired pseudometric.
 \end{remark}