Consolidating.

src/dyn/top/def.tex

@@ -66,6 +66,12 @@
     Let $G$ be an infinite group, $G \curvearrowright X$ be a $G$-flow, and $x \in X$, then $x$ is \textbf{recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is infinite.
 \end{definition}
 
+\begin{definition}[Uniformly Recurrent]
+\label{definition:tds-uniformly-recurrent}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{uniformly recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is syndetic. 
+\end{definition}
+
+
 \begin{definition}[Syndetic]
 \label{definition:group-syndetic}
     Let $G$ be a group and $S \subset G$, then $S$ is \textbf{(left) syndetic} if there exists $F \subset G$ finite $FS = G$. 
@@ -94,7 +100,7 @@
     \begin{enumerate}
         \item $G \curvearrowright X$ admits no proper subflows.
         \item Every point in $X$ is topologically transitive. 
-        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. 
+        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. In particular, $x$ is uniformly recurrent. 
     \end{enumerate}
     
 
@@ -119,10 +125,25 @@
 \end{proof}
 
 
+\begin{definition}[Minimal Point]
+\label{definition:tds-minimal-point}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $G \curvearrowright \ol{Gx}$ is a minimal $G$-subflow.
+        \item $x$ is uniformly recurrent. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): See \autoref{definition:tds-minimal}. 
+
+    (2) $\Rightarrow$ (1): Let $Y = \ol{Gx}$ and $y \in Y$. For any $U \in \cn_X(x)$, since $X$ is a compact Hausdorff space, there exists $V \in \cn_X(x)$ with $\ol{V} \subset U$. As $N(x, V)$ is syndetic in $G$, there exists $F \subset G$ finite such that $G = F^{-1}N(x, V)$. 
+
+    Since $\ol{Gx} = Y$, there exists a filter $\fF \subset 2^G$ such that $\fF x \to y$. As $G = \bigcup_{f \in F}N(x, V)$, there exists $f \in F$ such that $f^{-1}N(x, V) \in \fF$. Thus $y = \lim \fF x \subset \ol{f^{-1}N(x, V)} \subset \ol{f^{-1}V}$. Equivalently, $fy \in \ol{V}$, and $Gy \cap U \supset Gy \cao f^{-1}\ol{V} \ne \emptyset$. As this holds for all $U \in \cn_X(x)$, $x \in \ol{Gy}$, and $\ol{Gy} = \ol{Gx} = Y$. Therefore every point in $Y$ is transitive, and $G \curvearrowright Y$ is a minimal subflow. 
+\end{proof}
 
 \begin{theorem}[Existence of Minimal Subflows]
 \label{theorem:tds-minimal-existence}
-    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. 
+    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. In which case, every point in $Y$ is a minimal. 
 \end{theorem}
 % Proof omitted. 
 

src/op/banach/definitions.tex

@@ -86,3 +86,5 @@
 \end{proof}
 
 
+
+

src/op/c-star/involution.tex

@@ -42,3 +42,21 @@
 \end{proof}
 
 
+\begin{definition}[$M_n(A)$]
+\label{definition:matrix-cstar}
+    Let $A$ be a $C^*$-algebra, $M_n(A) = M_n(\complex) \otimes A$ be the space of $n \times n$ matrices over $A$, equipped with the product
+    \[
+    M_n(A) \times M_n(A) \quad (M \otimes a)(N \otimes b) = MN \otimes ab
+    \]
+    and the involution
+    \[
+    *: M_n(A) \to M_n(A) \quad M \otimes a \mapsto M^* \otimes a^*
+    \]
+
+    Let $H$ be a complex Hilbert space and $\pi: A \to B(H)$ be a faithful representation of $A$, then $\pi$ induces a faithful representation $M_n(A) \otimes A \to B(H^n)$. Under the identification that $M_n(A) \subset B(H^n)$, the inherited norm\footnote{Which is equal to the projective norm because $M_n(\complex)$ is finite-dimensional and hence nuclear.} makes $M_n(A)$ a $C^*$-algebra. 
+\end{definition}
+
+\begin{definition}[Completely]
+\label{definition:cstar-completely}
+    Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a function, then $\phi$ induces mappings $\phi_n = \text{Id}_n \otimes \phi: M_n(A) \to M_n(B)$. A property of $\phi$ holds \textbf{completely} if it holds for $\phi_n$ for all $n \in \natp$. 
+\end{definition}

src/op/system/definitions.tex

@@ -1,9 +1,98 @@
 \section{Operator Systems}
 \label{section:operator-systems}
 
-\begin{definition}[Operator System]
+\begin{definition}[Concrete Operator System]
-\label{definition:operator-system}
+\label{definition:concrete-operator-system}
-    Let $A$ be a $C^*$-algebra and $S \subset A$, then $S$ is an \textbf{operator system} if $S$ is a linear subspace of $A$ and $S = S^*$. 
+    Let $A$ be a $C^*$-algebra and $S \subset A$, then $S$ is a \textbf{concrete operator system} if:
+    \begin{enumerate}
+        \item $1_A \in S$.
+        \item $S = S^*$. 
+        \item $S$ is a linear subspace of $A$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Self-Adjoint Part]
+\label{definition:operator-system-sa}
+    Let $A$ be a $C^*$-algebra and $S \subset A$ be an operator system, then $S_{sa} = \bracsn{x \in S|x = x^*}$ is the \textbf{self-adjoint part} of $S$. 
+\end{definition}
+
+\begin{definition}[Positive Cone]
+\label{definition:operator-system-positive-cone}
+    Let $A$ be a $C^*$-algebra and $S \subset A$ be an operator system, then $S_{+} = \bracsn{x \in S_{sa}|x \ge 0}$ is the \textbf{positive cone} of $S$. 
 \end{definition}
 
 
+\begin{proposition}
+\label{proposition:operator-system-positive-span}
+    Let $A$ be a $C^*$-algebra, $S \subset A$ be an operator system, and $x \in S_{sa}$, then there exists positive elements $p, q \in S_{sa}$ such that $x = p - q$. 
+\end{proposition}
+\begin{proof}
+    Let $p = (\norm{x}_A + x)/2$ and $q = (\norm{x}_A - x)/2$, then $q$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
+\end{proof}
+
+\begin{definition}[Positive]
+\label{definition:operator-system-positive-map}
+    Let $A, B$ be $C^*$-algebra, $S \subset A$ and $T \subset B$ be operator systems, and $\phi: S \to T$ be a linear map, then $\phi$ is \textbf{positive} if $\phi(S_+) \subset \phi(T_+)$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:positive-automatic-continuity}
+    Let $A, B$ be $C^*$-algebra, $S \subset A$ and $T \subset B$ be operator systems, and $\phi: S \to T$ be a positive map, then
+    \[
+    \norm{\phi}_{L(S; T)} \le 2\norm{\phi(1_A)}_B
+    \]
+\end{proposition}
+\begin{proof}
+    Let $x \in T_+$, then $0 \le x \le \norm{x}_A$, and $0 \le \phi(x) \le \norm{x}_A \cdot \norm{\phi(1_A)}_B$, so $\norm{\phi(x)}_B \le \norm{x}_A \cdot \norm{\phi(1_A)}_B$. 
+
+    For any $x, y \in T_+$, $x - y \le x$ and $y - x \le y$. As such, 
+    \[
+    \norm{x - y}_{B} \le \norm{x}_B \vee \norm{y}_B
+    \]
+
+    In particular, for any $s \in S_{sa}$, 
+    \begin{align*}
+        \phi(s) &= \frac{1}{2}\braks{\phi(\norm{s}_A + s) - \phi(\norm{s}_A - s)} \\
+        \norm{\phi(s)}_{B} &\le \frac{1}{2}\braks{\norm{\phi(\norm{s}_A + s)}_B \vee \norm{\phi(\norm{s}_A - s)}_B} \\
+        &\le \norm{s}_A \norm{\phi(1_A)}_B
+    \end{align*}
+    
+    Finally, for any general $s \in S$, 
+    \begin{align*}
+        \phi(s) &= \phi(\text{Re}(s)) + i\phi(\text{Im}(s)) \\
+        \norm{\phi(s)}_B &\le \norm{\phi(\text{Re}(s))}_B + \norm{\phi(\text{Im}(s))}_{B} \\
+        &\le \norm{\phi(1_A)}_B \cdot (\norm{\text{Re}(s)}_A + \norm{\text{Im}(s)}_A) \le 2\norm{\phi(1_A)}_B\norm{s}_A
+    \end{align*}
+\end{proof}
+
+
+\begin{example}[Arveson]
+    \label{example:positive-automatic-continuity}
+
+    Let $\mathbb{T} = \bracsn{z \in \complex|\ |z| = 1}$, and $S = \text{span}\bracsn{1, z, \ol z}  \subset C(\mathbb{T}; \complex)$, then $S$ is an operator system in $C(\mathbb{T} ; \complex)$. Let
+    \[
+    \phi: S \to M_2(\complex) \quad (a + bz + c\ol z) = \begin{bmatrix}
+        a & 2b \\
+        2c & a
+    \end{bmatrix}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $\phi$ is a positive map.
+        \item $\norm{\phi}_{L(C(\mathbb{T} ; \complex); M_2(\complex))} = 2$. 
+    \end{enumerate}
+\end{example}
+\begin{proof}[Proof, {{\cite[Example 2.2]{Paulsen}}}. ]
+    (1): Let $a + bz + c \ol z \in S$, then the following are equivalent:
+    \begin{enumerate}
+        \item $a + bz + c\ol z \ge 0$. 
+        \item $c = \ol b$ and $a \ge 2\text{Re}(b)$. 
+        \item $c = \ol b$ and $a \ge 2|b|$. 
+    \end{enumerate}
+
+    In addition, since a $2 \times 2$ matrix is positive if and only if its diagonal entries and its determinant are non-negative. Thus $\phi$ is positive. 
+
+    (2): $\norm{\phi(z)}_{M_2(\complex)} = 2$. 
+    
+\end{proof}

src/topology/main/compactify.tex

@@ -79,7 +79,7 @@
     U_A = \bracsn{\fU \in \beta X| A \in \fU}
     \]
 
-    Equip $\beta X$ with the topology generated by $\fU := \bracsn{U_A|\emptyset \ne A \subset X}$. For each $x \in X$, let $e(x)$ be the principal ultrafilter at $x$, then
+    Equip $\beta X$ with the topology generated by $\fB := \bracsn{U_A|\emptyset \ne A \subset X}$. For each $x \in X$, let $e(x)$ be the principal ultrafilter at $x$, then
     \begin{enumerate}
         \item $(\beta X, e)$ is a compactification of $X$. 
         \item[(U1)] For any $f \in C(X; [0, 1])$, there exists a unique $\beta f \in C(\beta X; [0, 1])$ such that the following diagram commutes:
@@ -94,11 +94,11 @@
     In other words, $(\beta X, e)$ as constructed above is the Stone-Čech compactification of $X$. 
 \end{theorem}
 \begin{proof}
-    (1): Firstly, $\beta X = U_{X}$. For any $A, B \subset X$, $U_{A \cap B} = U_A \cap U_B$. Thus $\fU$ is a base for the topology that it generates. 
+    (1): Firstly, $\beta X = U_{X}$. For any $A, B \subset X$, $U_{A \cap B} = U_A \cap U_B$. Thus $\fB$ is a base for the topology that it generates. 
 
     Let $\mathfrak{U}, \mathfrak{V} \in \beta X$ be distinct, then there exists $A \subset X$ with $A \in \mathfrak{U}$ and $A^c \in \mathfrak{V}$, so $\mathfrak{U} \in U_A$, $\mathfrak{V} \in U_{A^c}$, and $U_A \cap U_{A^c} = \emptyset$. Thus $\beta X$ is Hausdorff. 
 
-    Let $\mathcal{U} \subset 2^{\beta X}$ be an open cover of $X$. Since $\fU$ is a base for $\beta X$, assume without loss of generality that $\mathcal{U}$ is of the form $\bracsn{U_A|A \in \mathcal{S}}$, where $\mathcal{S} \subset 2^X \setminus \bracsn{\emptyset}$. Suppose for contradiction that $\mathcal{U}$ admits no finite subcover, then for any $F \subset \mathcal{S}$ finite,
+    Let $\mathcal{U} \subset 2^{\beta X}$ be an open cover of $X$. Since $\fB$ is a base for $\beta X$, assume without loss of generality that $\mathcal{U}$ is of the form $\bracsn{U_A|A \in \mathcal{S}}$, where $\mathcal{S} \subset 2^X \setminus \bracsn{\emptyset}$. Suppose for contradiction that $\mathcal{U}$ admits no finite subcover, then for any $F \subset \mathcal{S}$ finite,
     \[
     \bigcap_{A \in F}U_{A^c} = \bigcap_{A \in F}U_A^c = \braks{\bigcup_{A \in F}U_A}^c \ne \emptyset
     \]