From 6341e8d81c3014f0d7e54301f0d389d8f2b075cb Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 5 Oct 2026 22:23:08 -0400 Subject: [PATCH] Emergency push! --- src/dyn/top/def.tex | 24 +++ src/dyn/top/index.tex | 1 + src/dyn/top/stone-cech.tex | 54 ++++++ src/op/example/bb.tex | 39 ++++ src/op/system/complete-upgrade.tex | 298 +++++++++++++++++++++++++++++ src/op/system/completely.tex | 202 ++++--------------- src/op/system/index.tex | 1 + src/op/vn/commutative.tex | 6 +- src/topology/main/compactify.tex | 10 + 9 files changed, 468 insertions(+), 167 deletions(-) create mode 100644 src/dyn/top/stone-cech.tex create mode 100644 src/op/system/complete-upgrade.tex diff --git a/src/dyn/top/def.tex b/src/dyn/top/def.tex index 6952c71..66d364c 100644 --- a/src/dyn/top/def.tex +++ b/src/dyn/top/def.tex @@ -4,6 +4,16 @@ \begin{definition}[Topological Dynamical System] \label{definition:topological-dynamical-system} Let $G$ be a group and $X$ be a (non-empty) compact Hausdorff space, then a \textbf{$G$-flow/topological dynamical system} is a homomorphism $\alpha: G \to \text{Aut}(X)$. In which case, the system is denoted by $G \curvearrowright X$. + + Let $G \curvearrowright X$ and $G \curvearrowright$ be $G$-flows, then a \textbf{morphism of $G$-flows} is a continuous map $\pi: X \to Y$ such that the following diagram commutes + \[ + \xymatrix{ + X \ar@{->}[r]^{\pi} \ar@{->}[d]_{g} & Y \ar@{->}[d]^{g} \\ + X \ar@{->}[r]_{\pi} & Y + } + \] + + for all $g \in G$. \end{definition} \begin{definition}[Invariant] @@ -59,6 +69,20 @@ \end{example} % Proof omitted. + +\begin{definition}[Pointed Flow] +\label{definition:tds-pointed} + Let $G$ be a group, then a \textbf{pointed $G$-flow} is a pair $(X, x_0)$, where $X$ is a $G$-flow, and $x_0 \in X$ is a distinguished point. + + Let $(G \curvearrowleft X, x_0)$ and $(G \curvearrowright Y, y_0)$ be pointed $G$-flows, then a \textbf{morphism of pointed $G$-flows} is a morphism of $G$-flows $\pi: X \to Y$ such that $\pi(x_0) = y_0$. +\end{definition} + +\begin{definition}[Ambit] +\label{definition:tds-ambit} + Let $G$ be a group and $(G \curvearrowright X, x_0)$ be a pointed $G$-flow, then $(G \curvearrowright X, x_0)$ is an \textbf{ambit} if $x_0$ is a topologically transitive point. +\end{definition} + + \begin{definition}[Bernoulli Shift] \label{definition:bernoulli-shift} Let $G$ be a group, $X = 2^G$ equipped with the product topology, and diff --git a/src/dyn/top/index.tex b/src/dyn/top/index.tex index 76a6106..8bdc4ff 100644 --- a/src/dyn/top/index.tex +++ b/src/dyn/top/index.tex @@ -4,3 +4,4 @@ \input{./def.tex} \input{./minimal.tex} \input{./recurrence.tex} +\input{./stone-cech.tex} diff --git a/src/dyn/top/stone-cech.tex b/src/dyn/top/stone-cech.tex new file mode 100644 index 0000000..92be03f --- /dev/null +++ b/src/dyn/top/stone-cech.tex @@ -0,0 +1,54 @@ +\section{$G \curvearrowright \beta G$} +\label{section:stone-cech-flow} + +\begin{definition}[Greatest Ambit] +\label{definition:stone-cech-flow} + Let $G$ be a discrete group, and + \[ + \beta L: G \to \text{Aut}(\beta G) \quad \beta L_g(\fU) = g\fU + \] + + then + \begin{enumerate} + \item $(\beta L: G \curvearrowright \beta G, 1_G)$ is an ambit. + \item[(U)] For any ambit $(G \curvearrowright X, x_0)$, there exists a unique morphism of pointed $G$-flows $\pi: (\beta G, 1_G) \to (X, x_0)$. Moreover, $\pi$ is surjective. + \end{enumerate} + + Any ambit $(Y, y_0)$ satisfying (U) is the \textbf{greatest $G$-ambit}. +\end{definition} +\begin{proof} + (1): Let + \[ + L: G \to \text{Perm}(G) \quad L_g(h) = gh + \] + + be the left translation action on itself. For each $g \in G$, the functoriality of the \hyperref[Stone-Čech compactification]{definition:stone-cech} implies that $L_g$ extends into a continuous map $\beta L_g: \beta G \to \beta G$. By uniqueness of continuous extensions, $\beta L_g(\fU) = g\fU$ for all $\fU \in \beta G$. In particular, $\text{Id}_{\beta G} = \beta L_{e}$. + + Moreover, for any $g, h \in G$, $\beta L_{gh} = \beta L_{g} \circ \beta L_{h}$, so $g \mapsto L_g$ is a homomorphism. Finally, for any $g \in G$, + \[ + \beta L_{g^{-1}} \circ \beta L_{g} = \beta L_{1_G} = \beta L_{g} \circ \beta L_{g^{-1}} + \] + + and $\beta L_g \in \text{Aut}(\beta G)$. + + Since $\bracsn{L_g1_G|g \in G} = G$ is dense in $\beta G$, $(G \curvearrowright \beta G, 1_G)$ is an ambit. + + (2): For each $g \in G$, let $\pi_0(g) = gx_0$, then $\pi_0 \in C(G; X)$, $\pi_0(1_G) = x_0$, and the following diagram commutes + \[ + \xymatrix{ + G \ar@{->}[r]^{\pi_0} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\ + G \ar@{->}[r]_{\pi_0} & X + } + \] + + for all $g \in G$. By universal property of $\beta G$, $\pi_0$ admits a unique continuous extension $\pi \in C(\beta G; X)$. By uniqueness of continuous extensions, the following diagram commutes + \[ + \xymatrix{ + \beta G \ar@{->}[r]^{\pi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\ + \beta G \ar@{->}[r]_{\pi} & X + } + \] + + for all $g \in G$. Finally, since $\beta G$ is compact and $X$ is Hausdorff, $\pi(\beta G)$ is \hyperref[also compact]{proposition:compact-extensions} and \hyperref[hence closed]{proposition:compact-closed}. As $(X, x_0)$ is an ambit and $\pi(\beta G) \supset Gx_0$, $\pi(\beta G) \supset \ol{Gx_0} = X$. +\end{proof} + diff --git a/src/op/example/bb.tex b/src/op/example/bb.tex index 159fb60..a42de99 100644 --- a/src/op/example/bb.tex +++ b/src/op/example/bb.tex @@ -55,16 +55,55 @@ (4): $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is a commutative unital $C^*$-algebra. \end{proof} +\begin{lemma} +\label{lemma:linfty-multiply-bound} + Let $(X, \cm, \mu)$ be a measure space, $f: X \to \complex$ be a measurable function, and $C \ge 0$ such that $\norm{fg}_{L^2(X; \complex)} \le C\norm{g}_{L^2(X; \complex)}$ for all $g \in L^2(X; \complex) \cap \Sigma(X; \complex)$, then $f \in L^\infty(X; \complex)$ with $\norm{f}_{L^\infty(X; \complex)} \le C$. +\end{lemma} +\begin{proof} + For any $h \in L^1(X; \complex) \cap L^\infty(X; \complex)$, + \begin{align*} + \normn{f^2h}_{L^1(X; \complex)} &= \int_X |f^2| \cdot |h| = \int_X |f||h|^{1/2} \cdot |f||h|^{1/2} = \normn{f|h|^{1/2}}_{L^2(X; \complex)}^2 \\ + &\le C^2 \normn{|h|^{1/2}}_{L^2(X; \complex)}^2 = C^2 \norm{h}_{L^1(X; \complex)} + \end{align*} + + By \autoref{theorem:lp-dual-function}, $f^2 \in L^\infty(X; \complex)$ with $\normn{f^2}_{L^\infty(X; \complex)} \le C^2$. Hence $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le C$. +\end{proof} + \begin{proposition} \label{proposition:linfty-von-neumann-algebra} Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^\infty(X; \complex)$ act on $L^2(X; \complex)$ by multiplication, then: \begin{enumerate} \item The weak* topology on $L^\infty(X; \complex)$ is equal to the weak operator topology of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$. \item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra. + \item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is maximal abelian. \end{enumerate} \end{proposition} \begin{proof} (2): Let $A \subset B(L^2(X; \complex))$ be the von Neumann algebra generated by $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$. By \autoref{theorem:lp-duality}, $L^\infty(X; \complex)$ is the dual of $L^1(X; \complex)$, so (1) and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu} imply that the closed unit ball of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is weak-operator closed. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_{L^\infty(X; \complex)}(0, 1)} = \ol{B_A(0, 1)}$. Therefore $L^\infty(X; \complex) = A$. + + (3): Using \autoref{proposition:maximal-commutative-vn}, it is sufficient to show that $L^\infty(X; \complex)' = L^\infty(X; \complex)$. For each $g \in L^\infty(X; \complex)$, denote $M_g \in B(L^2(X; \complex))$ as the multiplication map by $g$. + + First suppose that $\mu(X) < \infty$ and let $f = T\one_X$, then for each $g \in L^\infty(X; \complex)$, + \[ + M_fg = M_g f= M_g T\one_X = TM_g\one_X = Tg + \] + + By \autoref{lemma:linfty-multiply-bound}, $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le \norm{T}_{B(L^2(X; \complex))}$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex) \subset L^\infty(X; \complex)$ is dense in $L^2(X; \complex)$, so $M_fg = Tg$ for all $g \in L^2(X; \complex)$. + + Now suppose that $\mu$ is an arbitrary localisable measure. For each $A \in \cm$ with $\mu(A) < \infty$, the finite case shows that there exists $f_A \in L^\infty(X; \complex)$ such that + \begin{enumerate}[label=(\roman*)] + \item $f_A|_{X \setminus A} = 0$ almost everywhere. + \item $Tg = M_{f_A}g$ for all $g \in L^2(X; \complex)$ with $f_A|_{X \setminus A} = 0$ almost everywhere. + \end{enumerate} + + By (ii), $f_A|_{A \cap B} = f_B|_{A \cap B}$ for all $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$. By the \hyperref[gluing lemma for meaasurable functions]{lemma:gluing-measurable}, there exists $f \in L^\infty(X; \complex)$ such that $f|_A = f_A|_A$ for all $A \in \cm$ with $\mu(A) < \infty$. + + For each $g \in L^2(X; \complex)$, $\bracsn{g \ne 0}$ is $\sigma$-finite. As such, there exists $\seq{A_n} \subset \cm$ such that $\bracsn{g \ne 0} = \bigsqcup_{n \in \natp}A_n$ and $\mu(A_n) < \infty$ for all $n \in \natp$. In which case, + \[ + Tg = \sum_{n = 1}^\infty T(\one_{A_n}g) = \sum_{n = 1}^\infty M_{f_{A_n}} \one_{A_n}g = M_fg + \] + + By \autoref{lemma:linfty-multiply-bound}, $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le \norm{T}_{B(L^2(X; \complex))}$. Therefore $T = M_f$, and $L^\infty(X; \complex)$ is maximal abelian. \end{proof} \begin{theorem}[Order Structure of $L^\infty$] diff --git a/src/op/system/complete-upgrade.tex b/src/op/system/complete-upgrade.tex new file mode 100644 index 0000000..856891d --- /dev/null +++ b/src/op/system/complete-upgrade.tex @@ -0,0 +1,298 @@ +\section{Sufficient Conditions for Complete Positivity} +\label{section:completely-upgrade} + + +\begin{proposition} +\label{proposition:homomorphism-completely-positive} + Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a *-homomorphism, then $\phi$ is completely positive. +\end{proposition} +\begin{proof} + For each $n \in \natp$, the induced map $\phi_n: M_n(A) \to M_n(B)$ is a *-homomorphism. +\end{proof} + + +\begin{example} + \label{example:multiply-completely-bounded} + + Let $A$ be a $C^*$-algebra, $y, z \in A$, $\phi: A \to A$ be defined by $x \mapsto yxz$, then + \begin{enumerate} + \item $\phi$ is completely bounded. + \item If $y = z^*$, then $\phi$ is completely positive. + \end{enumerate} +\end{example} +\begin{proof} + (1): For each $n \in \natp$ and $T \in M_n(A)$, + \[ + \phi_n(T) = (1_{M_n(\complex)} \otimes y)T(1_{M_n(\complex)} \otimes z) + \] +\end{proof} + +The following is a slight generalisation of one form of the above example, but stating it helps greatly. + +\begin{proposition} +\label{proposition:multiply-completely-bounded} + Let $A$ be a $C^*$-algebra, $S, T \subset A$ be operator systems, and $\phi: S \to T$. If there exists $x \in A$ such that $\phi(s) = x\phi(s)x^*$ for all $s \in S$, then $\phi$ is completely positive. +\end{proposition} + + + +\begin{proposition} +\label{proposition:completely-positive-norm} + Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then + \[ + \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B + \] +\end{proposition} +\begin{proof} + After taking a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $B$ is unital. + + It is sufficient to show that $\norm{\phi}_{\text{cb}} \le \norm{\phi(1_A)}_B$. To this end, let $T \in M_n(S)$ with $\norm{T}_{M_n(S)} \le 1$, then + \[ + \begin{bmatrix} + 1_{M_n(A)} &T \\ + T^* &1_{M_n(A)} + \end{bmatrix} \ge 0 + \] + + in $M_{2}(M_n(S)) = M_{2n}(S)$ by \autoref{lemma:norm-matrix-test}. Since $\phi$ is completely positive, + \[ + \phi_{2n}\begin{bmatrix} + 1_{M_n(A)} &T \\ + T^* &1_{M_n(A)} + \end{bmatrix} = \begin{bmatrix} + \phi_n(1_{M_n(A)}) &\phi_n(T) \\ + \phi_n(T)^* & \phi_n(1_{M_n(A)}) + \end{bmatrix} \ge 0 + \] + + First suppose that $\phi_n(1_{M_n(A)})$ is invertible, then + + \begin{align*} + \phi_n(1_{M_n(A)}) &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot 1_{M_n(B)} \\ + 1_{M_n(B)} &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot \phi_n(1_{M_n(A)})^{-1} \\ + \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1} + \end{align*} + + By the \hyperref[Schur complement condition]{lemma:schur-complement-positive}, + \begin{align*} + \phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\ + &\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\ + \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^2 &\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)} = \normn{\phi_n(T)}_{M_n(B)}^2 + \end{align*} + + and $\normn{\phi_n(T)}_{M_n(B)} \le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}} \le \normn{\phi(1_A)}_B$. + + Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_B\psi$ is also a completely positive map. In which case, + \[ + \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B + \] +\end{proof} + +\begin{proposition} +\label{proposition:commutative-mn-facts} + Let $X$ be a compact Hausdorff space, then: + \begin{enumerate} + \item $C(X; M_n(\complex))$ equipped with pointwise product, pointwise involution, and the norm + \[ + \norm{f}_{C(X; M_n(\complex))} = \sup_{x \in X}\norm{f(x)}_{M_n(\complex)} + \] + + is a $C^*$-algebra. + \item The mapping + \[ + M_n(C(X; \complex)) \to C(X; M_n(\complex)) \quad M \otimes f \mapsto M \cdot f + \] + + is a *-isomorphism. + \item For any $f \in C(X; M_n(\complex))$, $f \ge 0$ if and only if $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): For any $f \in C(X; M_n(\complex))$, + \[ + \sup_{x \in X}\norm{f(x)}_{M_n(\complex)}^2 = \sup_{x \in X}\norm{f(x)^*f(x)}_{M_n(\complex)} = \norm{f^*f}_{C(X; M_n(\complex))} + \] + + (3): If $f \ge 0$, then there exists $g \in C(X; M_n(\complex))$ such that $f(x) = g^*(x)g(x)$ for all $x \in X$, so $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$. + + On the other hand, if $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$, then $g(x) = f(x)^{1/2}$ is defined for all $x \in X$. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $g$ is continuous from $X$ to $M_n(\complex)$. Therefore there exists $g \in C(X; M_n(\complex))$ such that $f = g^*g$. +\end{proof} + + +\begin{theorem} +\label{theorem:bounded-functional-completely-bounded} + Let $A$ be a unital $C^*$-algebra, $S \subset A$ be a subspace, $X$ be a compact Hausdorff space, and $\phi \in L(S; C(X; \complex))$, then: + \begin{enumerate} + \item $\norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; C(X; \complex))}$. + \item If $S$ is an operator system and $\phi$ is positive, then $\phi$ is completely positive. + \end{enumerate} +\end{theorem} +\begin{proof}[Proof, {{\cite[Proposition 3.8]{Paulsen}}}. ] + By \autoref{proposition:commutative-mn-facts}, the norm and order relation on $C(X; M_n(\complex)) \iso M_n(C(X; \complex))$ are pointwise. As such, assume without loss of generality that $X$ is a single point and $C(X; \complex) = \complex$. + + Let $a = (a_{ij}) \in M_n(S)$. For each $\xi, \eta \in \complex^n$, + \begin{align*} + \dpn{\phi_n(a)\xi, \eta}{\complex^n} &= \sum_{i, j = 1}^n \xi_j \ol{\eta_i} \dpn{a_{ij}, \phi}{S} = \angles{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}, \phi}_S + \end{align*} + + where + \begin{align*} + \sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i} &= (\one \otimes \ol{\eta}) \cdot a \cdot (\xi \otimes \one) + \end{align*} + + (1): + \begin{align*} + |\dpn{\phi_n(a)\xi, \eta}{\complex^n}| &\le \norm{\phi}_{S^*} \cdot \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S} \\ + \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S} &\le \norm{\xi}_{\complex^n}\norm{\eta}_{\complex^n}\norm{a}_{M_n(S)} \\ + \norm{\phi_n(a)}_{M_n(\complex)} &\le \norm{\phi}_{S^*}\norm{a}_{M_n(S)} \\ + \norm{\phi}_{\text{cb}} &\le \norm{\phi}_{S^*} + \end{align*} + + (2): If $\xi = \eta$, then + \[ + \sum_{i, j = 1}^n a_{ij}\eta_j\ol{\eta_i} = (\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one) \ge 0 + \] + + so + \[ + \dpn{\phi_n(a)\xi, \eta}{\complex^n} = \dpn{(\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one), \phi}{S} \ge 0 + \] +\end{proof} + + +\begin{theorem}[Stinespring] +\label{theorem:stinespring-commutative-completely-positive} + Let $X$ be a compact Hausdorff space, $B$ be a $C^*$-algebra, $\phi: C(X; \complex) \to B$ be a positive map, then $\phi$ is completely positive. +\end{theorem} +\begin{proof} + Let $P \in C(X; M_n(\complex))$ with $P \ge 0$ and $\eps > 0$. By \autoref{proposition:uniform-continuous-compact} and \autoref{theorem:lch-paracompact}, there exists a partition of unity $\seqf{\psi_j} \subset C(X; [0, 1])$ and positive matrices $\seqf{p_j} \subset M_n(\complex)$ such that + \[ + \norm{P(x) - \sum_{j = 1}^n p_j\psi_j(x)}_{M_n(\complex)} < \eps + \] + + for all $x \in X$. For each $1 \le j \le n$, $\phi_n(p_j\psi_j) = p_j\phi(\psi_j) \ge 0$. As $\eps > 0$ is arbitrary, $\phi_n(P) \ge 0$. +\end{proof} + + + +\begin{definition}[Matrix Unit] +\label{definition:choi-matrix-unit} + Let $n \in \natp$, $\seqf{e_j}$ be the standard orthonormal basis for $\complex^n$, and $1 \le i, j \le n$, then the \textbf{$(i, j)$-th matrix unit} is the operator $E_{ij} = e_j \otimes e_i$. +\end{definition} + +\begin{lemma} +\label{lemma:cstar-positive-matrix-decompose-sum} + Let $A$ be a $C^*$-algebra and $P \in M_n(A)$ be positive, then there exists $\bracsn{a_{k, j}}_{j, k = 1}^n$ such that: + \begin{enumerate} + \item $P = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j})$. + \item For each $1 \le k \le n$, + \[ + \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j}) = \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^*\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}} \ge 0 + \] + \end{enumerate} +\end{lemma} +\begin{proof} + Since $P \ge 0$, there exists $\bracsn{a_{j, k}}_{j, k = 1}^n$ such that + \begin{align*} + P &= \braks{\sum_{j, k = 1}^n E_{jk} \otimes a_{j, k}}^*\braks{\sum_{j, k = 1}^n E_{jk} \otimes a_{j, k}} \\ + &= \sum_{i, j, k, l = 1}^n (E_{ij} \otimes a_{i, j})^*(E_{kl} \otimes a_{k, l}) = \sum_{i, j, k, l = 1}^n (E_{ji}E_{kl}) \otimes (a_{i, j}^* a_{k, l}) + \end{align*} + + For each $1 \le i, j, k, l \le n$, $E_{ji}E_{kl} = E_{il}$ if $i = k$, and $E_{ji}E_{kl} = 0$ otherwise. Thus after relabeling, + \[ + P = \sum_{j, k, l = 1}^nE_{jl} \otimes (a_{k, j}^*a_{k, l}) = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j}) + \] + + Finally, for each $1 \le k \le n$, undoing the product shows that + \begin{align*} + \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j}) &= \sum_{i, j = 1}^n (E_{ik}E_{kj}) \otimes (a_{k, i}^*a_{k, j}) \\ + &= \sum_{i, j = 1}^n (E_{ki} \otimes a_{k, i})^* (E_{kj} \otimes a_{k, j}) \\ + &= \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^*\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}} \ge 0 + \end{align*} + +\end{proof} + + +\begin{theorem}[Choi] +\label{theorem:choi-matrix-positive} + Let $n \in \natp$, $B$ be a $C^*$-algebra, and $\phi: M_n(\complex) \to B$ be a linear map. For each $1 \le i, j \le n$, let $E_{ij}$\footnote{$e_j \otimes e_i \in \complex^n \otimes \complex^n$.} be the standard matrix unit for $M_n(\complex)$, then $C_\phi = (\phi(E_{ij})) \in M_n(B)$ is the \textbf{Choi matrix} of $\phi$. The mapping $\phi \mapsto C_\phi$ is an isomorphism from $L(M_n(\complex); B)$ to $M_n(B)$, and the following are equivalent: + \begin{enumerate} + \item $\phi$ is completely positive. + \item $\phi$ is $n$-positive. + \item $C_\phi$ is positive in $M_n(B)$. + \end{enumerate} +\end{theorem} +\begin{proof} + (2) $\Rightarrow$ (3): For each $1 \le i, j \le n$, $E_{ij} \otimes E_{ij}$ is a positive multiple of a rank one projection, and hence positive. Since $C_\phi = \sum_{i, j = 1}^n \phi(E_{ij} \otimes E_{ij})$ is a sum of positive elements, $C_\phi$ is positive. + + (3) $\Rightarrow$ (1): By \autoref{lemma:cstar-positive-matrix-decompose-sum}, there exist $\bracsn{b_{i, j}}_{i, j = 1}^n \subset B$ such that + \[ + \sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) = C_\phi = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes b_{k, i}^* b_{k, j} + \] + + So for any $1 \le i, j \le n$, + \[ + \phi(E_{ij}) = \sum_{k = 1}^n b_{k, i}^* b_{k, j} + \] + + For each $1 \le k \le n$, let $S_k = \sum_{l = 1}^n E_{kl} \otimes a_{k, l}^*$. Let $1 \le i, j \le n$, then + \begin{align*} + \sum_{k = 1}^n S_k (E_{ij} \otimes 1_B) S_k^* &= \sum_{k = 1}^n \sum_{s, t = 1}^n (E_{ks} \otimes a_{k, s}^*) (E_{ij} \otimes 1_B) (E_{kt} \otimes a_{k, t}^*)^* \\ + &= \sum_{k = 1}^n \sum_{s, t = 1}^n (E_{ks} \otimes a_{k, s}^*) (E_{ij} \otimes 1_B) (E_{tk} \otimes a_{k, t}) \\ + &= \sum_{k = 1}^n E_{kk} \otimes a_{k, i}^* a_{k, j}^* = \sum_{k = 1}^n E_{kk} \otimes \phi(E_{ij}) + \end{align*} + + By linearity, for each $T \in M_n(\complex)$, + \[ + \sum_{k = 1}^n E_{kk} \otimes \phi(T) = \sum_{k = 1}^n S_k(T \otimes 1_B)S_k^* + \] + + What the above shows, is that, under the identification that + \[ + M_n(\complex) \iso M_n(\complex) \otimes 1_B \subset M_n(B) + \] + + + and + \[ + B \iso \bracs{\sum_{k = 1}^n E_{kk} \otimes b \bigg |b \in B} = \bracs{\text{Diag}(b, \cdots, b)|b \in B} \subset M_n(B) + \] + + The map $\phi$ takes the form + \[ + \phi(T) = \sum_{k = 1}^n S_kTS_k^* \quad \forall T \in M_n(\complex) + \] + + Therefore $\phi$ is completely positive by \autoref{proposition:multiply-completely-bounded}. + + +\end{proof} + +\begin{remark} +\label{remark:choi-positive} + The Choi matrix yields a correspondence between linear maps and operator-valued matrices, and provides a way to check complete positivity. However, it is significantly harder to verify positivity through the matrix. +\end{remark} + +\begin{example}[Reduction Map] +\label{example:matric-reduction-map} + Let $\phi: M_n(\complex) \to M_n(\complex)$ be defined by + \[ + \phi(x) = \text{Tr}(x)1_{M_n(\complex)} - x + \] + + then $\phi$ is positive, but not completely positive. +\end{example} +\begin{proof} + (Positive): Let $x \in M_n(\complex)$ with $x \ge 0$, then $\norm{x}_{M_n(\complex)}$ is the largest eigenvalue of $x$, which is bounded by the trace of $x$. As such, $\text{Tr}(x)1_{M_n(\complex)} \ge x$, and $\phi(x)$ is positive. + + (Not Completely Positive): For each $1 \le i, j \le n$, $\text{Tr}(E_{ij}) = 1$ if and only if $i = j$. So the Choi matrix is given by + \begin{align*} + C_{ij} &= \phi_n\paren{\sum_{i, j = 1}^n E_{ij} \otimes E_{ij}} = \sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) \\ + &= \sum_{i, j = 1}^n E_{ij} \otimes (\one_{\bracsn{i = j}} \otimes 1_{M_n(\complex)} - E_{ij}) \\ + &= \sum_{i = 1}^n E_{ii} \otimes 1_{M_n(\complex)} - \sum_{i, j = 1}^n E_{ij} \otimes E_{ij} \\ + &= 1_{M_n(\complex)} \otimes 1_{M_n(\complex)}- \sum_{i, j = 1}^n E_{ij} \otimes E_{ij} + \end{align*} + + which is not positive. By \hyperref[Choi's Theorem]{theorem:choi-matrix-positive}, $\phi$ is not completely positive. +\end{proof} + diff --git a/src/op/system/completely.tex b/src/op/system/completely.tex index d3081b3..d35c3eb 100644 --- a/src/op/system/completely.tex +++ b/src/op/system/completely.tex @@ -298,58 +298,6 @@ Since the middle factor is positive, the left hand side is also positive. Therefore $x - yz^{-1}y^* \ge 0$ as well. \end{proof} -\begin{proposition} -\label{proposition:completely-positive-norm} - Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then - \[ - \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B - \] -\end{proposition} -\begin{proof} - After taking a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $B$ is unital. - - It is sufficient to show that $\norm{\phi}_{\text{cb}} \le \norm{\phi(1_A)}_B$. To this end, let $T \in M_n(S)$ with $\norm{T}_{M_n(S)} \le 1$, then - \[ - \begin{bmatrix} - 1_{M_n(A)} &T \\ - T^* &1_{M_n(A)} - \end{bmatrix} \ge 0 - \] - - in $M_{2}(M_n(S)) = M_{2n}(S)$ by \autoref{lemma:norm-matrix-test}. Since $\phi$ is completely positive, - \[ - \phi_{2n}\begin{bmatrix} - 1_{M_n(A)} &T \\ - T^* &1_{M_n(A)} - \end{bmatrix} = \begin{bmatrix} - \phi_n(1_{M_n(A)}) &\phi_n(T) \\ - \phi_n(T)^* & \phi_n(1_{M_n(A)}) - \end{bmatrix} \ge 0 - \] - - First suppose that $\phi_n(1_{M_n(A)})$ is invertible, then - - \begin{align*} - \phi_n(1_{M_n(A)}) &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot 1_{M_n(B)} \\ - 1_{M_n(B)} &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot \phi_n(1_{M_n(A)})^{-1} \\ - \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1} - \end{align*} - - By the \hyperref[Schur complement condition]{lemma:schur-complement-positive}, - \begin{align*} - \phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\ - &\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\ - \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^2 &\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)} = \normn{\phi_n(T)}_{M_n(B)}^2 - \end{align*} - - and $\normn{\phi_n(T)}_{M_n(B)} \le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}} \le \normn{\phi(1_A)}_B$. - - Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_B\psi$ is also a completely positive map. In which case, - \[ - \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B - \] -\end{proof} - \begin{example} \label{example:transpose-not-completely-positive} Let $T: M_2(\complex) \to M_2(\complex)$ be the transpose map, then $T$ is positive, but not $2$-positive. @@ -378,135 +326,61 @@ is invertible, with negative determinant, so $T_2(C)$ is not positive. \end{proof} -\begin{example} - \label{example:homomorphism-completely-positive} - Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a *-homomorphism, then $\phi$ is completely positive. -\end{example} -\begin{proof} - For each $n \in \natp$, the induced map $\phi_n: M_n(A) \to M_n(B)$ is a *-homomorphism. -\end{proof} - -\begin{example} - \label{example:multiply-completely-bounded} - - Let $A$ be a $C^*$-algebra, $y, z \in A$, $\phi: A \to A$ be defined by $x \mapsto yxz$, then - \begin{enumerate} - \item $\phi$ is completely bounded. - \item If $y = z^*$, then $\phi$ is completely positive. - \end{enumerate} -\end{example} -\begin{proof} - (1): For each $n \in \natp$ and $T \in M_n(A)$, - \[ - \phi_n(T) = (1_{M_n(\complex)} \otimes y)T(1_{M_n(\complex)} \otimes z) - \] -\end{proof} - -\begin{proposition} -\label{proposition:commutative-mn-facts} - Let $X$ be a compact Hausdorff space, then: - \begin{enumerate} - \item $C(X; M_n(\complex))$ equipped with pointwise product, pointwise involution, and the norm - \[ - \norm{f}_{C(X; M_n(\complex))} = \sup_{x \in X}\norm{f(x)}_{M_n(\complex)} - \] - - is a $C^*$-algebra. - \item The mapping - \[ - M_n(C(X; \complex)) \to C(X; M_n(\complex)) \quad M \otimes f \mapsto M \cdot f - \] - - is a *-isomorphism. - \item For any $f \in C(X; M_n(\complex))$, $f \ge 0$ if and only if $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$. - \end{enumerate} -\end{proposition} -\begin{proof} - (1): For any $f \in C(X; M_n(\complex))$, - \[ - \sup_{x \in X}\norm{f(x)}_{M_n(\complex)}^2 = \sup_{x \in X}\norm{f(x)^*f(x)}_{M_n(\complex)} = \norm{f^*f}_{C(X; M_n(\complex))} - \] - - (3): If $f \ge 0$, then there exists $g \in C(X; M_n(\complex))$ such that $f(x) = g^*(x)g(x)$ for all $x \in X$, so $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$. - - On the other hand, if $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$, then $g(x) = f(x)^{1/2}$ is defined for all $x \in X$. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $g$ is continuous from $X$ to $M_n(\complex)$. Therefore there exists $g \in C(X; M_n(\complex))$ such that $f = g^*g$. -\end{proof} +\begin{definition}[Module Map] +\label{definition:module-map} + Let $A, B, C$ be $C^*$-algebras with $C \subset A, B$ and $1_A = 1_B = 1_C$, then $A$ and $B$ are $C$-bimodules with respect to multiplication. + + Let $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{$C$-bimodule map} if $\phi(c_1ac_2) = c_1\phi(a)c_2$ for all $a \in A$ and $c_1, c_2 \in C$. +\end{definition} \begin{theorem} -\label{theorem:bounded-functional-completely-bounded} - Let $A$ be a unital $C^*$-algebra, $S \subset A$ be a subspace, $X$ be a compact Hausdorff space, and $\phi \in L(S; C(X; \complex))$, then: +\label{theorem:completely-positive-bimodule} + Let $A, B$ be unital $C^*$-algebras, $\phi: A \to B$ be a unital completely positive map, then: \begin{enumerate} - \item $\norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; C(X; \complex))}$. - \item If $S$ is an operator system and $\phi$ is positive, then $\phi$ is completely positive. + \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$. The space $R = \bracsn{a \in A|\phi(a)^*\phi(a) = \phi(a^*a)}$ is a subalgebra of $A$, and $\phi|_R$ is a homomorphism. + \item For any $a \in A$, $\phi(a)\phi(a)^* = \phi(aa^*)$ if and only if $\phi(ab) = \phi(a)\phi(b)$ for all $b \in A$. The space $L = \bracsn{a \in A| \phi(a)\phi(a)^* = \phi(aa^*)}$ is a subalgebra of $A$, and $\phi|_L$ is a homomorphism. + \item $L \cap R$ is a $C^*$-subalgebra of $A$, and $\phi|_{L \cap R}$ is a *-homomorphism. \end{enumerate} \end{theorem} -\begin{proof}[Proof, {{\cite[Proposition 3.8]{Paulsen}}}. ] - By \autoref{proposition:commutative-mn-facts}, the norm and order relation on $C(X; M_n(\complex)) \iso M_n(C(X; \complex))$ are pointwise. As such, assume without loss of generality that $X$ is a single point and $C(X; \complex) = \complex$. - - Let $a = (a_{ij}) \in M_n(S)$. For each $\xi, \eta \in \complex^n$, - \begin{align*} - \dpn{\phi_n(a)\xi, \eta}{\complex^n} &= \sum_{i, j = 1}^n \xi_j \ol{\eta_i} \dpn{a_{ij}, \phi}{S} = \angles{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}, \phi}_S - \end{align*} - - where - \begin{align*} - \sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i} &= (\one \otimes \ol{\eta}) \cdot a \cdot (\xi \otimes \one) - \end{align*} +\begin{proof} + (1, $\Leftarrow$): Suppose that $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$, then $\phi(a^*) = \phi(a)^*\phi(a)$ after plugging in $b = a^*$. - (1): + (1, $\Rightarrow$): Suppose that $\phi(a^*a) = \phi(a)^*\phi(a)$. By the \hyperref[Kadison-Schwarz inequality]{proposition:cauchy-schwarz-2-positive}, \begin{align*} - |\dpn{\phi_n(a)\xi, \eta}{\complex^n}| &\le \norm{\phi}_{S^*} \cdot \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S} \\ - \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S} &\le \norm{\xi}_{\complex^n}\norm{\eta}_{\complex^n}\norm{a}_{M_n(S)} \\ - \norm{\phi_n(a)}_{M_n(\complex)} &\le \norm{\phi}_{S^*}\norm{a}_{M_n(S)} \\ - \norm{\phi}_{\text{cb}} &\le \norm{\phi}_{S^*} + \phi_2\begin{bmatrix} + a & b^* \\ + 0 & 0 + \end{bmatrix}^*\phi_2\begin{bmatrix} + a & b^* \\ + 0 & 0 + \end{bmatrix} \le \phi_2\begin{bmatrix} + a^*a & a^*b^* \\ + ba & bb^* + \end{bmatrix} \end{align*} - (2): If $\xi = \eta$, then - \[ - \sum_{i, j = 1}^n a_{ij}\eta_j\ol{\eta_i} = (\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one) \ge 0 - \] - so \[ - \dpn{\phi_n(a)\xi, \eta}{\complex^n} = \dpn{(\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one), \phi}{S} \ge 0 + \begin{bmatrix} + \phi(a^*)\phi(a) & \phi(a^*)\phi(b) \\ + \phi(b)\phi(a) & \phi(b)\phi(b^*) + \end{bmatrix} \le + \begin{bmatrix} + \phi(a^*a) & \phi(a^*b) \\ + \phi(ba) & \phi(bb^*) + \end{bmatrix} \] -\end{proof} - -\begin{theorem}[Stinespring] -\label{theorem:stinespring-commutative-completely-positive} - Let $X$ be a compact Hausdorff space, $B$ be a $C^*$-algebra, $\phi: C(X; \complex) \to B$ be a positive map, then $\phi$ is completely positive. -\end{theorem} -\begin{proof} - Let $P \in C(X; M_n(\complex))$ with $P \ge 0$ and $\eps > 0$. By \autoref{proposition:uniform-continuous-compact} and \autoref{theorem:lch-paracompact}, there exists a partition of unity $\seqf{\psi_j} \subset C(X; [0, 1])$ and positive matrices $\seqf{p_j} \subset M_n(\complex)$ such that + and \[ - \norm{P(x) - \sum_{j = 1}^n p_j\psi_j(x)}_{M_n(\complex)} < \eps + \begin{bmatrix} + 0 & \phi(a^*b^*) - \phi(a^*)\phi(b) \\ + \phi(b)\phi(a) - \phi(ba) & \phi(b)\phi(b^*) - \phi(bb^*) + \end{bmatrix} \ge 0 \] - for all $x \in X$. For each $1 \le j \le n$, $\phi_n(p_j\psi_j) = p_j\phi(\psi_j) \ge 0$. As $\eps > 0$ is arbitrary, $\phi_n(P) \ge 0$. -\end{proof} - -\begin{theorem}[Choi] -\label{theorem:choi-matrix-positive} - Let $n \in \natp$, $B$ be a $C^*$-algebra, and $\phi: M_n(\complex) \to B$ be a linear map. For each $1 \le i, j \le n$, let $E_{ij}$ be the standard matrix units for $M_n(\complex)$, then $C_\phi = (\phi(E_{ij})) \in M_n(B)$ is the \textbf{Chi matrix} of $\phi$. The mapping $\phi \mapsto C_\phi$ is an isomorphism from $L(M_n(\complex); B)$ to $M_n(B)$, and the following are equivalent: - \begin{enumerate} - \item $\phi$ is completely positive. - \item $\phi$ is $n$-positive. - \item $C_\phi$ is positive in $M_n(B)$. - \end{enumerate} -\end{theorem} -\begin{proof} - (2) $\Rightarrow$ (3): The matrix $E = (E_{ij})$ is positive in $M_n(\complex)$. - - (3) $\Rightarrow$ (1): - - - - $\phi, \psi \in S(G)$, $K \subset G$ be compact, and $V \in \cn_G(1_G)$ be relatively compact, then - \begin{align*} - - \end{align*} + Since the top left entry is zero, $\phi(b)\phi(a) - \phi(ba) = 0$. \end{proof} diff --git a/src/op/system/index.tex b/src/op/system/index.tex index ab42cd4..e29ed63 100644 --- a/src/op/system/index.tex +++ b/src/op/system/index.tex @@ -4,3 +4,4 @@ \input{./dilation.tex} \input{./os.tex} \input{./completely.tex} +\input{./complete-upgrade.tex} diff --git a/src/op/vn/commutative.tex b/src/op/vn/commutative.tex index 743e9ed..4823ad8 100644 --- a/src/op/vn/commutative.tex +++ b/src/op/vn/commutative.tex @@ -8,12 +8,12 @@ \begin{proposition} \label{proposition:maximal-commutative-vn} - Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a maximal commutative von Neumann algebra if and only if $A = A'$. + Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a maximal abelian von Neumann algebra if and only if $A = A'$. \end{proposition} \begin{proof} - ($\Leftarrow$): Let $B \supset A$ be a commutative von Neumann algebra, then $B \subset A' = A$. + ($\Leftarrow$): Let $B \supset A$ be a abelian von Neumann algebra, then $B \subset A' = A$. - ($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is commutative. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa} + i(A')_{sa} \subset A$. + ($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is abelian. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa} + i(A')_{sa} \subset A$. \end{proof} diff --git a/src/topology/main/compactify.tex b/src/topology/main/compactify.tex index bf9eccf..4f8d4c5 100644 --- a/src/topology/main/compactify.tex +++ b/src/topology/main/compactify.tex @@ -125,7 +125,17 @@ For each $x \in X$, let $U \in \cn_X(x)$ be a compact neighbourhood. Since $Y$ is a compact Hausdorff space, $\varphi(U)$ is closed by \autoref{proposition:compact-closed}. As $\varphi \in C(X; Y)$ is an embedding, there exists $V \in \cn_Y(\varphi(x))$ such that $\varphi(U) = \varphi(X) \cap V$. Given that $\varphi(X)$ is dense in $Y$, $\varphi(U) = \ol{\varphi(X) \cap V} \supset V$. Therefore $\varphi(U) \in \cn_{Y}(\varphi(x))$, and $\varphi(X)$ is open in $Y$. \end{proof} +\begin{proposition} +\label{proposition:stone-cech-not-metrisable} + Let $X$ be an infinite discrete topological space, then $\beta X$ is not second countable and hence not metrisable. +\end{proposition} +\begin{proof} + Let $\fB \subset 2^X$ be a base for the topology on $\beta X$, then for each $A \subset \beta X$ clopen, there exists $\mathcal{F} \subset \fB$ finite such that $A = \bigcup_{U \in \mathcal{F}}U$. Should $\fB$ be countable, then there are at most countably many clopen subsets of $\beta X$. + However, for each $A \subset X$, let $U_A = \bracsn{\fU \in \beta X|A \in \fU}$, then $\bracsn{U_A|A \subset X}$ represents uncountably many clopen sets. Therefore $\fB$ cannot be countable, $\beta X$ is not second countable. + + If $\beta X$ was metrisable, then it is second countable by compactness and \autoref{proposition:separable-metric-space}. Therefore $\beta X$ is not metrisable. +\end{proof} \begin{definition}[One-Point Compactification] \label{definition:alexandroff-compactification}