diff --git a/.vscode/tasks.json b/.vscode/tasks.json index 3e378ec..74c83ef 100644 --- a/.vscode/tasks.json +++ b/.vscode/tasks.json @@ -1,57 +1,57 @@ { - "version": "2.0.0", - "tasks": [ - { - "label": "Build", - "type": "shell", - "command": "npx spec compile", - "windows": { - "options": { - "shell": { - "executable": "cmd.exe", - "args": ["/d", "/c"] - } - } - } - }, - { - "label": "Serve", - "type": "shell", - "command": "npx spec serve", - "windows": { - "options": { - "shell": { - "executable": "cmd.exe", - "args": ["/d", "/c"] - } - } - } - }, - { - "label": "Watch", - "type": "shell", - "command": "npx spec watch", - "windows": { - "options": { - "shell": { - "executable": "cmd.exe", - "args": ["/d", "/c"] - } - } - } - }, - { - "label": "Conservative", - "type": "shell", - "command": "npx spec watch --conservative", - "windows": { - "options": { - "shell": { - "executable": "cmd.exe", - "args": ["/d", "/c"] - } + "version": "2.0.0", + "tasks": [ + { + "label": "Build", + "type": "shell", + "command": "npx spec compile", + "windows": { + "options": { + "shell": { + "executable": "cmd.exe", + "args": ["/d", "/c"] } } } - ] - } \ No newline at end of file + }, + { + "label": "Serve", + "type": "shell", + "command": "npx spec serve", + "windows": { + "options": { + "shell": { + "executable": "cmd.exe", + "args": ["/d", "/c"] + } + } + } + }, + { + "label": "Watch", + "type": "shell", + "command": "npx spec watch", + "windows": { + "options": { + "shell": { + "executable": "cmd.exe", + "args": ["/d", "/c"] + } + } + } + }, + { + "label": "Conservative", + "type": "shell", + "command": "npx spec watch --conservative", + "windows": { + "options": { + "shell": { + "executable": "cmd.exe", + "args": ["/d", "/c"] + } + } + } + } + ] +} \ No newline at end of file diff --git a/src/dg/complex/derivative.tex b/src/dg/complex/derivative.tex index 777915d..c55f730 100644 --- a/src/dg/complex/derivative.tex +++ b/src/dg/complex/derivative.tex @@ -212,7 +212,7 @@ \] \end{corollary} \begin{proof} - By \autoref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, + By \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, \begin{align*} D^kf(z_0) &= \frac{k!}{2\pi i}\int_{\omega_{z_0, r}} \frac{f(z)}{(z - z_0)^{k+1}}dz \\ [D^kf(z_0)]_E &\le \frac{k!}{2\pi i}\int_0^{2\pi}\frac{[f(z)]_E}{|z - z_0|^{k+1}}dz \\ diff --git a/src/dg/complex/space.tex b/src/dg/complex/space.tex index f8c2283..ed88fee 100644 --- a/src/dg/complex/space.tex +++ b/src/dg/complex/space.tex @@ -11,7 +11,7 @@ Let $E$ be a complete separated locally convex space over $\complex$ and $U \subset \complex$ be open, then $H(U; E)$ is complete. \end{proposition} \begin{proof} - By \hyperref[corollary:cauchy-estimate]{Cauchy's estimate}, uniform convergence on compact sets is equivalent to uniform convergence of derivatives of all orders on compact sets. Since $U$ is locally compact, uniform limits of holomorphic functions are holomorphic by \autoref{theorem:differentiable-uniform-limit}. + By \hyperref[Cauchy's estimate]{corollary:cauchy-estimate}, uniform convergence on compact sets is equivalent to uniform convergence of derivatives of all orders on compact sets. Since $U$ is locally compact, uniform limits of holomorphic functions are holomorphic by \autoref{theorem:differentiable-uniform-limit}. \end{proof} \begin{theorem}[Montel] diff --git a/src/fa/lc/compact.tex b/src/fa/lc/compact.tex new file mode 100644 index 0000000..37fb45e --- /dev/null +++ b/src/fa/lc/compact.tex @@ -0,0 +1,85 @@ +\section{Compact Convex Sets} +\label{section:compact-convex} + +\begin{definition}[Extreme Point] +\label{definition:extreme-point} + Let $E$ be a vector space over $\real$, $K \subset E$, and $x \in K$, then $x$ is \textbf{extremal} if there exists no $y, z \in K$ such that $x \in (y, z) \subset K$. +\end{definition} + +\begin{definition}[Extreme Subset] +\label{definition:extreme-subset} + Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$, then $A$ is \textbf{extreme set} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well. +\end{definition} + + + +\begin{lemma} +\label{lemma:extremal-face} + Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^*$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha} \cap K$ is a non-empty extreme subset of $K$. +\end{lemma} +\begin{proof} + Since $K$ is compact, $\alpha < \infty$ and $A$ is non-empty by \autoref{proposition:compact-extensions}. Let $x \in A$ and $y, z \in K$ such that $x \in (y, z)$. By definition of $\alpha$, $\dpn{y, \phi}{E} = \dpn{z, \phi}{E} = \alpha$. Thus $y, z \in A$ as well. +\end{proof} + +\begin{theorem}[Krein-Milman] +\label{theorem:krein-milman} + Let $E$ be a separated locally convex space over $\real$ and $K \subset E$ be a compact convex set, then $K$ is the closed convex hull of its extreme points. +\end{theorem} +\begin{proof}[Proof, {{\cite[Theorem 1.12.5]{Bogachev}}}. ] + Assume without loss of generality that $K \ne \emptyset$. + + Let $K_0 \subset K$ be a closed, extreme subset of $K$, and $\mathcal{E}(K_0) \subset 2^K$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_0$. Since $K_0 \in \mathcal{E}(K_0)$, $\mathcal{E}(K_0) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C} \subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme. + + By Zorn's lemma, there exists a minimal element $A$ of $\mathcal{E}(K_0)$. Let $x, y \in A$, $\phi \in E^*$, and $\alpha = \sup_{z \in A}\dpn{z, \phi}{E}$, then $\bracs{\phi = \alpha} \cap K$ is a non-empty, closed, and extreme subset of $K$ by \autoref{lemma:extremal-face}, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $A$ consists of exactly one point. In which case, for any $y, z \in E$ with $A \subset (y, z) \subset K$, $y = z \in A$. Therefore there exists an extreme point of $K$ in $K_0$. + + Since $K$ itself is an extreme subset, a minimal element of $\mathcal{E}(K)$ represents an extreme point, so $K$ admits at least one extreme point. + + Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $x \in K$ and $\phi \in E^*$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by \autoref{lemma:extremal-face}, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. +\end{proof} + +\begin{theorem}[Markov-Kakutani Fixed Point Theorem] +\label{theorem:markov-kakutani} + Let $E$ be a separated topological vector space over $\real$, $K \subset E$ be a compact convex set, and $\cf \subset C(K; K)$ such that: + \begin{enumerate}[label=(\alph*)] + \item For any $f, g \in \cf$, $f \circ g = g \circ f$. + \item For each $f \in \cf$, $x, y \in K$, and $t \in [0, 1]$, + \[ + f(tx + (1 - t)y) = tf(x) + (1 - t)f(y) + \] + \end{enumerate} + + then there exists $x_0 \in K$ such that $f(x_0) = x_0$ for all $f \in \cf$. +\end{theorem} +\begin{proof}[Proof, {{\cite[Theorem 1.12.10]{Bogachev}}}. ] + For each $f \in \cf$ and $n \in \natp$, define $f^{(n)} = \frac{1}{n}\sum_{k = 0}^{n - 1}f^k$, then $f^{(n)} \in C(K; K)$ as well. Via a closure operation, assume without loss of generality that: + \begin{enumerate} + \item For any $f \in \cf$ and $n \in \natp$, $f^{(n)} \in \cf$. + \item For any $f, g \in \cf$, $f \circ g \in \cf$. + \end{enumerate} + + Let $\bracsn{f_j}_1^N \subset \cf$ and $\bracsn{n_j}_1^N \subset \natp$, then + \[ + \bigcap_{j = 1}^n f_j^{(n_j)}(K) \supset \braks{\prod_{j = 1}^n f_j^{(n_j)}}(K) \ne \emptyset + \] + + thus any finite intersections of elements in + \[ + \mathcal{K} = \bracsn{f^{(n)}(K)| f \in \cf, n \in \natp} + \] + + is non-empty. Since $E$ is separated, by \autoref{proposition:compact-extensions} and \autoref{proposition:compact-closed}, $\mathcal{K}$ is a family of closed sets satisfying the finite intersection property. Hence $\bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K) \ne \emptyset$. + + Now, let $x \in \bigcap_{n \in \natp}\bigcap_{f \in \cf}f^{(n)}(K)$ and $U \in \cn_E(0)$, then there exists $N \in \natp$ such that $NU \supset K$. In which case, there exists $y \in K$ such that + \[ + x = \frac{1}{N}\sum_{j = 0}^{N-1}f^j(y) \quad f(x) = \frac{1}{N}\sum_{j = 1}^{N}f^j(y) + \] + + In which case, + \[ + f(x) - x = \frac{1}{N}(f^N(y) - y) \in \frac{1}{N}(K - K) \subset U - U + \] + + As this holds for all $U \in \cn_E(0)$ and $E$ is separated, $f(x) = x$. + +\end{proof} + diff --git a/src/fa/lc/index.tex b/src/fa/lc/index.tex index 1a65c95..b605d8a 100644 --- a/src/fa/lc/index.tex +++ b/src/fa/lc/index.tex @@ -4,6 +4,7 @@ \input{./convex.tex} \input{./continuous.tex} +\input{./compact.tex} \input{./barrel.tex} \input{./bornologic.tex} \input{./quotient.tex} diff --git a/src/topology/main/cube.tex b/src/topology/main/cube.tex index 895adfa..ff15a3d 100644 --- a/src/topology/main/cube.tex +++ b/src/topology/main/cube.tex @@ -59,5 +59,23 @@ (2) $\Rightarrow$ (1): By \autoref{definition:embedding-in-cube}, $X$ embeds into $[0, 1]^{C(X; [0, 1])}$, which is a uniform space. The subspace uniformity on $X$ then induces the topology on $X$. \end{proof} +\begin{theorem}[Uryson Metrisation Theorem] +\label{theorem:urysohn-metrisation} + Let $X$ be a second countable regular space, then $X$ is metrisable. +\end{theorem} +\begin{proof} + By \autoref{proposition:second-countable-regular}, $X$ is normal. Let $\cb \subset 2^X$ be a countable base for $X$, and let + \[ + \mathcal{S} = \bracsn{(E, F) \in \mathcal{B}^2 | \ol{E} \subset F} + \] + + By \hyperref[Urysohn's Lemma]{lemma:urysohn}, for each $(E, F) \in \mathcal{S}$, there exists $f_{EF} \in C(X; [0, 1])$ such that $f|_E = 1$ and $f|_{F^c} = 0$. For any $x \in X$ and $U \in \cn^o_X(x)$, there exists $E, F \in \mathcal{B}$ such that $x \in E \subset \ol{E} \subset F \subset U$. Thus $f_{EF}(x) = 1$ and $f_{EF}|_{U^c} = 0$. Therefore + \[ + \cf = \bracsn{f_{EF}|(E, F) \in \mathcal{S}} \subset C(X; [0, 1]) + \] + + is a countable family of continuous functions that separate points and closed sets. By \autoref{definition:embedding-in-cube}, $X$ embeds into $[0, 1]^{\cf}$, which is metrisable by \autoref{proposition:countable-metric}. +\end{proof} + diff --git a/src/topology/main/definition.tex b/src/topology/main/definition.tex index 13935dd..66531d5 100644 --- a/src/topology/main/definition.tex +++ b/src/topology/main/definition.tex @@ -77,6 +77,18 @@ By definition, $\topo(\cb)$ satisfies (O3). \end{proof} +\begin{definition}[First Countable] +\label{definition:first-countable} + Let $X$ be a topological space, then $X$ is \textbf{first countable} if for every $x \in X$, there exists a countable fundamental system of neighbourhoods at $x$. +\end{definition} + +\begin{definition}[Second Countable] +\label{definition:second-countable} + Let $X$ be a topological space, then $X$ is \textbf{second countable} if it admits a countable base. +\end{definition} + + + \begin{definition}[Generated Topology] \label{definition:generated-topology} Let $X$ be a set and $\ce \subset 2^X$ such that $\bigcup_{U \in \ce}U = X$, then the smallest topology on $X$ containing $\ce$ is given by diff --git a/src/topology/main/regular.tex b/src/topology/main/regular.tex index 5c09ba6..5104c74 100644 --- a/src/topology/main/regular.tex +++ b/src/topology/main/regular.tex @@ -37,3 +37,25 @@ as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous. \end{proof} + +\begin{proposition} +\label{proposition:second-countable-regular} + Let $X$ be a second countable regular space, then $X$ is normal. +\end{proposition} +\begin{proof} + Let $\cb \subset 2^X$ be a base for $X$ and $A, B \subset X$ be disjoint closed sets. For each $x \in A$ and $y \in B$, there exists $U_x, V_y \in \cb$ such that $x \in \ol{U_x} \subset B^c$ and $y \in \ol{V_y} \subset B^c$. Since $\cb$ is countable, let $\seq{U_n}$ and $\seq{V_n}$ be enumerations of $\bracs{U_x|x \in A}$ and $\bracs{V_y|y \in B}$, respectively. + + For each $n \in \natp$, let + \[ + U_n' = U_n \setminus \bigcup_{j = 1}^n \ol{V_j} \quad V_n' = V_n \setminus \bigcup_{j = 1}^n \ol{V_j} + \] + + then $U_n'$ and $V_n'$ are both open. Let + \[ + U = \bigcup_{n \in \natp}U_n' \quad V = \bigcup_{n \in \natp}V_n' + \] + + then $U \in \cn_X(A)$ and $V \in \cn_X(B)$. For each $m, n \in \natp$ with $m \le n$, $V_n \cap \bigcup_{j = 1}^n U_j = \emptyset$, so $U_m \cap V_n = \emptyset$. Likewise, if $m \ge n$, then $U_m \cap V_n = \emptyset$ as well. Therefore $U \cap V = \emptyset$. +\end{proof} + + diff --git a/src/topology/metric/metric.tex b/src/topology/metric/metric.tex index a1d7709..4d598a0 100644 --- a/src/topology/metric/metric.tex +++ b/src/topology/metric/metric.tex @@ -3,7 +3,7 @@ \begin{definition}[Metric Space] \label{definition:metric} - Let $X$ be a set and $d: X \times X$, then $d$ is a \textbf{metric} if: + Let $X$ be a set and $d: X \times X \to [0, \infty]$, then $d$ is a \textbf{metric} if: \begin{enumerate} \item[(PM1)] For any $x \in X$, $d(x, x) = 0$. \item[(M)] For any $x, y \in X$ with $x \ne y$, $d(x, y) > 0$. @@ -41,10 +41,23 @@ (3) $\Rightarrow$ (1): Let $\seq{x_n} \subset X$ be a countable dense subset. Let $x \in X$ and $k \in \natp$, then there exists $x_n \in \natp$ such that $d(x, x_n) < 1/(2k)$. In which case, $x \in B(x_n, 1/(2k)) \subset B(x_n, 1/k)$. Therefore $\bracs{B(x_n, 1/k)|n, k \in \natp}$ forms a countable basis for $X$. \end{proof} +\begin{proposition} +\label{proposition:countable-metric} + Let $\seq{(X_n, d_n)}$ be metrisable spaces, then $\prod_{n \in \natp}X_n$ is also metrisable. +\end{proposition} +\begin{proof} + For each $n \in \natp$, let + \[ + d_n': \braks{\prod_{n \in \natp}X_n}^2 \to [0, \infty] \quad (x, y) \mapsto d_n(\pi_n(x), \pi_n(y)) + \] + + then $d_n'$ is a pseudometric on $X$, and $\bracsn{d_n'}_1^\infty$ induces the product uniformity on $\prod_{n \in \natp}X_n$. By \autoref{theorem:uniform-metrisable}, $\prod_{n \in \natp}X_n$ is also metrisable. +\end{proof} + \begin{theorem}[Banach's Fixed Point Theorem] \label{theorem:banach-fixed-point} - Let $(X, d)$ be a metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that + Let $(X, d)$ be a complete metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that \[ d(f(x), f(y)) \le Cd(x, y) \quad \forall x, y \in X \]