From 307f23ad57a4175bc4ea6eb4ed3d94fb7a298708 Mon Sep 17 00:00:00 2001 From: Bokuan Li <47512608+Jerry-licious@users.noreply.github.com> Date: Sat, 17 Jan 2026 23:11:19 -0500 Subject: [PATCH] Me when I forget to commit. --- preamble.sty | 2 + refs.bib | 10 + src/dg/derivative/derivative.tex | 153 +++++++++-- src/dg/derivative/sets.tex | 17 ++ src/fa/norm/normed.tex | 10 +- src/fa/tvs/bounded.tex | 4 +- src/fa/tvs/spaces-of-linear.tex | 61 ++++- src/measure/index.tex | 1 + src/measure/measurable-maps/index.tex | 5 + .../measurable-maps/measurable-maps.tex | 26 ++ src/measure/measurable-maps/product.tex | 52 ++++ src/measure/measure/index.tex | 1 + src/measure/measure/kolmogorov.tex | 76 ++++++ src/measure/measure/regular.tex | 11 + src/topology/functions/index.tex | 1 + src/topology/main/interiorclosureboundary.tex | 9 + src/topology/main/metric.tex | 50 ---- src/topology/main/neighbourhoods.tex | 1 - src/topology/main/normal.tex | 2 +- src/topology/uniform/cauchy.tex | 98 +++++++ src/topology/uniform/complete.tex | 97 ------- src/topology/uniform/index.tex | 1 + src/topology/uniform/metric.tex | 245 ++++++++++++++++++ 23 files changed, 751 insertions(+), 182 deletions(-) create mode 100644 src/dg/derivative/sets.tex create mode 100644 src/measure/measurable-maps/index.tex create mode 100644 src/measure/measurable-maps/measurable-maps.tex create mode 100644 src/measure/measurable-maps/product.tex create mode 100644 src/measure/measure/kolmogorov.tex delete mode 100644 src/topology/main/metric.tex create mode 100644 src/topology/uniform/metric.tex diff --git a/preamble.sty b/preamble.sty index 4144208..9585830 100644 --- a/preamble.sty +++ b/preamble.sty @@ -165,6 +165,7 @@ \newcommand{\cn}{\mathcal{N}} \newcommand{\lms}{\mathcal{L}} \newcommand{\ch}{\mathcal{H}} +\newcommand{\cg}{\mathcal{H}} \renewcommand{\cl}{\mathcal{L}} \newcommand{\cc}{\mathcal{C}} \newcommand{\ct}{\mathcal{T}} @@ -172,6 +173,7 @@ \newcommand{\cy}{\mathcal{Y}} \newcommand{\cs}{\mathcal{S}} \newcommand{\cd}{\mathcal{D}} +\newcommand{\calr}{\mathcal{R}} \newcommand{\scp}{\mathscr{P}} % Jokes diff --git a/refs.bib b/refs.bib index 0ffc38b..20c3f3e 100644 --- a/refs.bib +++ b/refs.bib @@ -64,3 +64,13 @@ year={1999}, publisher={Wiley} } +@book{Baudoin, + title={Diffusion Processes and Stochastic Calculus}, + author={Baudoin, F.}, + isbn={9783037191330}, + lccn={2014395958}, + series={EMS textbooks in mathematics}, + url={https://books.google.ca/books?id=ov4kcKkzTP4C}, + year={2014}, + publisher={European Mathematical Society} +} diff --git a/src/dg/derivative/derivative.tex b/src/dg/derivative/derivative.tex index 3224565..63cadf6 100644 --- a/src/dg/derivative/derivative.tex +++ b/src/dg/derivative/derivative.tex @@ -1,6 +1,52 @@ -\section{Derivatives} +\section{Derivatives and Remainders} \label{section:derivative} +\begin{definition}[Derivatives and Remainders] +\label{definition:derivative-system} + Let $E, F$ be TVSs over $K \in \RC$ and $\ch(E; F), \calr(E; F) \subset F^E$ be vector subspaces, then $(\ch, \calr) = (\ch(E; F), \calr(E; F))$ is a pair of \textbf{derivatives} and \textbf{remainders} if + \begin{enumerate} + \item[(T)] For any $T \in \ch$, if there exists $V \in \cn_E(0)$ and $r \in \calr$ such that $T|_V = r|_V$, then $T = 0$. + \end{enumerate} +\end{definition} + +\begin{definition}[$(\ch, \calr)$-Differentiability] +\label{definition:space-differentiability} + Let $E, F$ be TVSs over $K \in \RC$, $(\ch, \calr)$ be a pair of derivatives and remainders, $U \subset E$ be open, $f: U \to F$ be a function, and $x_0 \in U$, then $f$ is \textbf{$(\ch, \calr)$-differentiable} at $x_0$ if there exists $V \in \cn_E(0)$, $T \in \ch$, and $r \in \calr$ such that + \[ + f(x_0 + h) = f(x_0) + Th + r(h) + \] + for all $h \in V$. In which case, $T = D_{(\ch, \calr)}f(x_0)$ is the unique element of $\ch$ satisfying the above, known as the \textbf{$(\ch, \calr)$-derivative} of $f$ at $x_0$. +\end{definition} +\begin{proof} + Let $S, T \in \ch$, $r, s \in \calr$, and $V \in \cn_E(0)$ such that + \[ + f(x_0 + h) - f(x_0) = Sh + r(h) = Th + s(h) + \] + for all $h \in V$, then $(S - T)(h) = (s - r)(h)$. By (T), $S - T = 0$. Hence $S = T$. +\end{proof} + +\begin{proposition} +\label{proposition:derivative-linearity} + Let $E, F$ be TVSs over $K \in \RC$, $(\ch, \calr)$ be a pair of derivatives and remainders, $U \subset E$ be open, $f, g: U \to F$ be functions, and $x_0 \in U$. If $f, g$ are $(\ch, \calr)$-differentiable at $x_0$, then for any $\lambda \in K$, $\lambda f + g$ is $(\ch, \calr)$-differentiable at $x_0$, and + \[ + D_{(\ch, \calr)}(\lambda f + g)(x_0) = \lambda D_{(\ch, \calr)}f(x_0) + D_{(\ch, \calr)}g(x_0) + \] +\end{proposition} +\begin{proof} + Let $V \in \cn_E(0)$ and $r, s \in \calr$ such that + \begin{align*} + f(x_0 + h) - f(x_0) &= D_{(\ch, \calr)}f(x_0)h + r(h) \\ + g(x_0 + h) - f(x_0) &= D_{(\ch, \calr)}g(x_0)h + s(h) + \end{align*} + then + \[ + (\lambda f + g)(x_0+h) - (\lambda f + g)(x_0) = \underbrace{[\lambda D_{(\ch, \calr)}f(x_0) + D_{(\ch, \calr)}g(x_0)]}_{\in \ch}h + \underbrace{(\lambda r + s)}_{\in \calr}(h) + \] +\end{proof} + + + + \begin{definition}[$o(t)$] \label{definition:little-o} Let $U \in \cn(0) \subset \real$ and $r: U \to \real$, then $r \in o(t)$ if @@ -11,13 +57,82 @@ \begin{definition}[Tangent to $0$] \label{definition:tangent-to-0} - Let $E, F$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, then $\varphi$ is \textbf{tangent to $0$} if for any $W \in \cn_F(0)$, there exists $V \in \cn_E(0)$ circled and $r \in o(t)$ such that + Let $E, F$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, then $\varphi$ is \textbf{tangent to $0$} if for any $W \in \cn_F(0)$, there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that \[ \varphi(tV) \subset r(t)W \] for sufficiently small $t \in \real$. \end{definition} +\begin{definition}[Linear Order at $0$] +\label{definition:linear-order-at-0} + Let $E, F$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, then $\varphi$ is of \textbf{linear order at $0$} if for any $W \in \cn_F(0)$, there exists $V \in \cn_E(0)$ such that + \[ + \varphi(tV) \subset tW + \] + for sufficiently small $t \in \real$. +\end{definition} + +\begin{lemma} +\label{lemma:tangent-linear-at-0} + Let $E, F, G$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, $V \in \cn_F(0)$, $\varphi, \psi: U \to F$, and $\rho: V \to G$. If $\varphi, \psi, \rho$ are of linear order at $0$, then + \begin{enumerate} + \item $\rho \circ \varphi$ is of linear order at $0$. + \item $\varphi + \psi$ is of linear order at $0$. + \item If one of $\varphi, \rho$ is tangent to $0$, then $\rho \circ \varphi$ is tangent to $0$. + \end{enumerate} + If $\varphi, \psi$ are tangent to $0$, then + \begin{enumerate} + \item[(4)] $\varphi$ is of linear order at $0$. + \item[(5)] $\varphi + \psi$ is tangent at $0$. + \end{enumerate} + Finally, suppose that $\varphi$ is linear. + \begin{enumerate} + \item[(6)] If $\varphi$ is continuous, then $\varphi$ is of linear order at $0$. + \item[(7)] If $\varphi$ is tangent to $0$ and $E$ is Hausdorff, then $\varphi = 0$. + \end{enumerate} +\end{lemma} +\begin{proof} + (1): Let $W \in \cn_G(0)$, then there exists $V_0 \in \cn_F(0)$ with $\rho(tV_0) \subset tW$ and $U_0 \in \cn_E(0)$ with $\varphi(tU_0) \subset tV_0$ for sufficiently small $t$. In which case, $\rho \circ \varphi(tU_0) \subset tW$. + + (2): Let $W \in \cn_F(0)$, then there exists $W_0 \in \cn_F(0)$ with $W_0 + W_0 \subset W$ and $V_0 \in \cn_E(0)$ such that $\varphi(tV_0), \psi(tV_0) \subset tW_0$ for sufficiently small $t$. In which case, + \[ + \varphi(tV_0) + \psi(tV_0) \subset tW_0 + tW_0 \subset tW + \] + + (3): Let $W \in \cn_G(0)$, then there exists $V_0 \in \cn_F(0)$, $U_0 \in \cn_E(0)$, and $r \in o(t)$ such that one of the following holds for sufficiently small $t$: + \begin{enumerate} + \item[(a)] $\varphi(tU_0) \subset tV_0$ and $\rho(tV_0) \subset o(t)W$. + \item[(b)] $\varphi(tU_0) \subset o(t)V_0$ and $\rho(tV_0) \subset tW$. + \end{enumerate} + In both cases, $\rho \circ \varphi(tU_0) \subset o(t)W$. + + (4): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. By assumption, there exists $V_0 \in \cn_E(0)$ and $r \in o(t)$ such that + \[ + \varphi(tV_0) \subset o(t)W = \frac{r(t)}{t} \cdot tW + \] + for sufficiently small $t$. Since $r \in o(t)$, $\abs{r(t)/t} \le 1$ for sufficiently small $t$. In which case, $\frac{r(t)}{t} \cdot tW \subset tW$. + + (5): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Let $V_0 \in \cn_E(0)$ and $r, r' \in o(t)$ such that $\varphi(tV_0) \subset r(t)W$ and $\psi(tV_0) \subset r'(t)W$ for sufficiently small $t$, then + \[ + \varphi(tV_0) + \psi(tV_0) \subset r(t)W + r'(t)W \subset [\abs{r(t)} + \abs{r'(t)}]W + \] + + (6): Let $W \in \cn_F(0)$, then there exists $V_0 \in \cn_E(0)$ such that $\varphi(V_0) \subset W$. In which case, $\varphi(tV_0) \subset tW$ for all $t \in \real$. + + (7): Let $x \in E$ and $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi$ is tangent to $0$ and linear, then there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that + \[ + \varphi(tV) \subset r(t)W \quad \varphi(V) \subset \frac{r(t)}{t}W + \] + for sufficiently small $t \in \real$. Since $W \in \cn_F(0)$, there exists $\lambda \in K$ such that $x \in \lambda V$. Thus + \[ + \varphi(x) \in \varphi(\lambda V) = \lambda\varphi(V) \subset \frac{\lambda r(t)}{t}W + \] + As $r \in o(t)$, there exists $t \in \real$ such that $\abs{\lambda r(t)/t} \le 1$, so $\varphi(x) \in \frac{\lambda r(t)}{t}W \subset W$. Since this holds for all $W \in \cn_F(0)$ and $F$ is Hausdorff, $\varphi(x) \in \ol{\bracs{0}} = \bracs{0}$ by \ref{proposition:tvs-closure} and \ref{lemma:t1}. +\end{proof} + + + \begin{lemma} \label{lemma:tangent-to-0} Let $E, F, G$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, $\psi: U \to F$ be tangent to $0$, then: @@ -54,15 +169,15 @@ As $r \in o(t)$, there exists $t \in \real$ such that $\abs{\lambda r(t)/t} \le 1$, so $\varphi(x) \in \frac{\lambda r(t)}{t}W \subset W$. Since this holds for all $W \in \cn_F(0)$ and $F$ is Hausdorff, $\varphi(x) \in \ol{\bracs{0}} = \bracs{0}$ by \ref{proposition:tvs-closure} and \ref{lemma:t1}. \end{proof} -\begin{definition}[Derivative] +\begin{definition}[(Strong Fréchet) Derivative] \label{definition:derivative} - Let $E, F$ be TVSs over $K \in \RC$ with $F$ being Hausdorff, $U \subset E$ open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{differentiable at} $x_0$ if there exists $\lambda \in L(E; F)$, $V \in \cn_E(0)$, and $\varphi: V \to F$ tangent to $0$ such that + Let $E, F$ be TVSs over $K \in \RC$ with $F$ being Hausdorff, $U \subset E$ open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{(strongly Fréchet) differentiable at} $x_0$ if there exists $\lambda \in L(E; F)$, $V \in \cn_E(0)$, and $\varphi: V \to F$ tangent to $0$ such that \[ f(x_0 + h) = f(x_0) + \lambda h + \varphi(h) \] - for all $h \in V$. In which case, $Df(x_0) = \lambda$ is the unique continuous linear map satisfying the above, and the \textbf{derivative} of $f$ \textbf{at} $x_0$. + for all $h \in V$. In which case, $Df(x_0) = \lambda$ is the unique continuous linear map satisfying the above, and the \textbf{(strong Fréchet) derivative} of $f$ \textbf{at} $x_0$. - If $f$ is differentiable at every $x_0 \in U$, then $f$ is \textbf{differentiable}, and $Df: U \to L(E; F)$ is the \textbf{derivative} of $f$. + If $f$ is differentiable at every $x_0 \in U$, then $f$ is \textbf{(strongly Fréchet) differentiable}, and $Df: U \to L(E; F)$ is the \textbf{(strong Fréchet) derivative} of $f$. \end{definition} \begin{proof} Let $\lambda, \mu \in L(E; F)$ and $\varphi, \psi: V \to F$ be tangent to $0$ such that @@ -74,7 +189,7 @@ \[ 0 = (\lambda - \mu)h + (\varphi - \psi)(h) \] - By \ref{lemma:tangent-to-0}, $(\lambda - \mu)$ is tangent to $0$ and thus equal to $0$. + By (7) of \ref{lemma:tangent-to-0}, $(\lambda - \mu)$ is tangent to $0$ and thus equal to $0$. \end{proof} \begin{proposition}[Chain Rule] @@ -88,25 +203,13 @@ By differentiability of $f$ and $g$, there exists $\varphi, \psi$ tangent to $0$ such that \begin{align*} (g \circ f)(x_0 + h) &= (g \circ f)(x_0) + Dg(f(x_0))[f(x_0 + h) - f(x_0)] + \varphi(f(x_0 + h) - f(x_0)) \\ - &= (g \circ f)(x_0) + Dg(f(x_0))[Df(x_0)(h) + \psi(h)] + \varphi(Df(x_0)(h) + \psi(h)) + &= (g \circ f)(x_0) + Dg(f(x_0))[Df(x_0)(h) + \psi(h)] + \varphi(Df(x_0)(h) + \psi(h)) \\ + &= (g \circ f)(x_0) + [Dg(f(x_0)) \circ Df(x_0)](h) \\ + &+ [Dg(f(x_0)) \circ \psi + \varphi \circ (Df(x_0) + \psi)](h) \end{align*} - Since $Dg(f(x_0)) \in L(F; G)$, $Dg(f(x_0)) \circ \psi$ is tangent to $0$ by \ref{lemma:tangent-to-0}. - - On the other hand, let $W \in \cn_G(0)$, then there exists $V_1 \in \cn_F(0)$ circled and $r_1 \in o(t)$ such that + Since $Dg(f(x_0)) \in L(F; G)$, $Dg(f(x_0)) \circ \psi$ is tangent to $0$ by (6) and (3) \ref{lemma:tangent-linear-at-0}. Similarly, $Df(x_0) + \psi$ is of linear order at $0$ by (4) and (2) of \ref{lemma:tangent-linear-at-0}, so $\varphi \circ (Df(x_0) + \psi)$ is tangent to $0$ by (3) of \ref{lemma:tangent-linear-at-0}. Thus \[ - \varphi(tV_1) \subset r_1(t)W + Dg(f(x_0)) \circ \psi + \varphi \circ (Df(x_0) + \psi) \] - for sufficiently small $t \in \real$. Let $V_2 \in \cn_F(0)$ such that $V_2 + V_2 \subset V_1$ and assume without loss of generality that $V_2$ is circled using \ref{proposition:tvs-good-neighbourhood-base}, then there exists $U_1 \in \cn_E(0)$ and $r_2 \in o(t)$ with - \[ - \psi(tU_1) \subset r_2(t)V_2 - \] - for sufficiently small $t \in \real$. In particular, since $V_2$ is circled, if $t$ is small enough, then - \[ - \psi(tU_1) \subset r_2(t)V_2 \subset tV_2 - \] - Thus if $U_2 = U_1 \cap Df(x_0)^{-1}(V_2)$, then - \[ - \varphi(Df(x_0)(tU_2) + \psi(tU_2)) \subset \varphi(tV_2 + tV_2) \subset \varphi(tV_1) \subset r_1(t)W - \] - so $\varphi(Df(x_0)(h) + \psi(h))$ is tangent to $0$ as well. + is tangent to $0$ by (5) of \ref{lemma:tangent-linear-at-0}. \end{proof} diff --git a/src/dg/derivative/sets.tex b/src/dg/derivative/sets.tex new file mode 100644 index 0000000..e930fd6 --- /dev/null +++ b/src/dg/derivative/sets.tex @@ -0,0 +1,17 @@ +\section{Differentiation With Respect to Set Systems} +\label{section:differentiation-set-system} + +\begin{definition}[Small] +\label{definition:differentiation-small} + Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed family of sets that contains all finite sets, $r: E \to F$, and $n \in \natz$, then the following are equivalent: + \begin{enumerate} + \item For each $A \in \sigma$, $r(th)/t^n \to 0$ uniformly on $A$. + \item If $r_t(x) = r(tx)/t^n$, then $r_t \to 0$ as $t \to 0$ with respect to the $\sigma$-uniform topology on $F^E$. + \item For each $A \in \sigma$, $\seq{a_k} \subset A$, and $\seq{t_k} \subset K \setminus \bracs{0}$ with $t_k \to 0$ as $n \to \infty$, $r(t_ka_k)/t_k^n \to 0$ as $n \to \infty$. + \end{enumerate} + If the above holds, then $r$ is \textbf{$\sigma$-small of order $n$}. +\end{definition} + +\begin{remark} + In \ref{definition:differentiation-small}, the system $\sigma$ can be chosedn based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain. +\end{remark} diff --git a/src/fa/norm/normed.tex b/src/fa/norm/normed.tex index 76c093b..6b9677a 100644 --- a/src/fa/norm/normed.tex +++ b/src/fa/norm/normed.tex @@ -11,13 +11,13 @@ \end{enumerate} then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that: \begin{enumerate} - \item $\sum_{n \in \natp}\norm{x_n}_E < C\norm{y}/(1 - \gamma)$. + \item $\sum_{n \in \natp}\norm{x_n}_E \le C\norm{y}_F/(1 - \gamma)$. \item $\sum_{n = 1}^\infty Tx_n = y$. \end{enumerate} - In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}/(1 - \gamma)$ and $Tx = y$. + In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}_F/(1 - \gamma)$ and $Tx = y$. \end{theorem} \begin{proof} - Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}} \le \gamma \norm{y_n}_F$. + Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}}_F \le \gamma \norm{y_n}_F$. For each $n \in \nat$, \[ @@ -25,11 +25,11 @@ \] Since $\norm{x_n}_E \le C\norm{y_n}_F$, \[ - \sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}}{1 - \gamma} + \sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}_F}{1 - \gamma} \] In addition, \[ - \norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}} \le \gamma^n \norm{y}_F + \norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}}_F \le \gamma^n \norm{y}_F \] so $\sum_{n = 1}^\infty Tx_n = y$. \end{proof} diff --git a/src/fa/tvs/bounded.tex b/src/fa/tvs/bounded.tex index 13f619d..815a428 100644 --- a/src/fa/tvs/bounded.tex +++ b/src/fa/tvs/bounded.tex @@ -3,12 +3,12 @@ \begin{definition}[Bounded] \label{definition:bounded} - Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$. + Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$. The collection $B(E)$ is the set of all bounded sets of $E$. \end{definition} \begin{proposition}[{{\cite[1.5.1]{SchaeferWolff}}}] \label{proposition:bounded-operations} - Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$ be bounded, then the following sets are bounded: + Let $E$ be a TVS over $K \in \RC$ and $A, B \in B(E)$, then the following sets are bounded: \begin{enumerate} \item Any $C \subset B$. \item The closure $\ol{B}$. diff --git a/src/fa/tvs/spaces-of-linear.tex b/src/fa/tvs/spaces-of-linear.tex index b193de2..3004060 100644 --- a/src/fa/tvs/spaces-of-linear.tex +++ b/src/fa/tvs/spaces-of-linear.tex @@ -1,4 +1,4 @@ -\section{Spaces of Vector-Valued Maps} +\section{Vector-Valued Function Spaces} \label{section:spaces-linear-map} \begin{proposition}[{{\cite[3.3.1]{SchaeferWolff}}}] @@ -37,6 +37,65 @@ for all $x \in S$. \end{proof} +\begin{definition}[Space of Bounded Linear Maps] +\label{definition:bounded-linear-map-space} + Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $L_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology. +\end{definition} + +\begin{proposition} +\label{proposition:multilinear-identify} + Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system that contains all singletons, and $k \in \natp$, then + \begin{enumerate} + \item The map + \[ + I: L_{\mathfrak{S}}^k(E; L_{\mathfrak{S}}(E; F)) \to L^{k+1}_{\mathfrak{S}}(E; F) + \] + defined by + \[ + (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1}) + \] + is an isomorphism. + \item The map + \[ + I: \underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to L^k_{\mathfrak{S}}(E; F) + \] + defined by + \[ + IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k) + \] + is an isomorphism. + \end{enumerate} + which allows the identification + \[ + \underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = L^k_{\mathfrak{S}}(E; F) + \] + under the map $I$ in (2). +\end{proposition} +\begin{proof} + (1): To see that $I$ is surjective, let $T \in L_{\mathfrak{S}}^{k+1}(E; F)$ and + \[ + I^{-1}T: E \to L_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot) + \] + Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case, + \[ + T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F) + \] + by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in L_{\mathfrak{S}}(E; F)$. + + In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $L_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(L_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in L^k_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; F))$. + + It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \mathfrak{S}$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous. + + On the other hand, let $S_1 \in \mathfrak{S}$ and $E(S_2, U)$ be an entourage of $L_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well. + + (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then + \[ + \underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = L^k_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; F)) = L^{k+1}_{\mathfrak{S}}(E; F) + \] + Thus (2) holds for all $k \in \natp$. +\end{proof} + + \begin{definition}[Strong Operator Topology] \label{definition:strong-operator-topology} diff --git a/src/measure/index.tex b/src/measure/index.tex index e7cd0a9..3816288 100644 --- a/src/measure/index.tex +++ b/src/measure/index.tex @@ -3,3 +3,4 @@ \input{./src/measure/sets/index.tex} \input{./src/measure/measure/index.tex} +\input{./src/measure/measurable-maps/index.tex} diff --git a/src/measure/measurable-maps/index.tex b/src/measure/measurable-maps/index.tex new file mode 100644 index 0000000..6a6448f --- /dev/null +++ b/src/measure/measurable-maps/index.tex @@ -0,0 +1,5 @@ +\chapter{Measurable Functions} +\label{chap:measurable-maps} + +\input{./src/measure/measurable-maps/measurable-maps.tex} +\input{./src/measure/measurable-maps/product.tex} diff --git a/src/measure/measurable-maps/measurable-maps.tex b/src/measure/measurable-maps/measurable-maps.tex new file mode 100644 index 0000000..3ab680f --- /dev/null +++ b/src/measure/measurable-maps/measurable-maps.tex @@ -0,0 +1,26 @@ +\section{Measurable Functions} +\label{section:measurable-maps} + +\begin{definition}[Measurable Function] +\label{definition:measurable-function} + Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces and $f: X \to Y$ be a mapping, then $f$ is \textbf{$(\cm, \cn)$-measurable} if $f^{-1}(E) \in \cm$ for all $E \in \cn$. +\end{definition} + +\begin{definition}[Borel Measurable] +\label{definition:borel-measurable-function} + Let $(X, \cm)$ be a measurable space, $Y$ be a topological space, and $f: X \to Y$ be a mapping, then $f$ is \textbf{Borel measurable} if $f$ is $(\cm, \cb_Y)$-measurable. +\end{definition} + +\begin{lemma} +\label{lemma:continuous-borel-measurable} + Let $X, Y$ be topological spaces and $f: X \to Y$ be continuous, then $f$ is Borel measurable. +\end{lemma} + +\begin{definition}[Generated $\sigma$-Algebra] +\label{definition:generated-sigma-algebra-function} + Let $X$ be a set, $\bracs{(Y_i, \cn_i)}_{i \in I}$ be measurable spaces, and $\seqi{f}$ with $f_i: X \to Y_i$ for each $i \in I$. The \textbf{$\sigma$-algebra generated by} $\seqi{f}$, + \[ + \sigma(\bracs{f_i| i \in I}) = \sigma\paren{f_i^{-1}(\cn_i)|i \in I} + \] + is the smallest $\sigma$-algebra on $X$ such that each $\seqi{f}$ is measurable. +\end{definition} diff --git a/src/measure/measurable-maps/product.tex b/src/measure/measurable-maps/product.tex new file mode 100644 index 0000000..9936c73 --- /dev/null +++ b/src/measure/measurable-maps/product.tex @@ -0,0 +1,52 @@ +\section{Product $\sigma$-Algebras} +\label{section:product-sigma-algebra} + +\begin{definition}[Product $\sigma$-Algebra] +\label{definition:product-sigma-algebra} + Let $\bracs{(X_i, \cm_i)}_{i \in I}$ be measurable spaces, then the \textbf{product} $\sigma$-algebra $\bigotimes_{i \in I}\cm_i$ on $\prod_{i \in I}X_i$ is the $\sigma$-algebra generated by $\seqi{\pi_i}$. +\end{definition} + +\begin{lemma} +\label{lemma:rectangles} + Let $\bracs{(X_i, \cm_i)}_{i \in I}$, let + \[ + \ce = \bracs{\bigcap_{j \in J}\pi_j^{-1}(E_j) \bigg | J \subset I \text{ finite}, E_j \in \cm_j} + \] + then $\ce$ is an elementary family, and $\sigma(\ce) = \bigotimes_{i \in I}\cm_i$. +\end{lemma} +\begin{proof} +\begin{enumerate} + \item[(P1)] For any $j \in I$, $\emptyset = \pi_j^{-1}(\emptyset) \in \ce$. + \item[(P2)] Let + \[ + \bigcap_{j \in J}\pi_j^{-1}(E_j), \bigcap_{j \in J'}\pi_j^{-1}(F_j) \in \ce + \] + Assume without loss of generality that $J = J'$, then + \[ + \bigcap_{j \in J}\pi_j^{-1}(E_j) \cap \bigcap_{j \in J'}\pi_j^{-1}(F_j) = \bigcap_{j \in J}\pi_j^{-1}(E_j \cap F_j) \in \ce + \] + \item[(E)] For any $j \in I$, $\prod_{i \in I}X_i = \pi_j^{-1}(X_j) \in \ce$. Thus it is sufficient to show that $\ce$ is closed under complements. Let $\bigcap_{j \in J}\pi_j^{-1}(E_j) \in \ce$, then + \[ + \braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^c = \bigcup_{j \in J}\pi_j^{-1}(E_j^c) = \bigsqcup_{\emptyset \subsetneq K \subset J} + \underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce} + \] +\end{enumerate} +\end{proof} + +\begin{proposition}[{{\cite[Proposition 1.5]{Folland}}}] +\label{proposition:product-sigma-algebra-metric} + Let $\seqf{X_j}$ be topological spaces, $X = \prod_{j = 1}^n X_j$, then: + \begin{enumerate} + \item $\bigotimes_{j = 1}^n \cb_{X_j} \subset \cb_{X}$. + \item If $\seq{X_j}$ are separable, then $\bigotimes_{j = 1}^n \cb_{X_j} = \cb_{X}$ + \end{enumerate} +\end{proposition} +\begin{proof} + (2): By \ref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$. + + Since $\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^n \cb_{X_j}$ and + \[ + U = \bigcup_{j \in \natp}\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) + \] + The open set $U$ is in $\bigotimes_{j = 1}^n \cb_{X_j}$. +\end{proof} diff --git a/src/measure/measure/index.tex b/src/measure/measure/index.tex index 996e074..f1a8f19 100644 --- a/src/measure/measure/index.tex +++ b/src/measure/measure/index.tex @@ -8,3 +8,4 @@ \input{./src/measure/measure/outer.tex} \input{./src/measure/measure/regular.tex} \input{./src/measure/measure/lebesgue-stieltjes.tex} +\input{./src/measure/measure/kolmogorov.tex} diff --git a/src/measure/measure/kolmogorov.tex b/src/measure/measure/kolmogorov.tex new file mode 100644 index 0000000..e613e86 --- /dev/null +++ b/src/measure/measure/kolmogorov.tex @@ -0,0 +1,76 @@ +\section{Kolmogorov's Extension Theorem} +\label{section:kolmogorov-extension} + + +\begin{definition}[Consistent] +\label{definition:consistent-measures} + Let $\bracs{(\Omega_i, \cf_i)}_{i \in I}$ be a family of measurable spaces. For each $J \subset J' \subset I$, denote $\pi_J: \prod_{i \in I}X_i \to \prod_{j \in J}X_j$ and $\pi_{J', J}: \prod_{j \in J'}X_j \to \prod_{j \in J}X_j$ as the projection maps. + + Let $\bracs{\mu_J|J \subset I \text{ finite}}$ such that each $\mu_J$ is a probability measure on $\prod_{j \in J}\Omega_j$, then $\bracs{\mu_J}$ is \textbf{consistent} if for all $J \subset J' \subset I$, $\mu_{J'} = \mu_J \circ \pi_{J', J}^{-1}$. +\end{definition} + +\begin{lemma}[{{\cite[Lemma 1.17]{Baudoin}}}] +\label{lemma:kolmogorov-compact-sequence} + Let $\seq{X_n}$ be topological spaces where for each $n \in \natp$, + \begin{enumerate} + \item[(a)] Every finite measure on $\prod_{j = 1}^n X_j$ is regular\footnote{A potential sufficient condition for this is that each $X_n$ is LCH where every open set is $\sigma$-compact. However, I have yet to verify if this condition persists over products.}. + \item[(b)] $X_n$ is Hausdorff. + \item[(c)] $X_n$ is separable. + \end{enumerate} + Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then for any $\seq{B_n}$ where: + \begin{enumerate} + \item[(d)] For each $n \in \nat$, $B_n \in \cb(\prod_{j = 1}^n X_j)$. + \item[(e)] For each $n \in \nat$, $B_{n+1} \subset B_n \times X_{n+1}$. + \item[(f)] There exists $\eps > 0$ such that $\mu_{[n]}(B_n) > \eps$ for all $n \in \natp$. + \end{enumerate} + Then there exists $\seq{K_n}$ such that for every $n \in \natp$, + \begin{enumerate} + \item $K_n \subset \prod_{j = 1}^n X_j$ is compact. + \item $K_n \subset B_n$. + \item $K_{n+1} \subset K_n \times X_{n+1}$. + \item $\mu(K_n) \ge \eps/2$. + \end{enumerate} +\end{lemma} +\begin{proof} + Let $n \in \natp$. By (c), $\mu_{[n]}$ is a Borel probability measure. Thus $\mu_{[n]}$ is regular by (a). Thus there exists $C_n \subset \prod_{j = 1}^n X_j$ compact such that $C_n \subset B_n$ and $\mu_{[n]}(B_n \setminus C_n) < \eps/2^{n+1}$. Let + \[ + K_n = \bigcap_{j = 1}^n \pi_{[n], [j]}^{-1}(C_n) + \] + Since each $X_j$ is Hausdorff, $K_n \subset \prod_{j = 1}^n X_j$ is compact with $K_n \subset B_n$ and $K_{n+1} \subset K_n \times X_{n+1}$. Moreover, + \[ + \mu_{[n]}(B_n \setminus K_n) \le \sum_{j = 1}^n\mu_{[n]}\braks{\pi_{[n], [j]}^{-1}(B_j \setminus C_j)} \le \sum_{j = 1}^n\mu_{[j]}(B_j \setminus C_j) \le \eps/2 + \] + Thus $\mu_{[n]}(K_n) \ge \eps/2$. +\end{proof} + +\begin{theorem}[Kolmogorov's Extension Theorem, {{\cite[Theorem 1.14]{Baudoin}}}] +\label{theorem:kolmogorov-extension} + Let $\seqi{X}$ be topological spaces where for each $J \subset I$ finite, + \begin{enumerate} + \item[(a)] Every finite measure on $\prod_{j \in J} X_j$ is regular. + \item[(b)] $X_j$ is Hausdorff. + \item[(c)] $X_j$ is separable. + \end{enumerate} + Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then there exists a unique probability measure $\mu: \bigotimes_{i \in I}\cb_{X_i} \to [0, 1]$ such that for any $J \subset I$ finite, $\mu = \mu_J \circ \pi_J^{-1}$. +\end{theorem} +\begin{proof} + Let + \[ + \alg = \bracs{\pi_J^{-1}(B)| B \in \bigotimes_{j \in J}\cb_{X_j}, J \subset I \text{ finite}} + \] + then $\alg$ is an algebra. For any $B \in \bigotimes_{j \in J}\cb_{X_j}$, define $\mu_0(\pi_J^{-1}(B)) = \mu_J(B)$, then $\mu_0: \alg \to [0, 1]$ is well-defined and finitely additive by the consistency of $\bracs{\mu_J}$. + + To show that $\mu_0$ is a premeasure, it is sufficient to show that for any $\{\pi_{J_n}^{-1}(B_n)\}_1^\infty$ with $\pi_{J_n}^{-1}(B_n) \downto \emptyset$, $\mu_0(\pi_{J_n}^{-1}(B_n)) \downto 0$. To this end, suppose for contradiction that $\limv{n}\mu_0(\pi_{J_n}^{-1}(B_n)) = \eps > 0$. + + By inserting additional elements into the sequence and relabeling the indices, assume without loss of generality that $J_n = [n]$ for all $n \in \natp$. By \ref{lemma:kolmogorov-compact-sequence}, there exists $\seq{K_n}$ such that: + \begin{enumerate} + \item $K_n \subset \prod_{j = 1}^n X_j$ is compact. + \item $K_n \subset B_n$. + \item $K_{n+1} \subset K_n \times X_{n+1}$. + \item $\mu(K_n) \ge \eps/2$. + \end{enumerate} + Let $N \in \natp$ and $x \in \prod_{j = 1}^N X_j$ such that $x \in \bigcap_{n \ge N}\pi_{[N]}(K_n)$. By compactness and (4), there exists $x_{N+1} \in X_{N+1}$ such that $(x_1, \cdots, x_N, x_{N+1}) \in \bigcap_{n > N}\pi_{[N+1]}(K_n)$. Thus there exists $x \in \prod_{i \in I}X_i$ such that $x \in \pi_{[n]}^{-1}(K_n)$ for all $n \in \natp$, and + \[ + x \in \bigcap_{n \in \natp}\pi_{[n]}^{-1}(K_n) \subset \bigcap_{n \in \natp}\pi_{[n]}^{-1}(B_n) \ne \emptyset + \] +\end{proof} diff --git a/src/measure/measure/regular.tex b/src/measure/measure/regular.tex index f25d387..55e8ac2 100644 --- a/src/measure/measure/regular.tex +++ b/src/measure/measure/regular.tex @@ -21,3 +21,14 @@ \label{definition:regular-measure} Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a measure, then $\mu$ is \textbf{regular} if it is inner regular and outer regular. \end{definition} + +\begin{theorem}[{{\cite[Theorem 7.8]{Folland}}}] +\label{theorem:sigma-finite-regular-measure} + Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure. If: + \begin{enumerate} + \item[(a)] $X$ is a LCH space. + \item[(b)] Every open set of $X$ is $\sigma$-compact. + \item[(c)] For any $K \subset X$ compact, $\mu(K) < \infty$. + \end{enumerate} + then $\mu$ is a regular measure. +\end{theorem} diff --git a/src/topology/functions/index.tex b/src/topology/functions/index.tex index ad9e6b6..bf46898 100644 --- a/src/topology/functions/index.tex +++ b/src/topology/functions/index.tex @@ -2,3 +2,4 @@ \label{chap:function-spaces} \input{./src/topology/functions/set-systems.tex} +\input{./src/topology/functions/uniform.tex} diff --git a/src/topology/main/interiorclosureboundary.tex b/src/topology/main/interiorclosureboundary.tex index 36481f9..4baa67a 100644 --- a/src/topology/main/interiorclosureboundary.tex +++ b/src/topology/main/interiorclosureboundary.tex @@ -78,6 +78,15 @@ Let $x \in \prod_{i \in I}X_i$ and $U \in \cn(x)$, then there exists $J \subset I$ finite and $\seqf{U_j}$ such that $U_j \in \cn(\pi_j(x))$ for each $j \in J$, and $x \in \bigcap_{j \in J}\pi_j^{-1}(U_j) \subset U$. By density of each $A_j$, there exists $y \in X$ such that $y_j \in U_j$ for each $j \in J$. Thus $y \in U$, and $A$ is dense. \end{proof} +\begin{proposition} +\label{proposition:separable-product} + Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable. +\end{proposition} +\begin{proof} + Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \ref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$. +\end{proof} + + \begin{lemma} \label{lemma:closurecomplement} Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$. diff --git a/src/topology/main/metric.tex b/src/topology/main/metric.tex deleted file mode 100644 index d5c03b3..0000000 --- a/src/topology/main/metric.tex +++ /dev/null @@ -1,50 +0,0 @@ -\section{Metric Spaces} -\label{section:metric} - -\begin{definition}[Metric] -\label{definition:metric} - Let $X$ be a set, a \textbf{pseudo-metric} is a mapping $d: X \times X \to [0, \infty)$ such that: - \begin{enumerate} - \item For any $x \in X$, $d(x, x) = 0$. - \item For any $x, y \in X$, $d(x, y) = d(y, x)$. - \item For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$. - \end{enumerate} - and the pair $(X, d)$ is a \textbf{pseudo-metric space}. If in addition, - \begin{enumerate} - \item[(4)] For any $x, y \in X$, $d(x, y) = 0$ if and only if $x = y$. - \end{enumerate} - then $d$ is a \textbf{metric}. -\end{definition} - -\begin{definition}[Induced Uniformity] -\label{definition:metric-uniformity} - Let $(X, d)$ be a metric space. For each $r > 0$, let - \[ - U_r = \bracs{(x, y) \in X \times Y| d(x, y) < r} - \] - then there exists a uniformity $\fU$ on $X$ such that the family $\fB = \bracs{U_r| r > 0}$ is a fundamental system of entourages for $\fU$, known as the \textbf{uniformity induced by} $d$. - - Let $\seqi{d}$ be pseudo-metrics on $X$. For each $i \in I$, let $X_i = X$ be equipped with the uniformity induced by $d_i$, then the \textbf{uniformity induced by} $\seqi{d}$ is the initial uniformity generated by the identity $\text{Id}_X: X \to X_i$. -\end{definition} - -\begin{lemma} -\label{lemma:uniform-first-countable} - Let $X$ be a uniform space with a countable fundamental system of entourages, then there exists a countable fundamental system of symmetric entourages $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ such that $U_{1/2^{n+1}} \circ U_{1/2^{n+1}} \subset U_{1/2^n}$ for all $n \in \nat^+$. -\end{lemma} -\begin{proof} - Let $\seq{V}$ be a countable fundamental system of entourages. By \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $U_{1/2} \subset V_1$. Suppose inductively that $U_{1/2^n}$ has been constructed, then by (U2) and \ref{lemma:symmetricfundamentalentourage}, there exists a symmetric entourage $U_{1/2^{n+1}} \subset U_{1/2^n} \cap V$. Thus $\fB = \bracs{U_{1/2^n}|n \in \nat^+}$ is the desired family. -\end{proof} - -% Work in progress: pseudo-metrisability of uniform spaces. - -\begin{lemma} -\label{lemma:uniform-urysohn} - Let $X$ be a uniform space with a countable fundamental system of entourages, $\mathbb{D}$ be the dyadic rational numbers in $[0, 1]$, then there exists a fundamental system of entourages $\fB = \bracs{}$ -\end{lemma} - - - -\begin{proposition} -\label{proposition:uniform-pseudometric} - Let -\end{proposition} diff --git a/src/topology/main/neighbourhoods.tex b/src/topology/main/neighbourhoods.tex index 7658445..7a4a68d 100644 --- a/src/topology/main/neighbourhoods.tex +++ b/src/topology/main/neighbourhoods.tex @@ -22,7 +22,6 @@ \label{lemma:openneighbourhood} Let $(X, \topo)$ be a topological space, then $U \subset X$ is open if and only if $U \in \cn_\topo(x)$ for all $x \in U$. \end{lemma} - \begin{proof} Suppose that $U \in \cn_\topo(x)$ for all $x \in U$. For each $x \in U$, there exists $V_x \in \topo$ with $x \in V_x \subset U$. Thus $U = \bigcup_{x \in U}V_x \in \topo$. diff --git a/src/topology/main/normal.tex b/src/topology/main/normal.tex index b9013bb..4af6bf5 100644 --- a/src/topology/main/normal.tex +++ b/src/topology/main/normal.tex @@ -82,6 +82,6 @@ Since $\real$ is complete, so is $BC(X; \real)$ by \ref{proposition:set-uniform-complete}. Using successive approximations (\ref{theorem:successive-approximation}), for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_A = f$ and \[ - \norm{F}_u \le \frac{1}{3} \cdot \frac{1}{1 - 2/3} \cdot \norm{f}_u = 1 + \norm{F}_u \le \frac{1}{3} \cdot \frac{1}{1 - 2/3} \cdot \norm{f}_u = \norm{f}_u \] \end{proof} diff --git a/src/topology/uniform/cauchy.tex b/src/topology/uniform/cauchy.tex index e69de29..e3fd2c9 100644 --- a/src/topology/uniform/cauchy.tex +++ b/src/topology/uniform/cauchy.tex @@ -0,0 +1,98 @@ +\section{Cauchy Filters} +\label{section:cauchy-filter} + +\begin{definition}[Small] +\label{definition:small} + Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A \subset X$, then $A$ is \textbf{$V$-small} if $A \times A \subset V$. +\end{definition} + +\begin{lemma}[{{\cite[Proposition 2.3.1]{Bourbaki}}}] +\label{lemma:small-intersect} + Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A, B \subset X$ such that: + \begin{enumerate} + \item $A, B$ are $V$-small. + \item $A \cap B \ne \emptyset$. + \end{enumerate} + then $A \cup B$ is $V \circ V$-small. +\end{lemma} +\begin{proof} + Since $V \circ V \supset V$, it is sufficient to consider the case where $x \in A$ and $y \in B$. By assumption (2), there exists $z \in A \cap B$. By assumption (1), $(x, z), (z, y) \in V$, so $(x, y) \in V \circ V$. +\end{proof} + +\begin{definition}[Cauchy Filter] +\label{definition:cauchyfilter} + Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small. +\end{definition} + +\begin{proposition}[Cauchy Criterion] +\label{proposition:cauchycriterion} + Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy. +\end{proposition} +\begin{proof} + Let $x \in X$ such that $\cn(x) \subset \fF$ and $V \in \fU$. By \ref{lemma:symmetricfundamentalentourage}, assume without loss of generality that $V \in \fU$. Since $\cn(x) \subset \fF$, then $V(x) \in \fU$ and $V(x) \times V(x) \subset V$. +\end{proof} + +\begin{definition}[Cauchy Continuous] +\label{definition:cauchy-continuous} + Let $X, Y$ be uniform spaces and $f: X \to Y$, then $f$ is \textbf{Cauchy continuous} if for any Cauchy filter base $\fB \subset 2^X$, $f(\fB) \subset 2^Y$ is Cauchy. +\end{definition} + + +\begin{proposition}[{{\cite[Proposition 2.3.3]{Bourbaki}}}] +\label{proposition:imagecauchy} + Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f$ is Cauchy continuous. +\end{proposition} +\begin{proof} + Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. For any Cauchy filter base $\fB \subset 2^X$, there exists $E \in \fB$ such that $E \times E \subset V'$, so $f(E) \times f(E) \subset V$. +\end{proof} + +\begin{definition}[Minimal Cauchy Filter] +\label{definition:minimalcauchy} + Let $X$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then $\fF$ is \textbf{minimal} if it is minimal with respect to inclusion. +\end{definition} + +\begin{proposition}[{{\cite[Proposition 2.3.5]{Bourbaki}}}] +\label{proposition:minimalcauchyexistence} + Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then: + \begin{enumerate} + \item There exists a unique minimal Cauchy filter $\fF_0$ such that $\fF_0 \subset \fF$. + \item If $\fB \subset \fF$ is a base of $\fF$, and $\fV \subset \fU$ is a fundamental system of symmetric entourages, then $\mathfrak{M} = \bracs{V(M): V \in \fV, M \in \fB}$ is a base for $\fF_0$. + \end{enumerate} +\end{proposition} +\begin{proof} + Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_1, V_2 \in \fV$ and $M_1, M_2 \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_0 \in \fV$ and $M_0 \in \fB$ such that $V_0 \subset V_1 \cap V_2$ and $M_0 \subset M_1 \cap M_2$. So $V_0(M_0) \subset V_1(M_1) \cap V_2(M_2)$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By \ref{proposition:filterbasecriterion}, $\mathfrak{M}$ is a filter base for + \[ + \fF_0 = \bracs{E \subset X| \exists V \in \fV, M \in \fB: V(M) \subset E} \subset \fF + \] + + To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ \subset $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and \ref{lemma:compositiongymnastics}. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base. + + Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_0$. +\end{proof} + +\begin{proposition}[{{\cite[Corollary 2.3.4]{Bourbaki}}}] +\label{proposition:cauchyinterior} + Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a minimal Cauchy filter, then $\fF$ admits a base consisting of open sets. +\end{proposition} +\begin{proof} + Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By \ref{proposition:goodentourages}, $\fV$ is a fundamental system of entourages. By \ref{proposition:minimalcauchyexistence}, $\mathfrak{M} = \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$. + + Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by \ref{lemma:openentourageneighbourhoods}. Thus $\mathfrak{M}$ consists of open sets. +\end{proof} + +\begin{proposition}[{{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}] +\label{proposition:cauchyfilterlimit} + Let $(X, \fU)$ be a uniform space, then: + \begin{enumerate} + \item For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter. + \item For each Cauchy filter $\fF \subset 2^X$ and $x \in X$, $x$ is an accumulation point of $\fF$ if and only if $\fF$ converges to $x$. + \item For any $x \in X$, filter $\fF \subset 2^X$ converging to $x$, and Cauchy filter $\mathfrak{G} \subset \fF$, $\mathfrak{G}$ converges to $x$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By \ref{proposition:minimalcauchyexistence}, $\bracs{U(x): U \in \fU} = \cn(x)$ is the unique minimal filter contained in $\fF$. + + (2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By \ref{definition:generatedfilter}, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$. + + (3): Since $\fF$ converges to $x$, $x$ is a cluster point of $\fF$ and $\mathfrak{G}$. By (2), $x$ is also a limit point of $\mathfrak{G}$. +\end{proof} diff --git a/src/topology/uniform/complete.tex b/src/topology/uniform/complete.tex index 80f79cd..08f4878 100644 --- a/src/topology/uniform/complete.tex +++ b/src/topology/uniform/complete.tex @@ -1,103 +1,6 @@ \section{Completeness} \label{section:completeness} - -\begin{definition}[Small] -\label{definition:small} - Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A \subset X$, then $A$ is \textbf{$V$-small} if $A \times A \subset V$. -\end{definition} - -\begin{lemma}[{{\cite[Proposition 2.3.1]{Bourbaki}}}] -\label{lemma:small-intersect} - Let $(X, \fU)$ be a uniform space, $V \in \fU$, and $A, B \subset X$ such that: - \begin{enumerate} - \item $A, B$ are $V$-small. - \item $A \cap B \ne \emptyset$. - \end{enumerate} - then $A \cup B$ is $V \circ V$-small. -\end{lemma} -\begin{proof} - Since $V \circ V \supset V$, it is sufficient to consider the case where $x \in A$ and $y \in B$. By assumption (2), there exists $z \in A \cap B$. By assumption (1), $(x, z), (z, y) \in V$, so $(x, y) \in V \circ V$. -\end{proof} - -\begin{definition}[Cauchy Filter] -\label{definition:cauchyfilter} - Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small. -\end{definition} - -\begin{proposition}[Cauchy Criterion] -\label{proposition:cauchycriterion} - Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy. -\end{proposition} -\begin{proof} - Let $x \in X$ such that $\cn(x) \subset \fF$ and $V \in \fU$. By \ref{lemma:symmetricfundamentalentourage}, assume without loss of generality that $V \in \fU$. Since $\cn(x) \subset \fF$, then $V(x) \in \fU$ and $V(x) \times V(x) \subset V$. -\end{proof} - -\begin{definition}[Cauchy Continuous] -\label{definition:cauchy-continuous} - Let $X, Y$ be uniform spaces and $f: X \to Y$, then $f$ is \textbf{Cauchy continuous} if for any Cauchy filter base $\fB \subset 2^X$, $f(\fB) \subset 2^Y$ is Cauchy. -\end{definition} - - -\begin{proposition}[{{\cite[Proposition 2.3.3]{Bourbaki}}}] -\label{proposition:imagecauchy} - Let $(X, \fU)$ and $(Y, \mathfrak{V})$ be uniform spaces and $f \in UC(X; Y)$, then $f$ is Cauchy continuous. -\end{proposition} -\begin{proof} - Let $V \in \mathfrak{V}$, then there exists $V' \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in V'$. For any Cauchy filter base $\fB \subset 2^X$, there exists $E \in \fB$ such that $E \times E \subset V'$, so $f(E) \times f(E) \subset V$. -\end{proof} - -\begin{definition}[Minimal Cauchy Filter] -\label{definition:minimalcauchy} - Let $X$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then $\fF$ is \textbf{minimal} if it is minimal with respect to inclusion. -\end{definition} - -\begin{proposition}[{{\cite[Proposition 2.3.5]{Bourbaki}}}] -\label{proposition:minimalcauchyexistence} - Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a Cauchy filter, then: - \begin{enumerate} - \item There exists a unique minimal Cauchy filter $\fF_0$ such that $\fF_0 \subset \fF$. - \item If $\fB \subset \fF$ is a base of $\fF$, and $\fV \subset \fU$ is a fundamental system of symmetric entourages, then $\mathfrak{M} = \bracs{V(M): V \in \fV, M \in \fB}$ is a base for $\fF_0$. - \end{enumerate} -\end{proposition} -\begin{proof} - Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_1, V_2 \in \fV$ and $M_1, M_2 \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_0 \in \fV$ and $M_0 \in \fB$ such that $V_0 \subset V_1 \cap V_2$ and $M_0 \subset M_1 \cap M_2$. So $V_0(M_0) \subset V_1(M_1) \cap V_2(M_2)$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By \ref{proposition:filterbasecriterion}, $\mathfrak{M}$ is a filter base for - \[ - \fF_0 = \bracs{E \subset X| \exists V \in \fV, M \in \fB: V(M) \subset E} \subset \fF - \] - - To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ \subset $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and \ref{lemma:compositiongymnastics}. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base. - - Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_0$. -\end{proof} - -\begin{proposition}[{{\cite[Corollary 2.3.4]{Bourbaki}}}] -\label{proposition:cauchyinterior} - Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a minimal Cauchy filter, then $\fF$ admits a base consisting of open sets. -\end{proposition} -\begin{proof} - Let $\fV \subset \fU$ be the set of all symmetric, open entourages. By \ref{proposition:goodentourages}, $\fV$ is a fundamental system of entourages. By \ref{proposition:minimalcauchyexistence}, $\mathfrak{M} = \bracs{U(M)| U \in \fV, M \in \fF}$ is a base for $\fF$. - - Let $V \in \fU$, then there exists $U \in \fV$ such that $U \subset V$. For any $M \in \fF$, $U(M)$ is open by \ref{lemma:openentourageneighbourhoods}. Thus $\mathfrak{M}$ consists of open sets. -\end{proof} - -\begin{proposition}[{{\cite[Corollary 2.3.1, 2.3.2]{Bourbaki}}}] -\label{proposition:cauchyfilterlimit} - Let $(X, \fU)$ be a uniform space, then: - \begin{enumerate} - \item For each $x \in X$, $\cn(x)$ is a minimal Cauchy filter. - \item For each Cauchy filter $\fF \subset 2^X$ and $x \in X$, $x$ is an accumulation point of $\fF$ if and only if $\fF$ converges to $x$. - \item For any $x \in X$, filter $\fF \subset 2^X$ converging to $x$, and Cauchy filter $\mathfrak{G} \subset \fF$, $\mathfrak{G}$ converges to $x$. - \end{enumerate} -\end{proposition} -\begin{proof} - (1): Let $\fF = \bracs{E \subset X: x \in E}$, then $\fF$ is a filter generated by the filter base $\fB = \bracs{\bracs{x}}$. By \ref{proposition:minimalcauchyexistence}, $\bracs{U(x): U \in \fU} = \cn(x)$ is the unique minimal filter contained in $\fF$. - - (2): Let $x \in X$ be an accumulation point of $\fF$, then $U(x) \cap E \ne \emptyset$ for all $U \in \fU$ and $E \in \fF$. By \ref{definition:generatedfilter}, there exists a filter $\fF' \supset \fF \cup \cn(x)$. Since $\fF$ is a Cauchy filter, so is $\fF'$. By minimality of $\cn(x)$, $\fF \supset \cn(x)$. - - (3): Since $\fF$ converges to $x$, $x$ is a cluster point of $\fF$ and $\mathfrak{G}$. By (2), $x$ is also a limit point of $\mathfrak{G}$. -\end{proof} - \begin{definition}[Complete Space] \label{definition:completespace} Let $(X, \fU)$ be a uniform space, then $X$ is \textbf{complete} if every Cauchy filter converges to at least one point. diff --git a/src/topology/uniform/index.tex b/src/topology/uniform/index.tex index 9001d95..7785386 100644 --- a/src/topology/uniform/index.tex +++ b/src/topology/uniform/index.tex @@ -3,6 +3,7 @@ \input{./src/topology/uniform/definition.tex} \input{./src/topology/uniform/uc.tex} +\input{./src/topology/uniform/metric.tex} \input{./src/topology/uniform/cauchy.tex} \input{./src/topology/uniform/complete.tex} \input{./src/topology/uniform/completion.tex} diff --git a/src/topology/uniform/metric.tex b/src/topology/uniform/metric.tex new file mode 100644 index 0000000..4ed7e06 --- /dev/null +++ b/src/topology/uniform/metric.tex @@ -0,0 +1,245 @@ +\section{Pseudometrics} +\label{section:pseudometric} + +The axioms of uniform spaces strongly resembles working in a metric space. In fact, any uniform space may arise from a family of uniformly continuous pseudometrics. This allows understanding uniform spaces in a more familiar languagge. + +\begin{definition}[Pseudometric] +\label{definition:pseudometric} + Let $X$ be a set, a \textbf{pseudometric} is a mapping $d: X \times X \to [0, \infty)$ such that: + \begin{enumerate} + \item[(PM1)] For any $x \in X$, $d(x, x) = 0$. + \item[(PM2)] For any $x, y \in X$, $d(x, y) = d(y, x)$. + \item[(PM3)] For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$. + \end{enumerate} +\end{definition} + +\begin{lemma} +\label{lemma:pseudometric-continuous} + Let $(X, \fU)$ be a uniform space and $d: X \times X \to [0, \infty)$ be a pseudometric, then the following are equivalent: + \begin{enumerate} + \item $d \in UC(X \times X; [0, \infty))$. + \item For each $r > 0$, $U_r = \bracs{(x, y) \in X \times X| d(x, y) < r} \in \fU$. + \end{enumerate} +\end{lemma} +\begin{proof} + (1) $\Rightarrow$ (2): Let $r > 0$, then there exists $V \in \fU$ symmetric such that for any $(x, x'), (y, y') \in V$, $\abs{d(x, y) - d(x', y')} < r$. In particular, for any $(x, y) \in V$, $(x, x), (x, y) \in V$. Thus $d(x, y) < d(x, x) + r = r$. Therefore $V \subset U_r$, and $U_r \in \fU$. + + (2) $\Rightarrow$ (1): Let $r > 0$, then for any $(x, x'), (y, y') \in U_{r/2}$, $\abs{d(x, y) - d(x', y')} < r$ by the triangle inequality. +\end{proof} + + +\begin{definition}[Pseudometric Uniformity] +\label{definition:pseudometric-uniformity} + Let $X$ be a set and $\seqi{d}$ be pseudometric on $X$. For each $i \in I$, $r > 0$, and $x \in X$, let + \[ + B_i(x, r) = \bracs{y \in X| d(x, y) < r} + \] + and + \[ + U_{i, r} = \bracs{(x, y) \in X \times X| d(x, y) < r} + \] + then there exists a uniformity $\fU$ on $X$ such that: + \begin{enumerate} + \item The family + \[ + \fB = \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset I \text{ finite}, r > 0} + \] + forms a fundamental system of entourages consisting of symmetric open sets. + \item For any $x \in X$, + \[ + \cb(x) = \bracs{\bigcap_{j \in J}B_j(x, r) \bigg | J \subset I \text{ finite}, r > 0} + \] + is a fundamental system of neighbourhoods at $x$. + \item For each $U \subset X$, $U$ is open if and only if for every $x \in U$, there exists $J \subset I$ finite and $r > 0$ such that $\bigcap_{j \in J}B_j(x, r) \subset U$. + \item For each $i \in I$, $d_i \in UC(X \times X; [0, \infty))$. + \item[(U)] For any other uniformity $\mathfrak{V}$ satisfying (4), $\mathfrak{U} \subset \mathfrak{V}$. + \end{enumerate} + The uniformity $\fU$ is the \textbf{pseudometric uniformity} induced by $\seqi{d}$, and the topology induced by $\fU$ is the \textbf{pseudometric topology} on $X$ induced by $\seqi{d}$. +\end{definition} +\begin{proof} + (1, fundamental system): To see that $\fB$ is a fundamental system of entourages for a uniformity on $X$, it is sufficient to verify the conditions of \ref{proposition:fundamental-entourage-criterion}. + \begin{enumerate} + \item[(FB1)] For any $J, J' \subset I$ finite and $r, r' > 0$, + \[ + \bigcap_{j \in J \cup J'}U_{j, \min(r, r')} \subset \paren{\bigcap_{j \in J}U_{j, r}} \cap \paren{\bigcap_{j \in J'}U_{j, r'}} + \] + \item[(UB1)] For each $i \in I$, $d(x, x) = 0$ for all $x \in X$. Thus for any $i \in I$ and $r > 0$, $U_{i, r}$ contains the diagonal. + \item[(UB2)] For each $J \subset I$ finite and $r > 0$, + \[ + \paren{\bigcap_{j \in J}U_{j, r/2}} \circ \paren{\bigcap_{j \in J}U_{j, r}} \subset \bigcap_{j \in J}U_{j, r/2} \circ U_{j, r/2} \subset \bigcap_{j \in J}U_{j, r} + \] + by the triangle inequality. + \end{enumerate} + + (2): Since $\fB$ is a fundamental system of entourages for $\fU$, + \[ + \cb(x) = \bracs{U(x)| U \in \fB} + \] + is a fundamental system of neighbourhoods at $x$. + + (3): By definition of the uniform topology, for any $U \subset X$, $U$ is open if and only if for any $x \in U$, there exists $V \in \fU$ such that $x \in V(x) \subset U$. As $\fB$ is a fundamental system of entourages for $\fU$, this is equivalent to the existence of $J \subset I$ finite and $r > 0$ such that + \[ + x \in \paren{\bigcap_{j \in J}U_{j, r}}(x) = \bigcap_{j \in J}B_j(x, r) \subset U + \] + + (1, symmetric open): Let $i \in I$ and $r > 0$. Since $d_i$ is symmetric, so is $U_{i, r}$. For any $(x, y) \in U_{i, r}$, let $s = r - d_i(x, y)$, then for any $(x', y') \in X \times X$ with $d_i(x, x') < s/2$ and $d_i(y, y') < s/2$, $d(x', y') < s + d_i(x, y) = r$. Thus $B_i(x, s/2) \times B_i(y, s/2) \subset U_{i, r}$. + + By (3), $B_i(x, s/2) \in \cn(x)$ and $B_i(y, s/2) \in \cn(y)$, so $B_i(x, s/2) \times B_i(y, s/2) \in \cn((x, y))$. As such, $U_{i, r}$ is open by \ref{lemma:openneighbourhood}. + + (4): By \ref{lemma:pseudometric-continuous}. + + (5): By \ref{lemma:pseudometric-continuous}, $U_{i, r} \in \mathfrak{V}$ for all $i \in I$ and $r > 0$, so $\mathfrak{V} \supset \mathfrak{B}$ by (F2), and $\mathfrak{V} \supset \mathfrak{U}$. +\end{proof} + +\begin{lemma} +\label{lemma:uniform-sequence-pseudometric} + Let $(X, \fU)$ be a uniform space and $\bracsn{U_n}_0^\infty \subset \fU$ such that: + \begin{enumerate} + \item[(a)] $U_{0} = X \times X$. + \item[(b)] For each $n \in \natz$, $U_n$ is symmetric. + \item[(c)] For each $n \in \natz$, $U_{n + 1} \circ U_{n+1} \subset U_n$. + \end{enumerate} + then there exists a pseudometric $d: X \times X \to [0, 1]$ such that + \[ + U_{n+1} \subset \bracs{(x, y)| d(x, y) < 2^{-n}} \subset U_{n-1} + \] + for each $n \in \natp$. +\end{lemma} +\begin{proof} + Let + \[ + \rho: X \times X \to [0, 1] \quad (x, y) \mapsto \inf\bracs{2^{-n}| n \in \natz, (x, y) \in U_n} + \] + and $d: X \times X \to [0, 1]$ by + \[ + d(x, y) = \inf\bracs{\sum_{j = 1}^n\rho(x_{j-1}, x_j) \bigg | \seqf{x_j} \subset X, x_0 = x, x_1 = y, n \in \natp} + \] + then + \begin{enumerate} + \item[(PM1)] For any $x \in X$, $x \in \bigcap_{n \in \natp}U_n$. Thus $\rho(x, x) = 0$ and $d(x, x) \le \rho(x, x) = 0$. + \item[(PM2)] Let $x, y \in X$. By assumption (b), $\rho(x, y) = \rho(y, x)$. Thus $d(x, y) = d(y, x)$ as well. + \item[(PM3)] Let $x, y, z \in X$, then for any $\seqf{x_j}$ and $\seqf[m]{y_j}$ with $x_0 = x$, $x_n = y = y_0$, and $y_m = z$, + \[ + d(x, z) \le \sum_{j = 1}^n \rho(x_{j - 1}, x_j) + \sum_{j = 1}^m \rho(y_{j - 1}, y_j) + \] + As this holds for all such $\seqf{x_j}$ and $\seqf[m]{y_j}$, $d(x, z) \le d(x, y) + d(y, z)$. + \end{enumerate} + so $d$ is a pseudometric. + + For any $(x, y) \in U_{n+1}$, $d(x, y) \le \rho(x, y) < 2^{-n}$, so $U_{n+1} \subset \bracs{(x, y)| d(x, y) \le 2^{-n}}$. + + Let $x, y \in X$ with $d(x, y) < 2^{-n}$. If $\rho(x, y) < 2^{-n}$, then the claim holds directly. Assume inductively that for any $x, y \in X$ with $d(x, y) < 2^{-n}$, if there exists $\seqf[m]{x_j} \subset X$ such that $x_0 = x$, $x_m = y$ and $\sum_{j = 1}^m \rho(x_{j - 1}, x_j) < 2^{-n}$, then $(x, y) \in U_{n - 1}$. + + Let $x, y \in X$ and $\seqf[m+1]{x_j} \subset X$ such that $x_0 = x$, $x_{m+1} = y$, and $\sum_{j = 1}^{m+1} \rho(x_{j - 1}, x_j) < 2^{-n}$. Let $1 \le k < m+1$ such that $\sum_{j = 1}^{k}\rho(x_{j - 1}, x_j) < 2^{-n-1}$ and $\sum_{j = 1}^{k+1}\rho(x_{j-1}, x_j) \ge 2^{-n-1}$, then $\sum_{j = k + 1}^{m+1}\rho(x_{j-1}, x_j) < 2^{-n-1}$ as well. By the inductive hypothesis, $(x, x_k), (x_{k+1}, y) \in U_{n+1} \subset U_n$. Given that $\rho(x_k, x_{k+1}) < 2^{-n}$, $(x_k, x_{k+1}) \in U_{n+1}$ too. Thus + \[ + (x, y) \subset U_{n+1} \circ U_{n + 1} \circ U_{n + 1} \subset U_n \circ U_n \subset U_{n-1} + \] +\end{proof} + +\begin{remark} + Unfortunately, it was hard to access sources for this proof online, so I could not provide a specific citation. As such, I followed an online PDF given by the link: https://krex.k-state.edu/server/api/core/bitstreams/1bdf2b14-3b5a-4962-a589-93ee1998950c/content +\end{remark} + +\begin{remark} + It may be tempting to construst 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} + + +\begin{theorem} +\label{theorem:uniform-pseudometric} + Let $(X, \fU)$ be a uniform space, then $\fU$ is the pseudometric uniformity induced by the family of all uniformly continuous pseudometrics on $X$. +\end{theorem} +\begin{proof} + Let $\seqi{d}$ be the family of all uniformly continuous pseudometrics on $X$, and $\mathfrak{V}$ be the pseudometric uniformity induced by $\seqi{d}$. By (U) of \ref{definition:pseudometric-uniformity}, $\fU \supset \mathfrak{V}$. + + On the other hand, let $U_1 \in \mathfrak{U}$. By (U3), there exists $\seq{U_n} \subset \mathfrak{U}$ such that $U_{n + 1} \circ U_{n + 1} \subset U_n$ for all $n \in \natp$. Let $U_0 = X \times X$, then $\bracsn{U_n}_0^\infty \subset \fU$ satisfies the hypothesis of \ref{lemma:uniform-sequence-pseudometric}. Thus there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for all $n \in \natp$, + \[ + U_{n + 1} \subset \bracs{(x, y) \in X \times X|d(x, y) < 2^{-n}} \subset U_{n-1} + \] + By \ref{definition:pseudometric-uniformity}, $d$ is a uniformly continuous pseudometric on $X$. Since $\bracs{(x, y) \in X \times X|d(x, y) < 1/4} \subset U_1$, $U_1 \in \mathfrak{V}$. Therefore $\fU = \mathfrak{V}$. +\end{proof} + +\begin{definition}[Equivalent Pseudometrics] +\label{definition:equivalent-pseudometrics} + Let $X$ be a set and $\seqi{d}, \seqj{d}$ be pseudometrics on $X$, then $\seqi{d}$ and $\seqj{d}$ are \textbf{equivalent} if their induced uniformities coincide. +\end{definition} + +\begin{lemma} +\label{lemma:pseudometric-clamp} + Let $X$ be a set and $d: X \times X \to [0, \infty)$ be a pseudometric, then the pseudometric + \[ + \td d: X \times X \to [0, \infty) \quad (x, y) \mapsto d(x, y) \wedge 1 + \] + is equivalent to $d$. +\end{lemma} +\begin{proof} + For any $r \in (0, 1]$, + \[ + \bracsn{(x, y) \in X \times X| d(x, y) < r} = \bracsn{(x, y) \in X \times X| \td d(x, y) < r} + \] + Since sets of the above form generate the uniformity induced by $d$ and $\td d$, their induced uniformities coincide. +\end{proof} + +\begin{proposition} +\label{proposition:countable-equivalent-pseudometrics} + Let $X$ be a set and $\seq{d_n}$ be pseudometrics on $X$, then there exists a pseudometric $d: X \times X \to [0, \infty)$ equivalent to $\seq{d_n}$. +\end{proposition} +\begin{proof} + Using \ref{lemma:pseudometric-clamp}, assume without loss of generality that for each $n \in \natp$, $d_n$ takes values in $[0, 1]$. Let + \[ + d(x, y) = \sum_{n \in \natp}\frac{d_n(x, y)}{2^n} + \] + then $d$ is a well-defined a pseudometric. + + For each $n \in \nat$ and $r > 0$, denote + \begin{align*} + U_{n, r} &= \bracs{(x, y) \in X \times X| d_n(x, y)< r} \\ + U_{r} &= \bracs{(x, y) \in X \times X|d(x, y) < r} + \end{align*} + + + Let $r > 0$, then there exists $n \in \natp$ such that $2^{-n} < r$. Take $s = r - 2^{-n}$, then $\bigcap_{k = 1}^nU_{k, s} \subset U_{r}$. On the other hand, for any $n \in \natp$ and $r > 0$, $U_{r/2^n} \subset \bigcap_{k = 1}^n U_{k, r}$. Therefore $\seq{d_n}$ and $d$ are equivalent. +\end{proof} + +\begin{theorem}[Metrisability of Uniform Spaces] +\label{theorem:uniform-metrisable} + Let $(X, \fU)$ be a uniform space, then the following are equivalent: + \begin{enumerate} + \item There exists a countable fundamental system of entourages for $X$. + \item There exists a pseudometric $d: X \times X \to [0, \infty)$ that induces the uniformity on $X$. + \item There exists a countable family $\seq{d_n}$ of pseudometrics on $X$ that induce the uniformity on $X$. + \end{enumerate} +\end{theorem} +\begin{proof} + (1) $\Rightarrow$ (2): Let $\seq{U_n} \subset \fU$ be a fundamental system of entourages for $X$ and $V_1 = U_n$. Assume inductively that $\bracs{V_k|1 \le k \le n} \subset \fU$ has been constructed such that + \begin{enumerate} + \item[(a)] For each $1 \le k \le n$, $V_k$ is symmetric. + \item[(b)] For each $1 \le k \le n$, $V_k \subset U_k$. + \item[(c)] For each $1 \le k < n$, $V_{k+1} \circ V_{k+1} \subset V_{k}$. + \end{enumerate} + Let $W = V_n \cap U_{n+1}$, then by \ref{lemma:symmetricfundamentalentourage}, there exists $V_{n+1} \in \fU$ symmetric such that $V_{n+1} \circ V_{n+1} \subset W$. Thus $\bracs{V_k|1 \le k \le n + 1} \subset \fU$ satisfies (a), (b), and (c) for $n + 1$. + + Let $V_0 = X \times X$, then by \ref{lemma:uniform-sequence-pseudometric}, there exists a pseudometric $d: X \times X \to [0, \infty)$ such that for each $n \in \natp$, + \[ + V_{n+1} \subset \bracs{(x, y) \in X \times X| d(x, y) < 2^{-n}} \subset V_{n-1} + \] + For any $U \in \fU$, there exists $n \in \nat$ such that + \[ + U \supset U_n \supset V_n \supset \bracs{(x, y) \in X \times X| d(x, y) < 2^{-n-1}} + \] + so $d$ induces the uniformity on $\fU$. + + (3) $\Rightarrow$ (1): By (1) of \ref{definition:pseudometric-uniformity}, + \[ + \fB = \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset \nat \text{ finite}, r > 0} + \] + where + \[ + U_{j, r} = \bracs{(x, y) \in X \times X| d_n(x, y) < r} + \] + is a fundamental system of entourages for $\fU$. Since for any $r > 0$, there exists $q \in \rational \cap (0, r)$, + \[ + \bracs{\bigcap_{j \in J}U_{j, r} \bigg | J \subset \nat \text{ finite}, r \in \rational, r > 0} + \] + is a countable fundamental system of entourages for $\fU$. +\end{proof}