From 156c9e872894bee5ee2b4156d7efe468125162ff Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sun, 22 Mar 2026 00:45:15 -0400 Subject: [PATCH] Added missing steps and fixed typos. --- src/cat/tricks/dyadic.tex | 6 +++--- src/fa/norm/separable.tex | 4 ++-- src/fa/tvs/metric.tex | 4 ++-- src/topology/metric/metric.tex | 30 ++++++++++++++++++++++++++++++ src/topology/uniform/metric.tex | 2 +- 5 files changed, 38 insertions(+), 8 deletions(-) diff --git a/src/cat/tricks/dyadic.tex b/src/cat/tricks/dyadic.tex index 61f3958..040d6f5 100644 --- a/src/cat/tricks/dyadic.tex +++ b/src/cat/tricks/dyadic.tex @@ -44,10 +44,10 @@ \item[(b)] For each $x, y \in G$, $x + y \ge x, y$. \end{enumerate} - For each $x \in \mathbb{D} \cap (0, 1)$, let $\phi(x) = \sum_{n \in M(x)}g_n$, then + For each $x \in \mathbb{D} \cap [0, 1)$, let $\phi(x) = \sum_{n \in M(x)}g_n$, then \begin{enumerate} - \item For any $x, y \in \mathbb{D} \cap (0, 1)$ such that $x + y \in (0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$. - \item For any $x, y \in \mathbb{D} \cap (0, 1)$ with $x \le y$, $\phi(x) \le \phi(y)$. + \item For any $x, y \in \mathbb{D} \cap [0, 1)$ such that $x + y \in (0, 1)$, $\phi(x) + \phi(y) \le \phi(x + y)$. + \item For any $x, y \in \mathbb{D} \cap [0, 1)$ with $x \le y$, $\phi(x) \le \phi(y)$. \end{enumerate} \end{proposition} \begin{proof} diff --git a/src/fa/norm/separable.tex b/src/fa/norm/separable.tex index ebc9bf4..8ed9b6b 100644 --- a/src/fa/norm/separable.tex +++ b/src/fa/norm/separable.tex @@ -11,11 +11,11 @@ T_N: S \to \real^N \quad \phi \mapsto (\dpn{x_1, \phi}{E}, \cdots, \dpn{x_N, \phi}{E}) \] - then since $\real^N$ is separable, there exists $\bracs{\phi_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_N\phi_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$. + Since $\real^N$ is separable, $T_N(S)$ is separable by \autoref{proposition:separable-metric-space}. Thus there exists $\bracs{\phi_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_N\phi_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$. Let $\phi \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, \[ - |\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{n} + |\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{N} \] Thus for each $N \in \natp$, $\dpn{x_n, \phi_{N, k_N}}{E} \to \dpn{x_n, \phi}{E}$ as $N \to \infty$. Since $\phi_{N, k_N} \to \phi$ pointwise on a dense subset of $E$, and $\bracsn{\phi_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $\phi_{N, k_N} \to \phi$ in the weak*-topology by \autoref{proposition:strong-operator-dense}. diff --git a/src/fa/tvs/metric.tex b/src/fa/tvs/metric.tex index 920869b..5a1e1e1 100644 --- a/src/fa/tvs/metric.tex +++ b/src/fa/tvs/metric.tex @@ -76,7 +76,7 @@ $(4) \Rightarrow (1)$: By \autoref{definition:tvs-pseudonorm-topology}, for each $r > 0$, $\rho^{-1}([0, r)) \in \cn_E(0)$. Thus for any $x, y \in E$, if $x - y \in \rho^{-1}([0, r))$, then $\abs{\rho(x) - \rho(y)} \le r$. Therefore $\rho \in UC(E; [0, \infty))$. \end{proof} -\begin{lemma}[{{\cite[Theorem I.6.1]{SchaeferWolff}}}] +\begin{lemma}[] \label{lemma:tvs-sequence-pseudonorm} Let $E$ be a vector space over $K \in \RC$, $\seq{U_n} \subset 2^E$ such that \begin{enumerate} @@ -89,7 +89,7 @@ \] \end{lemma} -\begin{proof} +\begin{proof}[Proof {{\cite[Theorem I.6.1]{SchaeferWolff}}}.] For each $H \subset \natp$ finite, let \[ U_H = \sum_{n \in H}V_n \quad \rho_H = \sum_{n \in H}2^{-n} diff --git a/src/topology/metric/metric.tex b/src/topology/metric/metric.tex index 7fbd73e..17fce1a 100644 --- a/src/topology/metric/metric.tex +++ b/src/topology/metric/metric.tex @@ -12,3 +12,33 @@ \end{enumerate} The pair $(X, d)$ is a \textbf{metric space}, which comes with the metric uniformity induced by $d$, and the corresponding topology. \end{definition} + +\begin{proposition} +\label{proposition:separable-metric-space} + Let $(X, d)$ be a metric space, then the following are equivalent: + \begin{enumerate} + \item $X$ is second countable. + \item $X$ is Lindelöf. + \item $X$ is separable. + \end{enumerate} + + In particular, since second countability is hereditary, separability and the Lindelöf property are both hereditary in metric spaces. + +\end{proposition} +\begin{proof} + (1) $\Rightarrow$ (2): Let $\seqi{U} \subset 2^X$ be an open cover of $X$. Let $\seq{V_n} \subset 2^X$ be a countable base for the topology on $X$, and + \[ + K = \bracs{n \in \natp| \exists i \in I: V_n \subset U_i} + \] + + For any $x \in X$, there exists $i \in I$ such that $x \in U_i$, and $n \in \natp$ such that $x \in V_n \in U_i$. Therefore $\bigcup_{k \in K}V_k = X$. + + Let $J \subset I$ be countable such that for each $k \in K$, there exists $j \in J$ with $V_n \subset U_j$, then $X = \bigcup_{k \in K}V_k = \bigcup_{j \in J}V_j$. + + % TODO: This stuff may be moved to more general spaces. + + (2) $\Rightarrow$ (3): Let $n \in \nat$, then $\bracs{B(x, 1/n)|x \in X}$ is an open cover of $X$, so there exists $\seq{x_{n, k}} \subset X$ such that $X = \bigcup_{k \in \natp}B(x_k, 1/n)$. Let $D = \bracs{x_{n, k}| n, k \in \natp}$, then for any $x \in X$ and $n \in \natp$, there exists $k \in \natp$ such that $x \in B(x_{n, k}, 1/n)$, so $x_{n, k} \in B(x, 1/n)$. Therefore $D$ is dense in $X$, and $X$ is separable. + + (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} + diff --git a/src/topology/uniform/metric.tex b/src/topology/uniform/metric.tex index d52b3bc..7fd9fa0 100644 --- a/src/topology/uniform/metric.tex +++ b/src/topology/uniform/metric.tex @@ -128,7 +128,7 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa and $d: X \times X \to [0, 1]$ by \[ - d(x, y) = \inf\bracs{\sum_{j = 1}^n\rho(x_{j-1}, x_j) \bigg | \seqf{x_j} \subset X, x_0 = x, x_1 = y, n \in \natp} + d(x, y) = \inf\bracs{\sum_{j = 1}^n\rho(x_{j-1}, x_j) \bigg | \seqf{x_j} \subset X, x_0 = x, x_n = y, n \in \natp} \] then