From 2e160500ccd1de48698243053edc3ff4a2f9f352 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sun, 15 Mar 2026 14:44:36 -0400 Subject: [PATCH] Added Riemann's rearrangement theorem. --- preamble.sty | 11 ++-- src/fa/norm/absolute.tex | 109 +++++++++++++++++++++++++++++++++++++++ src/fa/norm/index.tex | 1 + 3 files changed, 116 insertions(+), 5 deletions(-) create mode 100644 src/fa/norm/absolute.tex diff --git a/preamble.sty b/preamble.sty index b567d48..a9281ac 100644 --- a/preamble.sty +++ b/preamble.sty @@ -8,12 +8,13 @@ % ------------- Block Environmets --------------- -\newtheorem{theorem}[subsection]{Theorem} -\newtheorem{proposition}[subsection]{Proposition} -\newtheorem{lemma}[subsection]{Lemma} +\newtheorem{theorem}{Theorem}[section] +\newtheorem{proposition}[theorem]{Proposition} +\newtheorem{corollary}[theorem]{Corollary} +\newtheorem{lemma}[theorem]{Lemma} -\newtheorem{definition}[subsection]{Definition} -\newtheorem{example}[subsection]{Example} +\newtheorem{definition}[theorem]{Definition} +\newtheorem{example}[theorem]{Example} % \newtheorem{exercise}[subsection]{Exercise} % \newtheorem{situation}[subsection]{Situation} diff --git a/src/fa/norm/absolute.tex b/src/fa/norm/absolute.tex new file mode 100644 index 0000000..8697c37 --- /dev/null +++ b/src/fa/norm/absolute.tex @@ -0,0 +1,109 @@ +\section{Conditional and Absolute Convergence} +\label{section:conditional-absolute-convergence} + +\begin{definition}[Absolute Convergence] +\label{definition:absolute-convergence} + Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges absolutely} if $\sum_{n \in \natp}\norm{x_n}_E < \infty$. +\end{definition} + +\begin{definition}[Unconditional Convergence] +\label{definition:unconditional-convergence} + Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges unconditionally} if for any bijection $\sigma: \natp \to \natp$, + \[ + \sum_{n = 1}^\infty x_n = \sum_{n = 1}^\infty x_{\sigma(n)} + \] +\end{definition} + +\begin{theorem}[Riemann's Rearrangement Theorem] +\label{theorem:riemann-rearrangement} + Let $\seq{x_n} \subset \real$ and $N = P \sqcup N$ such that $x_n \ge 0$ for all $n \in P$ and $x_n \le 0$ for all $n \in N$, then + \begin{enumerate} + \item If $\sum_{n \in P}x_n = \infty$ and $\sum_{n \in N}x_n = -\infty$, then there exists bijections $\sigma, \tau: \natp \to \natp$ such that + + \[ + \sum_{n = 1}^\infty x_{\sigma(n)} = \infty \quad \sum_{n = 1}^\infty x_{\tau(n)} = -\infty + \] + + \item If $\sum_{n \in P}x_n = \infty$, $\sum_{n \in N}x_n = -\infty$, and $\sum_{n =1}^\infty x_n$ converges, then for any $x \in \ol{\real}$, there exists a bijection $\sigma: \natp \to \natp$ such that $\sum_{n = 1}^\infty x_{\sigma(n)} = x$. + \item If $\sum_{n \in P}x_n = \infty$ but $\sum_{n \in N}x_n > -\infty$, then $\sum_{n = 1}^\infty x_n$ converges to $\infty$ unconditionally. + \item If $\sum_{n \in N}x_n = -\infty$ but $\sum_{n \in P}x_n < \infty$, then $\sum_{n = 1}^\infty x_n$ converges to $-\infty$ unconditionally. + \item If $\sum_{n \in \natp}|x_n| < \infty$, then $\sum_{n = 1}^\infty x_n$ converges unconditionally. + \end{enumerate} + In other words, a series in $\real$ converges unconditionally if and only if its positive parts or its negative parts are finite. +\end{theorem} +\begin{proof} + By inserting zeroes, assume without loss of generality that $P$ and $N$ are both infinite. Let $\seq{p_k} = P$ be an enumeration of $P$ and $\seq{n_k} = N$ be an enumeration of $N$. + + (1): By taking $\sum_{n \in P}-x_n$, assume without loss of generality that $\sum_{n \in P}x_n = \infty$. + + Let $k_0 = 0$. Let $n \in \natp$ and suppose inductively that $k_n$ has been constructed such that + \[ + \sum_{k = 1}^{k_n}x_{p_k} + \sum_{j = 1}^n x_{n_j} \ge n + \] + + Since $\sum_{n \in \natp}p_n = \infty$, there exists $k_{n+1} \in \natp$ such that $\sum_{k = k_{n}+1}^{k_{n+1}}x_{p_k} \ge x_{n_{n+1}} + 1$. + + Therefore there exists $\bracs{k_n}_1^K \subset \natp$ such that + \[ + \sum_{k = 1}^{k_n}x_{p_k} + \sum_{j = 1}^n x_{n_j} \ge n + \] + + for all $1 \le n < K$. In which case, if $\sigma: \natp \to \natp$ is defined by + \[ + (p_1, \cdots, p_{k_1}, n_1, p_1, \cdots, p_{k_2}, n_2, \cdots) + \] + + Let $n \in \natp$, then for any $K \ge k_{n} + n$, $\sum_{k = 1}^K x_{\sigma(k)} \ge n$, so $\sum_{k = 1}^\infty x_{\sigma(k)} = \infty$. + + (2): Assume without loss of generality that $x > 0$. Define $m_0 = 0$ and $n_0 = 0$. Let $n \in \natz$. If $n \ge 1$, suppose inductively that $m_n, n_n \in \natp$ has been constructed such that + \[ + x < \sum_{k = 1}^{m_{n}}x_{p_k} + \sum_{k = 1}^{n_{-1}}x_{n_k} < x + x_{m_{n}} + \] + and + \[ + x + x_{n_n} < \sum_{k = 1}^{m_n}x_{p_k} + \sum_{k = 1}^{n_n}x_{n_k} < x + \] + + Since $\sum_{k \in \natp}x_{p_k} = \infty$, there exists $m_{n+1} > m_n$ such that + \[ + x < \sum_{k = 1}^{m_{n+1}}x_{p_k} + \sum_{k = 1}^{n_n}x_{n_k} < x + x_{m_{n+1}} + \] + + As $\sum_{k \in \natp}x_{n_k} = -\infty$, there exists $n_{n+1} > n_n$ such that + \[ + x + x_{n_{n+1}} < \sum_{k = 1}^{m_{n+1}}x_{p_k} + \sum_{k = 1}^{n_{n+1}}x_{n_k} < x + \] + + Let $\sigma: \natp \to \natp$ be defined by + \[ + (p_1, \cdots, p_{m_1}, n_{1}, \cdots, n_{n_1}, p_{m_1 + 1}, \cdots, p_{m_2}, \cdots) + \] + + then for any $K \in [m_n + n_n, m_{n+1} + n_{n+1}]$, + \[ + x + x_{n_n} + x_{n_{n+1}} \le \sum_{k = 1}^K x_{\sigma(k)} \le x + x_{p_{n+1}} + \] + + Since $x_n \to 0$ as $n \to \infty$, $\sum_{k = 1}^K x_{\sigma(k)} \to x$ as $K \to \infty$. + + (3): Let $\sigma: \natp \to \natp$ be a bijection and $\alpha > 0$, then there exists $K \in \natp$ such that $\sum_{k = 1}^K x_{p_k} > \alpha - \sum_{k \in \natp}x_{n_k}$. Since $\sigma$ is a bijection, there exists $K' \in \natp$ such that $\sigma([1, K']) \supset \bracs{p_k|1 \le k \le K}$. In which case, for any $k \ge K'$, + \[ + \sum_{j = 1}^k x_{\sigma(j)} \ge \sum_{j = 1}^K x_{p_k} + \sum_{k \in \natp}x_{n_k} > \alpha + \] + + (4): By applying (3) to $\sum_{n = 1}^\infty -x_n$. + + (5): Let $\sigma: \natp \to \natp$ be a bijection, then for any $N \in \natp$, there exists $K \in \natp$ such that $\sigma([1, K]) \supset [1, N]$. In which case, + \[ + \abs{\sum_{n = 1}^N x_n - \sum_{n = 1}^K x_{\sigma(n)}} \le \sum_{n > N}|x_n| + \] + + so + \[ + \sum_{n = 1}^\infty x_n = \sum_{n = 1}^\infty x_{\sigma(n)} + \] +\end{proof} + + + + diff --git a/src/fa/norm/index.tex b/src/fa/norm/index.tex index 0a50f0b..e4904f8 100644 --- a/src/fa/norm/index.tex +++ b/src/fa/norm/index.tex @@ -2,5 +2,6 @@ \label{chap:normed-spaces} \input{./normed.tex} +\input{./absolute.tex} \input{./linear.tex} \input{./multilinear.tex}