Added Riemann's rearrangement theorem.

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}
+
+
+
+

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}