Finished two sections

chapters/sections/conv_of_series.tex

@@ -198,7 +198,7 @@ Then $\forall J \subset \natn \finite$
 Therefore $\series[n]{k} |x_k|$ is bounded and monotonic increasing, and hence it is converging. So $\series{k} |x_k| < \infty$.
 \end{proof}
 
-\begin{thm}
+\begin{thm}\label{259}
 Every absolutely convergent series converges and the limit does not depend on the order of summation.
 \end{thm}
 \begin{proof}

chapters/sections/metr_and_normed.tex

@@ -0,0 +1,142 @@
+\documentclass[../../script.tex] {subfiles}
+%! TEX root = ../../script.tex
+
+\begin{document}
+\section{Metric and Normed spaces}
+
+\begin{defi}[Metric space]
+    A metric space $(X, d)$ is an ordered pair consisting of a set $X$ and a mapping 
+    \[
+        d: X \times X \longrightarrow [0, \infty]
+    \] 
+    called metric. This mapping must fulfil the following conditions $\forall x, y, z \in X$:
+    \begin{itemize}
+        \item $d(x, y) \ge 0$ \text{ (Positivity)}
+        \item $d(x, y) = 0 \iff x = y$ \text{ (Definedness)}
+        \item $d(x, y) = d(y, x)$ \text{ (Symmetry)}
+        \item $d(x, y) \le d(x, z) + d(z, y)$ \text{ (Triangle inequality)}
+    \end{itemize}
+\end{defi}
+
+\begin{eg}
+    \begin{enumerate}[(i)]
+        \item Let $M$ be a set. Then 
+        \[
+            d(x, y) = \begin{cases}
+                1, & x \ne y \\
+                0, & \text{else}
+            \end{cases}
+        \]
+        is called the discrete metric.
+
+        \item Let $X$ be the set of edges of a graph.
+        \begin{align*}
+            d(x, y) := &\mbox{ Minimum amount of edges that have} \\
+            &\mbox{ to be passed to get from } x \mbox{ to } y
+        \end{align*}
+        
+        \begin{center}
+            \begin{tikzpicture}[scale=0.75]
+                \node[shape=circle, draw=black, fill=black] (A) at (0, 0) {};
+                \node[shape=circle, draw=black] (B) at (1, -4) {};
+                \node[shape=circle, draw=black] (C) at (3, -2) {};
+                \node[shape=circle, draw=black] (D) at (5, -3.5) {};
+                \node[shape=circle, draw=black] (E) at (6.5, -5.5) {};
+                \node[shape=circle, draw=black, fill=black] (F) at (7, -1) {};
+
+                \path [-] (A) edge node[left] {$1$} (B);
+                \path [-] (B) edge node[above left] {$2$} (C);
+                \path [-] (B) edge (D);
+                \path [-] (B) edge (E);
+                \path [-] (C) edge (D);
+                \path [-] (C) edge node[above] {$3$} (F);
+                \path [-] (E) edge (F);
+
+                \node[above=0.3cm] at (A) {$x$};
+                \node[above=0.3cm] at (F) {$y$};
+            \end{tikzpicture}
+        \end{center}
+
+        \item Let $X$ be the surface of a sphere.
+        \[
+            d(x, y) := \text{"Bee line"}
+        \]
+
+        \item Let $X$ be the set of points of the European street network.
+        \[
+            d(x, y) := \text{ Shortest route along this network}
+        \]
+
+        \item Let $(X, d_X)$, $(Y, d_Y)$ be metric spaces. Then 
+        \[
+            d_{X \times Y}((x_1, y_1), (x_2, y_2)) := d_X(x_1, x_2) + d_Y(y_1, y_2)
+        \]
+        defines a metric on $X \times Y$.
+    \end{enumerate}
+\end{eg}
+
+\begin{defi}[Normed space]
+    $(V, \dnorm)$ is said to be a normed space if $V$ is a vector space and 
+    \[
+        \dnorm: V \longrightarrow [0, \infty)
+    \]
+    is a mapping (called norm) with the following properties
+    \begin{itemize}
+        \item $\norm{x} \ge 0$ (Positivity)
+        \item $\norm{x} = 0 \iff x = 0$ (Definedness)
+        \item $\norm{\lambda x} = |\lambda|\norm{x}$ 
+        \item $\norm{x + y} \le \norm{x} + \norm{y}$ (Triangle inequality)
+    \end{itemize}
+    To every norm belongs a unique induced metric
+    \[
+        d(x, y) = \norm{x - y}
+    \]
+\end{defi}
+
+\begin{eg}[$\realn^n$ with Euclidian norm]
+    \begin{align*}
+        \dnorm: \realn^n &\longrightarrow [0, \infty) \\
+        (x_1, x_2, \cdots, x_n) &\longmapsto \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}
+    \end{align*}
+    Then $(\realn^n, \dnorm)$ is a normed space.
+\end{eg}
+
+\begin{eg}
+    \begin{enumerate}[(i)]
+        \item $(x_1, x_2, \cdots, x_n) \mapsto |x_1| + |x_2| + \cdots + |x_n|$ is also a norm on $\realn^n$.
+        \item On
+        \[
+            V = \set[f \text{ continuous}]{f: [0, 1] \longrightarrow \realn}
+        \]
+        we can define the supremum norm 
+        \[
+            \supnorm{f} = \sup \set[{x \in [0, 1]}]{|f(x)|}
+        \]
+        \item We can define sequence spaces as 
+        \[
+            \ell^p = \set[\series{n} |x_n|^p < \infty]{\anyseqdef{\cmpln^n}}
+        \]
+        with the norm 
+        \[
+            \norm{(x_n)}_p := \sqrt{\series{n} |x_n|^2}
+        \]
+        A special space is $\ell^2$, called Hilbert space
+    \end{enumerate}
+\end{eg}
+
+\begin{rem}
+    The Minkowski metric is not a metric in this sense.
+\end{rem}
+
+\begin{defi}[Balls and Boundedness]
+    Let $\metric$ be a metric space, and $x \in X, r > 0$. We then define 
+    \begin{align*}
+        \Oball(x) = \set[d(x, y) < r]{y \in X} && \text{Open ball} \\
+        \Cball(x) = \set[d(x, y) \le r]{y \in X} && \text{Closed ball}
+    \end{align*}
+    A subset $M \subset X$ is called bounded if
+    \[
+        \exists x \in X, r > 0: ~~M \subset \Oball(x)
+    \]
+\end{defi}
+\end{document}

chapters/sections/seq_ser_limits.tex

@@ -0,0 +1,188 @@
+\documentclass[../../script.tex] {subfiles}
+%! TEX root = ../../script.tex
+
+\begin{document}
+\section{Sequences, Series and Limits}
+
+\begin{defi}[Sequences and Convergence]
+    Let $\metric$ be a metric space. A sequence is a mapping $\natn \rightarrow X$. We write $\seq{x}_{n \in \natn}$ or $\seq{x}$.
+
+    The sequence $\seq{x}$ is said to be convergent to $x \in X$ if
+    \[
+        \forall \epsilon > 0 ~\exists N \in \natn ~\forall n \ge N: ~~d(x_n, x) < \epsilon
+    \]
+    $x$ is said to be the limit, and sequences that aren't convergent are called divergent.
+\end{defi}
+
+\begin{rem}
+    On $\realn$ the metric is the Euclidian metric $|\cdot|$, therefore this new definition of convergence is merely a generalization of the old one.
+\end{rem}
+
+\begin{thm}
+    Let $\seq{x}$ be a sequence in the metric space $\metric$ and $x \in X$. Then the following statements are equivalent:
+    \begin{enumerate}[(i)]
+        \item $\seq{x}$ converges to $x$
+        \item $\forall \epsilon > 0$ $\oball(x)$ contains all but finitely many elements of the sequence (almost every (a.e.) element)
+        \item $(d(x, x_n))$ is a null sequence
+    \end{enumerate}
+\end{thm}
+\begin{proof}
+    (ii) is merely a reformulation of (i), and $(ii) \iff (iii)$ follows from
+    \begin{equation}
+        d(x_n, x) = |d(x_n, x) - 0|
+    \end{equation}
+\end{proof}
+
+\begin{thm}
+    Let $\left(x^{(n)}\right) = (x_1^{(n)}, x_2^{(n)}, \cdots, x_d^{(n)}) \subset \realn^d$ and 
+    \[
+        x = (x_1, \cdots, x_d) \in \realn^d
+    \]
+    $\left(x^{(n)}\right)$ is said to converge to $x$ if and only if $x_i^{(n)}$ converges to $x_i$ for all $i$ in $\set{1, \cdots, d}$
+\end{thm}
+\begin{proof}
+    For $y = (y_1, \cdots, y_d) \in \realn^d$ we have
+    \begin{equation}
+        \norm{y_i} < \norm{y} ~~\forall i \in \set{1, \cdots, d}
+    \end{equation}
+    If $\left(x^{(n)}\right)$ converges to $x$, then
+    \begin{equation}
+        \abs{x_i^{(n)} - x_i} \le \norm{x^{(n)} - x} \conv{} 0
+    \end{equation}
+    If $(x_i^{(n)})$ converges to $x_i ~~\forall i \in \set{1, \cdots d}$, then
+    \begin{equation}
+        \forall \epsilon > 0 ~\exists N \in \natn ~\forall n > N: ~~\abs{x_i^{(n)} - x_i} < \frac{\epsilon}{\sqrt{d}} ~~\forall i \in \set{1, \cdots d}
+    \end{equation}
+    Thus
+    \begin{equation}
+    \begin{split}
+        \norm{x^{(n)} - x} &= \sqrt{(x_1^{(n)} - x_1)^2 + (x_2^{(n)} - x_2)^2 + \cdots + (x_d^{(n)} - x_d)^2} \\
+        &\le \sqrt{\frac{\epsilon^2}{d} + \frac{\epsilon^2}{d} + \cdots + \frac{\epsilon}{2}} \\
+        &= \epsilon
+    \end{split}
+    \end{equation}
+    So $\left(x^{(n)}\right)$ converges to $x$.
+\end{proof}
+
+\begin{thm}
+    Every convergent sequence has exactly one limit and is bounded.
+\end{thm}
+\begin{proof}
+    Assume that $x, y$ are limits of $(x_n)$ with $x \ne y$. Then $d(x, y) > 0$. There exists $N_1, N_2 \in \natn$, such that
+    \begin{subequations}
+    \begin{align}
+        d(x_n, x) &< \frac{d(x, y)}{2} ~~\forall n \ge N_1 \\
+        d(x_n, x) &< \frac{d(x, y)}{2} ~~\forall n \ge N_2
+    \end{align}
+    \end{subequations}
+    From this follows that
+    \begin{equation}
+        d(x, y) \le d(x, x_n) + d(x_n, y) < d(x, y) ~~\forall \max\set{N_1, N_2}
+    \end{equation}
+    which is a contradiction, thus sequences can have only one limit.
+
+    Now if $\seq{x}$ converges to $x$, then 
+    \begin{equation}
+        \exists N \in \natn ~\forall n \ge N: ~~d(x_n, x) < 1
+    \end{equation}
+    Then 
+    \begin{equation}
+        d(x_n, x) \le \max\set{d(x_1, x), d(x_2, x), \cdots, d(x_{N-1}, x), 1}
+    \end{equation}
+\end{proof}
+
+\begin{thm}
+    Let $\normed$ be a normed space over $\field$. Let $\seq{x}, \seq{y} \subset V$ be sequences with limits $x, y \in V$ 
+    and $\anyseqdef[\lambda]{\field}$ a sequence with limit $\lambda \in \field$. Then 
+    \begin{align*}
+        x_n + y_n \longrightarrow x + y && \lambda_n x_n \longrightarrow \lambda x
+    \end{align*}
+\end{thm}
+\begin{proof}
+    \reader
+\end{proof}
+
+\begin{defi}[Cauchy sequences and completeness]
+    A sequence $\seq{x}$ in a metric space $\metric$ is called Cauchy sequence if 
+    \[
+        \forall \epsilon > 0 ~\exists N \in \natn: ~~d(x_n, x_m) < \epsilon ~~\forall m, n \ge N
+    \]
+    A metric space is complete if every Cauchy sequence converges. A complete normed space is called Banach space.
+\end{defi}
+
+\begin{eg}
+    \item $(\realn, \abs{\cdot})$ and $(\cmpln, \abs{\cdot})$ are complete 
+    \item $(\ratn, \abs{\cdot})$ is not complete
+\end{eg}
+
+\begin{thm}
+    Every convering series is a Cauchy sequence
+\end{thm}
+\begin{proof}
+    Let $\seq{x} \conv{} x$. This means that 
+    \begin{equation}
+        \forall \epsilon > 0 ~\epsilon N \in \natn: ~~d(x_n, x) < \frac{\epsilon}{2} ~~\forall n \ge N
+    \end{equation}
+    Then 
+    \begin{equation}
+        d(x_n, x_m) \le d(x_n, x) + d(x, x_m) < \epsilon ~~\forall m, n \ge N
+    \end{equation}
+\end{proof}
+
+\begin{thm}
+    $\realn^n$ with the Euclidian norm is complete.
+\end{thm}
+\begin{proof}
+    Let $\left(x^{(n)}\right) \subset \realn^n$ be a Cauchy sequence. We know that 
+    \begin{equation}
+        \forall y \in \realn^n: ~~\abs{y_i} \le \norm{y} ~~\forall i \in \set{1, \cdots, n}
+    \end{equation}
+    We also know that $(x_i^{(n)})$ are Cauchy sequences because 
+    \begin{equation}
+        \abs{(x_i^{(n)} - x_i^{m})} \le \norm{x^{(n)} - x^{(m)}} ~~\forall i \in \set{1, \dots, n}
+    \end{equation}
+    Thus $x_i^{(n)} \conv{} x_i$ and therefore $\left(x^{(n)}\right) \conv{} x$.
+\end{proof}
+
+\begin{defi}[Series and (absolute) convergence]
+    Let $\normed$ be a normed space and $\anyseqdef{V}$. The series 
+    \[
+        \series{k} x_k
+    \]
+    is the sequence of partial sums
+    \[
+        s_n = \series[n]{k} x_k
+    \]
+    If the series converges then $\series{k} x_k$ also denotes the limit. 
+    The series is said to absolutely convergent if 
+    \[
+        \series{k} \norm{x_k} < \infty
+    \]
+\end{defi}
+
+\begin{thm}
+    In Banach spaces every absolutely convergent series is convergent.
+\end{thm}
+\begin{proof}
+    Let $\normed$, $\anyseqdef{V}$ and require $\series{n} \normed{x_n} < \infty$. We need to show that $s_n = \series[n]{k} x_k$ is a Cauchy sequence.
+    Let $\epsilon > 0$ and $t_n = \series[n]{k} \norm{x_k}$. $\seq{t}$ is convergent in $\realn$, and thus a Cauchy sequence.
+    I.e. 
+    \begin{equation}
+        \exists N \in \natn: ~~|t_n - t| < \epsilon ~~\forall m, n \ge N
+    \end{equation}
+    For $n > m > N$:
+    \begin{equation}
+        \norm{s_n - s_m} = \norm{\sum_{k=m+1}^n x_k} \le \sum_{k=m+1}^n \norm{x_k} = t_n - t_m = |t_n - t_m| < \epsilon
+    \end{equation}
+\end{proof}
+
+\begin{thm}
+    Let $\normed$ be a Banach space, $\series{k} x_k$ absolutely convergent and let $\sigma: \natn \rightarrow \natn$ be a bijective mapping. Then 
+    \[
+        \series{k} x_k = \series{k} x_{\sigma(k)}
+    \]
+\end{thm}
+\begin{proof}
+    Analogous to $\Cref{259}$
+\end{proof}
+\end{document}

chapters/topo_of_metr_spaces.tex

@@ -0,0 +1,9 @@
+\documentclass[../script.tex]{subfiles}
+%! TEX root = ../../script.tex
+
+\begin{document}
+    \chapter{Topology in Metric spaces}
+
+    \subfile{sections/metr_and_normed.tex}
+    \subfile{sections/seq_ser_limits.tex}
+\end{document}

script.tex

@@ -50,7 +50,7 @@
 \newcommand{\limz}{\lim_{n \rightarrow 0}}
 \newcommand{\limsupn}{\limsup_{n \rightarrow \infty}}
 \newcommand{\liminfn}{\liminf_{n \rightarrow \infty}}
-\newcommand{\seq}[1]{(#1_n)}
+\newcommand{\seq}[1]{\left(#1_n\right)}
 \newcommand{\nseqdef}[1]{\seq{#1} \subset \natn}
 \newcommand{\rseqdef}[1]{\seq{#1} \subset \realn}
 \newcommand{\cseqdef}[1]{\seq{#1} \subset \cmpln}
@@ -59,10 +59,19 @@
 \newcommand{\series}[2][\infty]{\sum_{#2 = 1}^{#1}}
 \newcommand{\finite}{\text{ finite}}
 \newcommand{\conj}[1]{\overline{#1}}
-\newcommand{\oball}[1][\epsilon]{B_{#1}}
 \newcommand{\closure}[1]{\overline{#1}}
 \newcommand{\must}[1][=]{\stackrel{!}{#1}}
 
+\newcommand{\metric}[1][X]{(#1, d)}
+\newcommand{\dnorm}{\norm{\cdot}}
+\newcommand{\normed}[1][V]{(#1, \dnorm)}
+\newcommand{\supnorm}[1]{\norm{#1}_{\infty}}
+
+\newcommand{\oball}[1][\epsilon]{B_{#1}}
+\newcommand{\cball}[1][\epsilon]{K_{#1}}
+\newcommand{\Oball}{\oball[r]}
+\newcommand{\Cball}{\cball[r]}
+
 \newcommand{\conv}[1]{\xrightarrow{\makebox[2em][c]{$\scriptstyle#1$}}}
 \newcommand{\convinf}{\conv{n \rightarrow \infty}}
 
@@ -141,5 +150,6 @@
 \subfile{chapters/real_analysis_1.tex}
 \subfile{chapters/linear_algebra.tex}
 \subfile{chapters/real_analysis_2.tex}
+\subfile{chapters/topo_of_metr_spaces.tex}
 
 \end{document}