diff --git a/chapters/sections/conv_of_series.tex b/chapters/sections/conv_of_series.tex index 468830d..d4058dc 100644 --- a/chapters/sections/conv_of_series.tex +++ b/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} diff --git a/chapters/sections/metr_and_normed.tex b/chapters/sections/metr_and_normed.tex new file mode 100644 index 0000000..c9bd77b --- /dev/null +++ b/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} \ No newline at end of file diff --git a/chapters/sections/seq_ser_limits.tex b/chapters/sections/seq_ser_limits.tex new file mode 100644 index 0000000..362870a --- /dev/null +++ b/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} \ No newline at end of file diff --git a/chapters/topo_of_metr_spaces.tex b/chapters/topo_of_metr_spaces.tex new file mode 100644 index 0000000..b6e1af2 --- /dev/null +++ b/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} \ No newline at end of file diff --git a/script.loe b/script.loe new file mode 100644 index 0000000..e69de29 diff --git a/script.pdf b/script.pdf index 7146ad2..9b4b55b 100644 Binary files a/script.pdf and b/script.pdf differ diff --git a/script.tex b/script.tex index c1995d6..d404560 100644 --- a/script.tex +++ b/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}