Finished two sections
This commit is contained in:
Robert
parent
42b06eda82
commit
5a9ecc2d55
|
|
@ -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}
|
||||
|
|
|
|||
142
chapters/sections/metr_and_normed.tex
Normal file
142
chapters/sections/metr_and_normed.tex
Normal file
|
|
@ -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}
|
||||
188
chapters/sections/seq_ser_limits.tex
Normal file
188
chapters/sections/seq_ser_limits.tex
Normal file
|
|
@ -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}
|
||||
9
chapters/topo_of_metr_spaces.tex
Normal file
9
chapters/topo_of_metr_spaces.tex
Normal file
|
|
@ -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}
|
||||
0
script.loe
Normal file
0
script.loe
Normal file
BIN
script.pdf
BIN
script.pdf
Binary file not shown.
14
script.tex
14
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}
|
||||
|
|
|
|||
Loading…
Reference in a new issue