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*.aux
*.fdb*
*.fls
*.log
*.out
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*.toc

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# Mathematics
This is my attempt at digitalizing (and translating) my math notes from uni. It's not finished yet, I'll update it bit by bit when I feel like it
The topics covered in this script will be:
1. Fundamentals and Notation
1.1 Logic
1.2 Sets and Functions
1.3 Numbers
2. Analysis: Part 1
2.1 Elementary Inequalities
2.2 Sequences and Limits
2.3 Convergence of Series
3. Linear Algebra
3.1 Vector spaces
3.2 Matrices and Gaussian elimination
3.3 The Determinant
3.4 Scalar Product
3.5 Eigenvalue problems
4. Analysis: Part 2
4.1 Limits of Functions
4.2 Differential Calculus
5. Topology in Metric spaces
5.1 Metric and Normed spaces
5.2 Sequences, Series and Limits
5.3 Open and Closed Sets
5.4 ????
5.5 Continuiuty
5.6 Convergence of Function Sequences
6. Differential Calculus for Functions with multiple Variables
6.1 Partial and Total Differentiability
6.2 Higher Derivatives
6.3 Function Sequences and Differentiability
6.4 The Banach Fixed-Point Theorem and the Implicit Function Theorem
7. Measures and Integrals
7.1 Contents and Measures
7.2 Integrals
7.3 Integrals over the real numbers
7.4 ????
7.5 Product Measures and the Fubini Theorem
7.6 The Transformation Theorem
8. Ordinary Differential Equations
8.1 Solution Methods
8.2 The Picard-Lindelöf Theorem
8.3 Linear Differential Equation Systems
9. Integration over Submanifolds
9.1 Line Integrals
9.2 Surface Integrals
9.3 Ingegral Theorems
10. Elements of Complex Analysis
10.1 Complex Differentiability
10.2 Complex Line Integrals
10.3 Identity Theorems and Analytic Continuation
10.4 Laurent Series
10.5 Residue Theorem
10.6 Application: Potential Theory
11. Fourier Transform and Basics of Distribution Theory
11.1 Fourier Transform on L¹(ℝᵈ)
11.2 Fourier Transform on L²(ℝᵈ)
11.3 Tempered Distributions
12. Operator Theory
12.1 Linear Operators
12.2 Dual Spaces
12.3 Hilbert Spaces
12.4 Orthonormal Sets
12.5 Adjoint Operators
13. Spectral Theory
13.1 Spectral Theory of Bounded Linear Operators
13.2 Spectral Representation of Bounded Self-Adjoint Operators I
13.3 Spectral Representation of Bounded Self-Adjoint Operators II
13.4 Compact Linear Operators
13.5 Unbounded Linear Operators
13.6 Spectral Representation of Unbounded Self-Adjoint Operaotrs
14. Curves in ℝ³
15. Differentiable Manifolds

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\documentclass[../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\chapter{Fundamentals and Notation}
\subfile{sections/logic.tex}
\subfile{sections/sets_and_functions.tex}
\subfile{sections/numbers.tex}
\end{document}

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\documentclass[../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\chapter{Linear Algebra}
\subfile{sections/vector_spaces.tex}
\subfile{sections/matrices.tex}
\end{document}

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\documentclass[../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\chapter{Real Analysis: Part I}
\subfile{sections/elem_ineqs.tex}
\subfile{sections/seq_and_lims.tex}
\subfile{sections/conv_of_series.tex}
\end{document}

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\documentclass[../../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\section{Convergence of Series}
\begin{defi}
Let $\rcseqdef{x}$. Then the series
\[
\series{k} x_k
\]
is the sequence of partial sums $\seq{s}$:
\[
s_n = \series[n]{k} x_k
\]
If the series converges, then $\series{k}$ denotes the limit.
\end{defi}
\begin{thm}\label{thm:seriesnull}
Let $\rcseqdef{x}$. Then
\[
\series{n} x_n \text{ converges} \implies \seq{x} \text{ null sequence}
\]
\end{thm}
\begin{proof}
Let $s_n = \series{n} x_n$. This is a Cauchy series. Let $\epsilon > 0$. Then
\begin{equation}
\exists N \in \natn ~\forall n \ge N: ~~|s_{n+1} - s_n| = |x_{n+1}| < \epsilon
\end{equation}
\end{proof}
\begin{eg}[Geometric series]
Let $x \in \realn$ (or $\cmpln$). Then
\[
\series{k} x^k
\]
converges if $|x| < 1$. (Why?)
\end{eg}
\begin{eg}[Harmonic series]
This is a good example of why the inverse of \Cref{thm:seriesnull} does not hold. Consider
\[
x_n = \frac{1}{n}
\]
This is a null sequence, but $\series{k}\frac{1}{k}$ does not converge. (Why?)
\end{eg}
\begin{lem}
Let $\rcseqdef{x}$. Then
\[
\series{k} x_n \text{ converges} \iff \sum_{k=N}^{\infty} x_n \text{ converges for some } N \in \natn
\]
\end{lem}
\begin{proof}
\reader
\end{proof}
\begin{thm}[Alternating series test]\label{thm:alttest}
Let $\seq{x} \subset [0, \infty)$ be a monotonic decreasing null sequence. Then
\[
\series{k} (-1)^k x_k
\]
converges, and
\[
\left| \series{k} (-1)^k x_k - \series[N]{k} (-1)^k x_k \right| \le x_{N+1}
\]
\end{thm}
\begin{proof}
Let $s_n = \series[n]{k} (-1)^k x_n$, and define the sub sequences $a_n = s_{2n}$, $b_n = s_{2n+1}$. Then
\begin{equation}
a_{n+1} = s_{2n} - \underbrace{(x_{2n+1} - x_{2n+2})}_{\ge 0} \le s_{2n} = a_n
\end{equation}
Hence, $\seq{a}$ is monotonic decreasing. By the same argument, $\seq{b}$ is monotonic decreasing. Let $m, n \in \natn$ such that $m \le n$. Then
\begin{equation}\label{eq:act}
b_m \le b_n = a_n - x_{2n+1} \le a_n \le a_m
\end{equation}
Therefore $\seq{a}$, $\seq{b}$ are bounded. By \Cref{thm:monotone}, these sequence converge
\begin{align}
\seq{a} &\convinf a & \seq{b} &\convinf b
\end{align}
Furthermore
\begin{equation}
b_n - a_n = -x_{2n+1} \convinf 0 \implies a = b
\end{equation}
From \cref{eq:act} we know that
\begin{equation}
b_m \le b = a \le a_m
\end{equation}
So therefore
\begin{align}
|s_{2n} - a| &= a_n - a \le a_n - b_n = x_{2n+1} \\
|s_{2n+1} - a| &= b - b_n \le a_{m+1} - b_n = x_{2n+2}
\end{align}
\end{proof}
\begin{eg}[Alternating harmonic series]
\[
\begin{split}
s = \series{k} (-1)^{k+1} \rec{k} &= 1 - \rec{2} + \rec{3} - \rec{4} + \rec{5} - \cdots \\
&= \left(1 - \rec{2}\right) - \rec{4} + \left(\rec{3} - \rec{6}\right) - \rec{8} + \left(\rec{5} - \rec{10}\right) - \rec{12} + \cdots \\
&= \rec{2} - \rec{4} + \rec{6} - \rec{8} + \rec{10} - \rec{12} + \cdots \\
&= \rec{2}\left(1 - \rec{2} + \rec{3} - \rec{4} + \rec{5} - \rec{6} + \cdots\right) \\
&= \rec{2}s
\end{split}
\]
But $s \in \left[\rec{2}, 1\right]$, this is an example on why rearranging infinite sums can lead to weird results.
\end{eg}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item The convergence behaviour does not change if we rearrange finitely many terms.
\item Associativity holds without restrictions
\[
\series{k} x_k = \series{k} (x_{2k} + x_{2k-1})
\]
\item Let $I$ be a set, and define
\begin{align*}
I &\longrightarrow \realn \\
i &\longmapsto a_i
\end{align*}
Consider the sum
\[
\sum_{i \in I} a_i
\]
If $I$ is finite, there are no problems. However if $I$ is infinite then the solution of that sum can depend on the order of summation!
\end{enumerate}
\end{rem}
\begin{defi}
Let $\rcseqdef{x}$. The series $\series{k} x_k$ is said to converge absolutely if $\series{k} |x_k|$ converges.
\end{defi}
\begin{rem}
Let $\seq{x} \subset [0, \infty)$. Then the sequence
\[
s_n = \series[n]{k} x_k
\]
is monotonic increasing. If $\seq{s}$ is bounded it converges, if it is unbounded it diverges properly. The notation for absolute convergence is
\[
\series{k} |x_k| < \infty
\]
\end{rem}
\begin{lem}\label{lem:absolutebounded}
Let $\series{k} x_k$ be a series. Then the following are all equivalent
\begin{enumerate}[(i)]
\item
\[
\series{k} x_k \text{ converges absolutely}
\]
\item
\[
\set[I \subset \natn \text{ finite}]{\sum_{k \in I} |x_k|} \text{ is bounded}
\]
\item
\[
\forall \epsilon > 0 ~\exists I \subset \natn \text{ finite} ~\forall J \subset \natn \text{ finite}: ~~\sum_{k \in J \setminus I} |x_k| < \epsilon
\]
\end{enumerate}
\end{lem}
\begin{proof}
To prove the equivalence of all of these statements, we will show that (i) $\implies$ (ii) $\implies$ (iii) $\implies$ (i). This is sufficient. First we prove (i) $\implies$ (ii). Let
\begin{equation}
\series{n} |x_n| = k \in [0, \infty)
\end{equation}
Let $I \subset \natn$ be a finite set, and let $N = \max I$. Then
\begin{equation}
\sum_{n \in I} |x_n| \le \series[N]{n} |x_n| \leexpl{Monotony of the partial sums} \series{n} |x_n|
\end{equation}
Now to prove (ii) $\implies$ (iii), set
\begin{equation}
K := \set[I \subset \natn \finite]{\sum_{k \in I} |x_k|}
\end{equation}
Let $\epsilon > 0$. Then by definition of $\sup$
\begin{equation}
\exists I \subset \natn \finite: ~~\sum_{k \in I} |x_k| > k - \epsilon
\end{equation}
Let $J \subset \natn \finite$. Then
\begin{equation}
k - \epsilon < \sum_{k \in I} |x_k| \le \sum_{k \in I \cup J} |x_k| \le K
\end{equation}
Hence
\begin{equation}
\sum_{k \in J \setminus I} |x_k| = \sum_{k \in I \cup J} |x_k| - \sum_{k \in I} |x_k| \le \epsilon
\end{equation}
Finally we show that (iii) $\implies$ (i). Choose $I \subset \natn \finite$ such that
\begin{equation}
\forall J \subset \natn \finite: ~~\sum_{k \in J \setminus I} |x_k| < 1
\end{equation}
Then $\forall J \subset \natn \finite$
\begin{equation}
\sum_{k \in J} |x_k| \le \sum_{k \in J \setminus I} |x_k| + \sum_{k \in I} |x_k| \le \sum_{k \in I} |x_k| + 1
\end{equation}
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}
Every absolutely convergent series converges and the limit does not depend on the order of summation.
\end{thm}
\begin{proof}
Let $\series{k} x_k$ be absolutely convergent and let $\epsilon > 0$. Choose $I \subset \natn \finite$ such that
\begin{equation}
\forall J \subset \natn: ~~\sum_{k \in I} |x_k| < \epsilon
\end{equation}
Choose $N = \max I$. Define the series
\begin{equation}
s_n = \series[n]{k} x_k
\end{equation}
Then for $n \le m \le N$
\begin{equation}
|s_n - s_m| \le \sum_{k=m+1}^n |x_k| \le \sum_{k \in \set{1, \cdots, n} \setminus I} |x_k| < \epsilon
\end{equation}
Hence $s_n$ is a Cauchy sequence, so it converges. Let $\phi: \natn \rightarrow \natn$ be a bijective mapping. According to \Cref{lem:absolutebounded} the series $\series{k} x_{\phi(n)}$ converges absolutely. Let $\epsilon > 0$. According to the same Lemma
\begin{equation}
\exists I \subset \natn \finite ~\forall J \subset \natn \finite: ~~\sum_{k \in J \setminus I} |x_k| < \frac{\epsilon}{2}
\end{equation}
Choose $N \in \natn$ such that
\begin{equation}
I \subset \set{1, \cdots, N} \cap \set{\phi(1), \phi(2), \cdots, \phi(n)}
\end{equation}
Then for $n \ge N$
\begin{equation}
\begin{split}
\left|\series{k} x_k - \series[n]{k} x_{\phi(k)}\right| &= \left| \sum_{k \in \set{1, \cdots, N} \setminus I} x_k - \sum_{k \in \set{\phi(1), \cdots, \phi(n)} \setminus I} x_k \right| \\
&\le \sum_{k \in \set{1, \cdots, N} \setminus I} |x_k| + \sum_{k \in \set{\phi(1), \cdots, \phi(n)} \setminus I} |x_k| < \epsilon
\end{split}
\end{equation}
Therefore
\begin{equation}
\limn\left( \series[n]{k} x_k - \series[n]{k} x_{\phi(k)} \right) = 0
\end{equation}
\end{proof}
\begin{thm}
Let $\series{k} x_k$ be a converging series. Then
\[
\left| \series{k} x_k \right| \le \series{k} |x_k|
\]
\end{thm}
\begin{proof}
\reader
\end{proof}
\begin{thm}[Direct comparison test]
Let $\series{k} x_k$ be a series. If a converging series $\series{k} y_k$ exists with $|x_k| \le y_k$ for all sufficiently large $k$, then $\series{k} x_k$ converges absolutely. If a series $\series{k} z_k$ diverges with $0 \le z_k \le x_k$ for all sufficiently large $k$, then $\series{k} x_k$ diverges.
\end{thm}
\begin{proof}
\begin{equation}
\series[n]{k} |x_k| \le \series[n]{k} y_k \implies \series[n]{k} x_k \text{ bounded} \implbl{\cref{lem:absolutebounded}} \series{k} |x_k| < \infty
\end{equation}
\begin{equation}
\series[n]{k} z_k \le \series[n]{k} x_k \implies \series{k} x_k \text{ unbounded}
\end{equation}
\end{proof}
\begin{cor}[Ratio test]
Let $\seq{x}$ be a sequence. If $\exists q \in (0, 1)$ such that
\[
\left| \frac{x_{n+1}}{x_n} \right| \le q
\]
for a.e. $n \in \natn$, then $\series{k} x_k$ converges absolutely. If
\[
\left| \frac{x_{n+1}}{x_n} \right| \ge 1
\]
then the series diverges.
\end{cor}
\begin{proof}
Let $q \in (0, 1)$ and choose $N \in \natn$ such that
\begin{equation}
\forall n \ge N: ~~\left|\frac{x_{n+1}}{x_n}\right| \le q
\end{equation}
Then
\begin{equation}
|x_{N+1}| \le q|x_N|, ~|x_{N+2}| \le q|x_{N+1}| \le q^2|x_N|, ~\cdots
\end{equation}
This means that
\begin{equation}
\series{k} |x_k| \le \series[N]{k} |x_k| + \sum_{k=N+1}^\infty q^{k-N} \cdot |x_N| < \infty
\end{equation}
Hence, $\series{k} x_k$ converges absolutely. Now choose $N \in \natn$ such that
\begin{equation}
\forall n \ge N: ~~\left|\frac{x_{n+1}}{x_n}\right| > 1
\end{equation}
However this means that
\begin{equation}
|x_{n+1}| \ge |x_{n}| ~~\forall n \ge N
\end{equation}
So $\seq{x}$ is monotonic increasing and therefore not a null sequence. Hence $\series{k} x_k$ diverges.
\end{proof}
\begin{cor}[Root test]
Let $\seq{x}$ be a sequence. If $\exists q \in (0, 1)$ such that
\[
\sqrt[n]{|x_n|} \le q
\]
for a.e. $n \in \natn$, then $\series{k} x_k$ converges absolutely. If
\[
\sqrt[n]{|x_n|} \ge 1
\]
for all $n \in \natn$ then $\series{k} x_k$ diverges.
\end{cor}
\begin{proof}
\reader
\end{proof}
\begin{rem}
The previous tests can be summed up by the formulas
\begin{align*}
\limn \left|\frac{x_{n+1}}{x_n}\right| &< 1 & \limn \sqrt[n]{|x_n|} &< 1 \\
\limn \left|\frac{x_{n+1}}{x_n}\right| &> 1 & \limn \sqrt[n]{|x_n|} &> 1
\end{align*}
for convergence and divergence respectively. If any of these limits is equal to $1$ then the test is inconclusive.
\end{rem}
\begin{eg}
Let $z \in \cmpln$. Then
\[
\exp(z) := \sum_{k=0}^\infty \frac{z^k}{k!}
\]
converges. To prove this, apply the ratio test:
\[
\frac{|z|^{k+1} k!}{(k+1)! |z|^k} = \frac{|z|}{k+1} \conv{} 0
\]
The function $\exp: \cmpln \rightarrow \cmpln$ is called the exponential function.
\end{eg}
\begin{rem}[Binomial coefficient]
The binomial coefficient is defined as
\begin{align*}
{n\choose 0} &:= 1 & {n\choose k+1} &= {n\choose k} \cdot \frac{n-k}{k+1}
\end{align*}
and represents the number of ways one can choose $k$ objects from a set of $n$ objects. Some rules are
\begin{enumerate}[(i)]
\item \[{n\choose k} = 0 ~~\text{ if } k > n\]
\item \[k \le n: ~~{n\choose k} = \frac{n!}{k!(n-k)!}\]
\item \[{n\choose k} + {n\choose k-1} = {n+1\choose k}\]
\item \[\forall x, y \in \cmpln: ~~(x+y)^n = \series[n]{k} {n\choose k} x^ky^{n-k}\]
\end{enumerate}
\end{rem}
\begin{thm}
\[
\forall u, v \in \cmpln: ~~\exp(u+v) = \exp(u)\cdot\exp(v)
\]
\end{thm}
\begin{proof}
\begin{equation}
\begin{split}
\exp(u)\cdot\exp(v) = \left(\sum_{n=0}^{\infty} \frac{u^n}{n!} \right) \cdot \left(\sum_{m=0}^{\infty} \frac{v^m}{m!} \right) &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{u^nv^m}{n!m!} \\
&= \sum_{l=0}^{\infty} \sum_{k=0}^{l} \frac{u^kv^{l-k}}{k!(l-k)!} \\
&=\sum_{l=0}^{\infty} \frac{(u+v)^l}{l!} \\
&= \exp(u+v)
\end{split}
\end{equation}
\end{proof}
\begin{rem}
We define Euler's number as
\[
e := \exp(1)
\]
We will also take note of the following rules $\forall x \in \cmpln, n \in \natn$
\[
\exp(0) = \exp(x)\exp(-x) = 1 \implies \exp(-x) = \frac{1}{\exp(x)}
\]
\[
\exp(nx) = \exp(x + x + x + \cdots + x) = \exp(x)^n
\]
\[
\exp(x)^{\frac{1}{n}} = \exp(\frac{x}{n})
\]
Alternatively we can write
\[
\exp(z) = e^z
\]
\end{rem}
\begin{thm}
Let $x, y \in \realn$.
\begin{enumerate}[(i)]
\item
\[
x < y \implies \exp(x) < \exp(y)
\]
\item
\[
\exp(x) > 0 ~~\forall x \in \realn
\]
\item
\[
\exp(x) \ge 1 + x ~~\forall x \in \realn
\]
\item
\[
\limn \frac{n^d}{\exp(n)} = 0 ~~\forall d \in \natn
\]
\end{enumerate}
\end{thm}
\begin{proof}\leavevmode
\begin{enumerate}[(i)]
\item \reader
\item For $x \ge 0$ this is trivial. For $x < 0$
\begin{equation}
\exp(x) = \frac{1}{\exp(-x)} > 0
\end{equation}
\item For $x \ge 0$ this is trivial. For $x < 0$
\begin{equation}
\sum_{k=0}^{\infty} \frac{x^k}{k!}
\end{equation}
is an alternating series, and therefore the statement follows from \Cref{thm:alttest}.
\item Let $d \in \natn$. Then $\forall n \in \natn$
\begin{equation}
0 < \frac{n^d}{\exp(n)} < \frac{n^d}{\sum_{k=0}^{d+1} \frac{n^k}{k!}} \convinf 0
\end{equation}
\end{enumerate}
\end{proof}
\begin{defi}
Define
\[
\sin, \cos: \realn \longrightarrow \realn
\]
as
\begin{align*}
\sin(x) &:= \Im(\exp(ix)) \\
\cos(x) &:= \Re(\exp(ix))
\end{align*}
\end{defi}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item Euler's formula
\[
\exp(ix) = \cos(x) + i\sin(x)
\]
\item $\forall z \in \cmpln: ~~\overline{\exp(z)} = \exp(\bar{z})$
\[
|\exp(ix)|^2 = \exp(ix) \cdot \overline{\exp(ix)} = \exp(ix) \cdot \exp(-ix) = 1
\]
Also:
\[
1 = \cos^2(x) + \sin^2(x)
\]
On the symmetry of $\cos$ and $\sin$:
\[
\cos(-x) + i\sin(-x) = \exp(-ix) = \overline{\exp(ix)} = \cos(x) - i\sin(x)
\]
\item From
\[
\exp(ix) = \sum_{k=0}^{\infty} \frac{(ix)^k}{k!} ~~~(i^0 = 1, i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, \cdots)
\]
follow the following series
\begin{align*}
\sin(x) &= \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!} & \cos(x) &= \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}
\end{align*}
\item For $x \in \realn$
\[
\begin{split}
\exp(i2x) &= \cos(2x) + i\sin(2x) \\
&= (\cos(x) + i\sin(x))^2 \\
&= \cos^2(x) - \sin^2(x) + 2i\sin(x)\cos(x)
\end{split}
\]
By comparing the real and imaginary parts we get the following identities
\begin{align*}
\cos(2x) &= \cos^2(x) - \sin^2(x) \\
\sin(2x) &= 2\sin(x)\cos(x)
\end{align*}
\item Later we will show that $\cos$ as exactly one root in the interval $[0, 2]$. We define $\pi$ as the number in the interval $[0, 4]$ such that $\cos(\frac{\pi}{2}) = 0$.
\[
\implies \sin(\frac{\pi}{2}) = \pm 1
\]
$\cos$ and $\sin$ are $2\pi$-periodic.
\end{enumerate}
\end{rem}
\begin{thm}
$\forall z \in \cmpln$
\[
\limn \left(1 + \frac{z}{n}\right)^n = \limn \left(1 - \frac{z}{n}\right)^{-n} = \exp(z)
\]
\end{thm}
\begin{proof}
Without proof.
\end{proof}
\end{document}

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\documentclass[../../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\section{Elementary Inequalities}
\begin{eg}\leavevmode
\begin{itemize}
\item $x \in \realn \implies x^2 \ge 0$
\item $x^2 - 2xy + y^2 = (x - y)^2 \ge 0 ~~\forall x, y \in \realn$
\item $x^2 + y^2 \ge 2xy$
\end{itemize}
\end{eg}
\begin{thm}[Absolute inequalities]\label{thm:abs}
Let $x \in \realn$, $c \in [0, \infty)$. Then
\begin{enumerate}[(i)]
\item $-|x| \le x \le |x|$
\item $|x| \le c \iff -c \le x \le c$
\item $|x| \ge c \iff x \le -c \vee c \le x$
\item $|x| = 0 \iff x = 0$
\end{enumerate}
\end{thm}
\begin{thm}[Triangle inequality]\label{thm:triangle}
Let $x, y \in \realn$. Then
\[
|x + y| \le |x| + |y|
\]
\end{thm}
\begin{proof}
From \Cref{thm:abs} follows $x \le |x|$ and $y \le |y|$.
\begin{equation}
\implies x + y \le |x| + |y|
\end{equation}
However, from the same theorem follows $-|x| \le x$ and $-|y| \le y$.
\begin{align}
&\implies -|x|-|y| = x + y \\
&\implies |x + y| \le |x| + |y|
\end{align}
\end{proof}
\begin{cor}
$n \in \natn$, $x_1, \cdots, x_n \in \realn$. Then
\[
\left| \sum_{i=1}^n x_i \right| \le \sum_{i=1}^n |x_i|
\]
\end{cor}
\begin{proof}
Proof by induction. Let $n = 1$:
\begin{equation}
|x_1| \le |x_1|
\end{equation}
This statement is trivially true. Now assume the corollary holds for $n \in \natn$. Then
\begin{equation}
\begin{split}
\left| \sum_{i=1}^{n+1} x_i \right| = \left| \sum_{i=1}^n x_i + x_{n+1} \right| &\le \left| \sum_{i=1}^n x_n \right| + |x_{n+1}| \\
&\le \sum_{i=1}^n |x_i| + |x_{n+1}| \\
&= \sum_{i=1}^{n+1} |x_i|
\end{split}
\end{equation}
\end{proof}
\begin{thm}[Bernoulli inequality]\label{thm:bernoulli}
Let $x \in [-1, \infty)$ and $n \in \natn$. Then
\[
(1 + x)^n \ge 1 + nx
\]
\end{thm}
\begin{proof}
Proof by induction. Let $n = 1$:
\begin{equation}
1 + x \ge 1 + 1\cdot x
\end{equation}
This is trivial. Now assume the theorem holds for $n \in \natn$. Then
\begin{equation}
\begin{split}
(1 + x)^{n+1} = (1+x)^n (1+x) &\ge (1 + nx)(1 + x) \\
&= 1 + (n+1)x + nx^2 \\
&\ge 1 + (n+1)x
\end{split}
\end{equation}
\end{proof}
\end{document}

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\documentclass[../../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\section{Logic}
\begin{defi}[Statements]
A statement is a sentence (mathematically or colloquially) which can be either true or false.
\end{defi}
\begin{eg}
Statements are
\begin{itemize}
\item Tomorrow is Monday
\item $x > 1$ where $x$ is a natural number
\item Green rabbits grow at full moon
\end{itemize}
No statements are
\begin{itemize}
\item What is a statement?
\item $x + 20y$ where $x, y$ are natural numbers
\item This sentence is false
\end{itemize}
\end{eg}
\begin{defi}[Connectives]
When $\Phi, \Psi$ are statements, then
\begin{enumerate}[(i)]
\item $\neg\Phi$ (not $\Phi$)
\item $\Phi \wedge \Psi$ ($\Phi$ and $\Psi$)
\item $\Phi \vee \Psi$ ($\Phi$ or $\Psi$)
\item $\Phi \implies \Psi$ (if $\Phi$ then $\Psi$)
\item $\Phi \iff \Psi$ ($\Phi$ if and only if (iff.) $\Psi$)
\end{enumerate}
are also statements. We can represent connectives with truth tables
\begin{center}
\begin{tabular}{ c|c||c|c|c|c|c }
$\Phi$ & $\Psi$ & $\neg\Phi$ & $\Phi \wedge \Psi$ & $\Phi \vee \Psi$ & $\Phi \implies \Psi$ & $\Phi \iff \Psi$ \\
\hline
t & t & f & t & t & t & t\\
t & f & f & f & t & f & f\\
f & t & t & f & t & t & f\\
f & f & t & f & f & t & t\\
\end{tabular}
\end{center}
\end{defi}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item $\vee$ is inclusive
\item $\Phi \implies \Psi$, $\Phi \impliedby \Psi$, $\Phi \iff \Psi$ are NOT the same
\item $\Phi \implies \Psi$ is always true if $\Phi$ is false (ex falso quodlibet)
\end{enumerate}
\end{rem}
\begin{defi}[Hierarchy of logical operators]
$\neg$ is stronger than $\wedge$ and $\vee$, which are stronger than $\implies$ and $\iff$.
\end{defi}
\begin{eg}\leavevmode
\begin{align*}
\neg\Phi \wedge \Psi ~&\cong~ (\neg\Phi) \wedge \Psi \\
\neg\Phi \implies \Psi ~&\cong~ (\neg\Phi) \wedge \Psi \\
\Phi \wedge \Psi \iff \Psi ~&\cong~ (\Phi \wedge \Psi) \iff \Psi \\
\neg\Phi \vee \neg\Psi \implies \neg\Psi \wedge \Psi ~&\cong~ ((\neg\Phi) \vee (\neg\Psi)) \implies ((\neg\Psi) \wedge \Psi)
\end{align*}
We avoid writing statements like $\Phi \wedge \Psi \vee \Theta$. A statement that is always true is called a tautology. Some important equivalencies are
\begin{align*}
\Phi ~&\text{equiv.}~ \neg(\neg\Phi)) \\
\Phi \implies \Psi ~&\text{equiv.}~ \neg\Psi \implies \neg\Phi \\
\Phi \iff \Psi ~&\text{equiv.}~ (\Phi \implies \Psi) \wedge (\Psi \implies \Phi) \\
\Phi \vee \Psi ~&\text{equiv.}~ \neg(\neg\Phi \wedge \neg\Psi)
\end{align*}
Logical operators are commutative, associative and distributive.
\end{eg}
\begin{defi}[Quantifiers]
Let $\Phi(x)$ be a statement depending on $x$. Then $\forall x ~\Phi(x)$ and $\exists x ~\Phi(x)$ are also statements. The interpretation of these statements is
\begin{itemize}
\item $\forall x ~\Phi(x)$: "For all $x$, $\Phi(x)$ holds."
\item $\exists x ~\Phi(x)$: "There is (at least one) $x$ s.t. $\Phi(x)$ holds."
\end{itemize}
\end{defi}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item $\forall x ~x \ge 1$ is true for natural numbers, but not for integers. We must specify a domain.
\item If the domain is infinite the truth value of $\forall x ~\Phi(x)$ cannot be algorithmically determined.
\item $\forall x ~\Phi(x)$ and $\forall y ~\Phi(y)$ are equivalent.
\item Same operators can be exchanged, different ones cannot.
\item $\forall x ~\Phi(x)$ is equivalent to $\neg\exists x ~\neg\Phi(x)$.
\end{enumerate}
\end{rem}
\end{document}

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\documentclass[../../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\section{Matrices and Gaussian elimination}
\begin{defi}
Let $a_{ij} \in \field$, with $i \in \set{1, \cdots, n}$, $j \in \set{1, \cdots, m}$. Then
\[
\begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1m} \\
a_{21} & a_{22} & \cdots & a_{2m} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nm}
\end{pmatrix}
\]
is called an $n \times m$-matrix. $(n, m)$ is said to be the dimension of the matrix. An alternative notation is
\[
A = (a_{ij}) \in \field^{n \times m}
\]
$\field^{n\times m}$ is the space of all $n \times m$-matrices. The following operations are defined for $A, B \in \field^{n \times m}$, $C \in \field^{m \times l}$:
\begin{enumerate}[(i)]
\item Addition
\[
A + B =
\begin{pmatrix}
a_{11} + b_{11} & \cdots & a_{1m} + b_{1m} \\
\vdots & \ddots & \vdots \\
a_{n1} + b_{n1} & \cdots & a_{nm} + b_{nm}
\end{pmatrix}
\]
\item Scalar multiplication
\[
\alpha \cdot A =
\begin{pmatrix}
\alpha a_{11} & \cdots & \alpha a_{1m} \\
\vdots & \ddots & \vdots \\
\alpha a_{n1} & \cdots & \alpha a_{nm}
\end{pmatrix}
\]
\item Matrix multiplication
\[
A \cdot C =
\begin{pmatrix}
a_{11}c_{11}+a_{12}c_{21}+\cdots+a_{1m}c_{m1} & \cdots & a_{11}c_{1l}+a_{12}c_{2l}+\cdots+a_{1m}c_{ml} \\
\vdots & \ddots & \vdots \\
a_{n1}c_{11}+a_{n2}c_{21}+\cdots+a_{nm}c_{m1} & \cdots & a_{n1}c_{1l}+a_{n2}c_{2l}+\cdots+a_{nm}c_{ml}
\end{pmatrix}
\]
or in shorthand notation
\[
(AC)_{ij} = \series[m]{k} a_{ik}c_{kj}
\]
\item Transposition
The transposed matrix $A^T \in \field^{m \times n}$ is created by writing the rows of $A$ as the columns of $A^T$ (and vice versa).
\item Conjugate transposition
\[
\conj{A} = \left(\overline{A}\right)^T
\]
\end{enumerate}
\end{defi}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item $\field^{n \times m}$ (for $n, m \in \natn$) is a vector space.
\item $A \cdot B$ is only defined if $A$ has as many columns as $B$ has rows.
\item $\field^{n \times 1}$ and $\field^{1 \times n}$ can be trivially identified with $\field^n$.
\item Let $A, B, C, D, E$ matrices of fitting dimensions and $\alpha \in \field$. Then
\begin{align*}
(A + B) C &= AC + BC \\
A(B + C) &= AB + AC \\
A(CE) &= (AC)E \\
\alpha (AC) &= (\alpha A) C = A (\alpha C)
\end{align*}
\begin{align*}
(A + B)^T &= A^T + B^T & \conj{(A + B)} &= \conj{A} + \conj{B} \\
(\alpha A)^T &= \alpha (A)^T & \conj{(\alpha A)} &= \overline{A} \conj{A} \\
(AC)^T &= C^T \cdot A^T & \conj{(AC)} &= \conj{C} \conj{A}
\end{align*}
\begin{proof}[Proof of associativity]
Let $A \in \field^{n \times m}, C \in \field^{m \times l}, E \in \field^{l \times p}$. Furthermore let $i \in \set{1, \cdots, n}, j \in \set{1, \cdots, p}$.
\begin{equation}
\begin{split}
\left((AC)E\right)_{ij} &= \sum_{k=1}^l (AC)_{ik} E_{kj} = \sum_{k=1}^l \left(\sum_{\tilde{k} = 1}^m a_{i\tilde{k}} c_{\tilde{k}k}\right) \cdot e_{kj} \\
&= \sum_{k=1}^l \sum_{\tilde{k} = 1}^m a_{i\tilde{k}} \cdot c_{\tilde{k}k} \cdot e_{kj} \\
&= \sum_{\tilde{k} = 1}^m a_{i\tilde{k}} \left( \sum_{k=1}^l c_{\tilde{k} k} e_{kj}\right) \\
&= \sum_{\tilde{k} = 1}^m a_{i \tilde{k}} \cdot (CE)_{\tilde{k}j} \\
&= (A(CE))_{ij}
\end{split}
\end{equation}
\begin{equation}
\implies A(CE) = A(CE)
\end{equation}
\end{proof}
\end{enumerate}
\end{rem}
\end{document}

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\documentclass[../../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\section{Numbers}
\begin{defi}
The real numbers are a set $\realn$ with the following structure
\begin{enumerate}[(i)]
\item Addition
\begin{align*}
+: \realn \times \realn \longrightarrow \realn
\end{align*}
\item Multiplication
\begin{align*}
\cdot: \realn \times \realn \longrightarrow \realn
\end{align*}
Instead of $+(x, y)$ and $\cdot(x, y)$ we write $x+y$ and $x \cdot y$.
\item Order relations
$\le$ is a relation on $\realn$, i.e. $x \le y$ is a statement.
\end{enumerate}
\end{defi}
\begin{defi}[Axioms of Addition]\leavevmode
\begin{enumerate}[label=A\arabic*:]
\item Associativity
\[
\forall a, b, c \in \realn: ~~(a + b) + c = a + (b + c)
\]
\item Existence of a neutral element
\[
\exists 0 \in \realn ~\forall x \in \realn: ~~x + 0 = x
\]
\item Existence of an inverse element
\[
\forall x \in \realn ~\exists (-x) \in \realn: ~~ x + (-x) = 0
\]
\item Commutativity
\[
\forall x, y \in \realn: ~~x + y = y + x
\]
\end{enumerate}
\end{defi}
\begin{thm}\label{thm:addition}
$x, y \in \realn$
\begin{enumerate}[(i)]
\item The neutral element is unique
\item $\forall x \in \realn$ the inverse is unique
\item $-(-x) = x$
\item $-(x + y) = (-x) + (-y)$
\end{enumerate}
\end{thm}
\begin{proof}\leavevmode
\begin{enumerate}[(i)]
\item Assume $a, b \in \realn$ are both neutral elements, i.e.
\begin{equation}
\forall x \in \realn: x + a = x = x + b
\end{equation}
This also implies that $a + b = a$ and $b + a = b$.
\begin{equation}
\implies b = b + a \stackrel{\text{A4}}{=} a + b = a
\end{equation}
Therefore $a = b$.
\item Assume $c, d \in \realn$ are both inverse elements of $x \in \realn$, i.e.
\begin{equation}
x + c = 0 = x + d
\end{equation}
\begin{equation}
c = 0 + c = x + d + c \stackrel{\text{A4}}{=} x + c + d = 0 + d = d
\end{equation}
Therefore $c = d$.
\item \reader
\item
\begin{equation}
\begin{split}
x + y + ((-x) + (-y)) &= x + y + (-x) + (-y) \\
&\eqlbl{A4} x + (-x) + y + (-y) = 0
\end{split}
\end{equation}
Therefore $(-x) + (-y)$ is the inverse element of $(x+y)$, i.e. $-(x + y) = (-x) + (-y)$.
\end{enumerate}
\end{proof}
\begin{defi}[Axioms of Multiplication]\leavevmode
\begin{enumerate}[label=M\arabic*:]
\item $\forall x, y, z \in \realn: ~~(xy)z = x(yz)$
\item $\exists 1 \in \realn ~\forall x \in \realn: ~~x1 = x$
\item $\forall x \in \realn \setminus \{0\} ~\exists \inv{x} \in \realn: ~~x\inv{x} = 1$
\item $\forall x, y \in \realn: ~~xy = yx$
\end{enumerate}
\end{defi}
\begin{defi}[Compatibility of Addition and Multiplication]\leavevmode
\begin{enumerate}[label=R\arabic*:]
\item Distributivity
\[
\forall x, y, z \in \realn: ~~ x\cdot(y + z) = (x \cdot y) + (x \cdot z)
\]
\item $0 \ne 1$
\end{enumerate}
\end{defi}
\begin{thm}
$x, y \in \realn$
\begin{enumerate}[(i)]
\item $x \cdot 0 = 0$
\item $-(x \cdot y) = x \cdot (-y) = (-x) \cdot y$
\item $(-x) \cdot (-y) = x \cdot y$
\item $\inv{(-x)} = -(\inv{x}) ~~(\text{only for } x \ne 0)$
\item $xy = 0 \implies x = 0 \vee y = 0$
\end{enumerate}
\end{thm}
\begin{proof}\leavevmode
\begin{enumerate}[(i)]
\item $x \in \realn$
\begin{equation}
x \cdot 0 \eqlbl{A2} x \cdot (0 + 0) \eqlbl{R1} x \cdot 0 + x \cdot 0
\end{equation}
\begin{equation}
\stackrel{\text{A3}}{\implies} 0 = x \cdot 0
\end{equation}
\item $x,y \in \realn$
\begin{equation}
xy + (-(xy)) \eqlbl{A3} 0 \eqlbl{(i)} x \cdot 0 = x(y + (-y)) \eqlbl{R1} xy + x(-y)
\end{equation}
\begin{equation}
\stackrel{\text{A3}}{\implies} -(xy) = x\cdot(-y)
\end{equation}
\item \reader
\item $x \in \realn$
\begin{equation}
x \cdot (-\inv{(-x)}) \eqlbl{(ii)} -(x \cdot \inv{(-x)}) \eqlbl{(ii)} (-x) \cdot \inv{(-x)} \eqlbl{M3} 1 \eqlbl{M3} x \cdot \inv{x}
\end{equation}
\begin{equation}
\stackrel{\text{M3}}{\implies} -\inv{(-x)} = \inv{x} \stackrel{\ref{thm:addition} (iii)}{\implies} \inv{(-x)} = -(\inv{x})
\end{equation}
\item $x, y \in \realn$ and $y \ne 0$. Then $\exists \inv{y} \in \realn$:
\begin{equation}
xy = 0 \implies xy\inv{y} \eqlbl{M3} x \cdot 1 \eqlbl{M2} x = 0 = 0 \cdot \inv{y}
\end{equation}
\end{enumerate}
\end{proof}
\begin{rem}
A structure that fulfils all the previous axioms is called a field. We introduce the following notation for $x, y \in \realn, ~y \ne 0$
\[
\frac{x}{y} = x\inv{y}
\]
\end{rem}
\begin{defi}[Order relations]\leavevmode
\begin{enumerate}[label=O\arabic*:]
\item Reflexivity
\[
\forall x \in \realn: ~~x \le x
\]
\item Transitivity
\[
\forall x, y, z \in \realn: ~~x \le y \wedge y \le z \implies x \le z
\]
\item Anti-Symmetry
\[
\forall x, y \in \realn: ~~x \le y \wedge y \le x \implies x = y
\]
\item Totality
\[
\forall x, y \in \realn: ~~x \le y \vee y \le x
\]
\item
\[
\forall x, y, z \in \realn: ~~x \le y \implies x + z \le y + z
\]
\item
\[
\forall x, y \in \realn: ~~0 \le x \wedge 0 \le y \implies 0 \le x \cdot y
\]
\end{enumerate}
We write $x < y$ for $x \le y \wedge x \ne y$
\end{defi}
\begin{thm}
$x, y \in \realn$
\begin{enumerate}[(i)]
\item $x \le y \implies -y \le -x$
\item $x \le 0 \wedge y \le 0 \implies 0 \le xy$
\item $0 \le 1$
\item $0 \le x \implies 0 \le \inv{x}$
\item $0 < x \le y \implies \inv{y} \le \inv{x}$
\end{enumerate}
\end{thm}
\begin{proof}\leavevmode
\begin{enumerate}[(i)]
\item
\begin{equation}
\begin{split}
x \le y &\implbl{O5} x + (-x) + (-y) \le y + (-x) + (-y) \\
&\iff -y \le -x
\end{split}
\end{equation}
\item With $y \le 0 \implbl{(i)} 0 \le -y$ and $x \le 0 \implbl{(i)} 0 \le -x$ follows from O6:
\begin{equation}
0 \le (-x)(-y) = xy
\end{equation}
\item Assume $0 \le 1$ is not true. From O4 we know that
\begin{equation}
1 \le 0 \implbl{(ii)} 0 \le 1 \cdot 1 = 1
\end{equation}
\item \reader
\item
\begin{equation}
0 \le \inv{x} \wedge 0 \le \inv{y} \implbl{O6} 0 \le \inv{x}\inv{y}
\end{equation}
From $x \le y$ follows $0 \le y - x$
\begin{align}
&\implbl{O6} 0 \le (y - x)\inv{x}\inv{y} \eqlbl{R1} y\inv{x}\inv{y} - x\inv{x}\inv{y} = \inv{x} - \inv{y} \\
&\implbl{O5} \inv{y} \le \inv{x}
\end{align}
\end{enumerate}
\end{proof}
\begin{rem}
A structure that fulfils all the previous axioms is called an ordered field.
\end{rem}
\begin{defi}
Let $A \subset \mathbb{R}$, $x \in \realn$.
\begin{enumerate}[(i)]
\item $x$ is called an upper bound of $A$ if $\forall y \in A: ~y \le x$
\item $x$ is called a maximum of $A$ if $x$ is an upper bound of $A$ and $x \in A$
\item $x$ is called supremum of $A$ is $x$ is an upper bound of $A$ and if for every other upper bound $y \in \realn$ the statement $x \le y$ holds. In other words, $x$ is the smallest upper bound of $A$.
\end{enumerate}
$A$ is called bounded above if it has an upper bound. Analogously, there exists a lower bound, a minimum and an infimum. We introduce the notation $\sup A$ for the supremum and $\inf A$ for the infimum.
\end{defi}
\begin{defi}
$a, b \in \realn$, $a < b$. We define
\begin{itemize}
\item $(a, b) := \{x \in \realn \setvert a < x \wedge x < b\}$
\item $[a, b] := \{x \in \realn \setvert a \le x \wedge x \le b\}$
\item $(a, \infty) := \{x \in \realn \setvert a < x\}$
\end{itemize}
\end{defi}
\begin{eg}
$(-\infty, 1)$ is bounded above ($1$, $2$, $1000$, $\cdots$ are upper bounds), but has no maximum. $1$ is the supremum.
\end{eg}
\begin{defi}[Completeness of the real numbers]
Every non-empty subset of $\realn$ with an upper bound has a supremum.
\end{defi}
\begin{defi}
A set $A \subset \realn$ is called inductive if $1 \in A$ and
\[
x \in A \implies x + 1 \in A
\]
\end{defi}
\begin{lem}
Let $I$ be an index set, and let $A_i$ be inductive sets for every $i \in I$. Then $\bigcap_{i \in I} A_i$ is also inductive.
\end{lem}
\begin{proof}
Since $A_i$ is inductive $\forall i \in I$, we know that $1 \in A_i$. Therefore
\begin{equation}
1 \in \bigcap_{i \in I} A_i
\end{equation}
Now let $x \in \bigcap_{i \in I} A_i$, this means that $x \in A_i ~~\forall i \in I$.
\begin{equation}
\implies x + 1 \in A_i ~~\forall i \in I \implies x + 1 \in \bigcap_{i \in I} A_i
\end{equation}
\end{proof}
\begin{defi}
The natural numbers are the smallest inductive subset of $\realn$. I.e.
\[
\bigcap_{A \text{ inductive}} A =: \natn
\]
\end{defi}
\begin{thm}[The principle of induction]
Let $\Phi(x)$ be a statement with a free variable $x$. If $\Phi(1)$ is true, and if $\Phi(x) \implies \Phi(x + 1)$, then $\Phi(x)$ holds for all $x \in \natn$.
\end{thm}
\begin{proof}
Define $A = \{x \in \realn \setvert \Phi(x)\}$. According to the assumptions, $A$ is inductive and therefore $\natn \subset A$. This means that $\forall n \in \natn: ~~\Phi(n)$.
\end{proof}
\begin{cor}
$m, n \in \natn$
\begin{enumerate}[(i)]
\item $m + n \in \natn$
\item $mn \in \natn$
\item $1 \le n ~~\forall n \in \natn$
\end{enumerate}
\end{cor}
\begin{proof}
We will only proof (i). (ii) and (iii) are left as an exercise for the reader. Let $n \in \natn$. Define $A = \{m \in \natn \setvert m + n \in \natn\}$. Then $1 \in A$, since $\natn$ is inductive. Now let $m \in A$, therefore $n + m \in \natn$.
\begin{align}
&\implies n + m + 1 \in \natn \\
&\iff m + 1 \in A
\end{align}
Hence $A$ is inductive, so $\natn \subset A$. From $A \subset N$ follows that $\natn = A$.
\end{proof}
\begin{thm}
$n \in \natn$. There are no natural numbers between $n$ and $n + 1$.
\end{thm}
\begin{hproof}
Show that $x \in \natn \cap (1, 2)$ implies that $\natn \setminus \{x\}$ is inductive. Now show that if $\natn \cap (n, n+1) = \varnothing$ and $x \in \natn \cap (n + 1, n + 2)$ then $\natn \setminus \{x\}$ is inductive.
\end{hproof}
\begin{thm}[Archimedian property]
\[
\forall x \in \realn ~\exists n \in \natn: ~~x<n
\]
\end{thm}
\begin{proof}
If $x < 1$ there is nothing to prove, so let $x \ge 1$. Define the set
\begin{equation}
A = \{n \in \natn \setvert n \le x\}
\end{equation}
$A$ is bounded above by definition. There exists the supremum $s = \sup A$. By definition, $s-1$ is not an upper bound of $A$, i.e. $\exists m \in A: ~~s-1 < m$. Therefore $s \le m + 1$.
\begin{equation}
m \in A \subset \natn \implies m + 1 \in \natn
\end{equation}
Since $s$ is an upper bound of $A$, this implies that $m+1 \not\subset A$, so therefore $m + 1 > x$.
\end{proof}
\begin{cor}\label{cor:minimum}
Every non-empty subset of $\natn$ has a minimum, and every non-empty subset of $\natn$ that is bounded above has a maximum.
\end{cor}
\begin{proof}
Let $A \subset \natn$. Propose that $A$ has no minimum. Define the set
\begin{equation}
\tilde{A} := \{n \in \natn \setvert \forall m \in A: ~n < m\}
\end{equation}
$1$ is a lower bound of $A$, but according to the proposition $A$ has no minimum, so therefore $1 \notin A$. This implies that $1 \in \tilde{A}$.
\begin{equation}
n \in \tilde{A} \implies n < m ~\forall m \in A
\end{equation}
But since there exists no natural number between $n$ and $n+1$, this means that $n+1$ is also a lower bound of $A$, and therefore
\begin{equation}
n+1 \le m ~\forall m \in A \implies n+1 \in \tilde{A}
\end{equation}
So $\tilde{A}$ is an inductive set, hence $\tilde{A} = \natn$. Therefore $A = \varnothing$.
\end{proof}
\begin{defi}
We define the following new sets:
\begin{align*}
&\intn := \{x \in \realn \setvert x \in \natn_0 \vee (-x) \in \natn_0\}\\
&\ratn := \left\{\frac{p}{q} \setvert p, q \in \intn \wedge q \ne 0\right\}
\end{align*}
$\intn$ are called integers, and $\ratn$ are called the rational numbers. $\natn_0$ are the natural numbers with the $0$ ($\natn_0 = \natn \cap \{0\}$).
\end{defi}
\begin{rem}
\begin{align*}
x, y \in \intn &\implies x+y, x\cdot y, (-x) \in \intn \\
x, y \in \ratn &\implies x+y, x\cdot y, (-x) \in \ratn \text{ and } \inv{x} \in \ratn \text{ if } x \ne 0
\end{align*}
The second statement implies that $\ratn$ is a field.
\end{rem}
\begin{cor}[Density of the rationals]
$x, y \in \realn, ~x < y$. Then
\[
\exists r \in \ratn: ~~x < r < y
\]
\end{cor}
\begin{proof}
This proof relies on the Archimedian property.
\begin{equation}
\exists q \in \natn: ~~ \frac{1}{y-x} < q \left( \iff \frac{1}{q} < y - x \right)
\end{equation}
Let $p \in \intn$ be the greatest integer that is smaller than $y \cdot q$. The existence of $p$ is ensured by corollary \Cref{cor:minimum}. Then $\frac{p}{q} < y$ and
\begin{equation}
p + 1 \ge y \cdot q \implies y \le \frac{p}{q} + \frac{1}{q} < \frac{p}{q} + (y - x)
\end{equation}
\begin{equation}
\implies x < \frac{p}{q} < y
\end{equation}
\end{proof}
\begin{defi}[Absolute values]
We define the following function
\begin{align*}
|\cdot|: \realn &\longrightarrow [0, \infty) \\
x &\longmapsto
\begin{cases}
x &, x \ge 0 \\
-x &, x < 0
\end{cases}
\end{align*}
\end{defi}
\begin{thm}
\[
x, y \in \realn \implies |xy| = |x||y|
\]
\end{thm}
\begin{proof}
\reader
\end{proof}
\begin{defi}[Complex numbers]
Complex numbers are defined as the set $\cmpln = \realn^2$. Addition and multiplication are defined as mappings $\cmpln \times \cmpln \rightarrow \cmpln$. Let $(x, y), (\tilde{x}, \tilde{y}) \in \cmpln$.
\begin{align*}
(x, y) + (\tilde{x}, \tilde{y}) &:= (x + \tilde{x}, y + \tilde{y}) \\
(x, y) \cdot (\tilde{x}, \tilde{y}) &:= (x\tilde{x} - y\tilde{y}, x\tilde{y} + \tilde{x}y)
\end{align*}
$\cmpln$ is a field. Let $z = (x, y) \in \cmpln$. We define
\begin{align*}
\real(z) = \Re(z) = x& ~~\text{ the real part} \\
\imag(z) = \Im(z) = y& ~~\text{ the imaginary part}
\end{align*}
\end{defi}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item We will not prove that $\cmpln$ fulfils the field axioms here, this can be left as an exercise to the reader. However, we will note the following statements
\begin{itemize}
\item Additive neutral element: $(0, 0)$
\item Additive inverse of $(x, y)$: $(-x, -y)$
\item Multiplicative neutral element: $(1, 0)$
\item Multiplicative inverse of $(x, y) \ne (0, 0)$: $\left( \frac{x}{x^2 + y^2}, -\frac{y}{x^2 + y^2} \right)$
\end{itemize}
\item Numbers with $y = 0$ are called real.
\item The imaginary unit is defined as $i = (0, 1)$
\[
(0, 1) \cdot (x, y) = (-y, x)
\]
Especially
\[
i^2 = (0, 1)^2 = (-1, 0) = -(1, 0) = -1
\]
\end{enumerate}
We also introduce the following notation
\[
(x, y) = (x, 0) + i\cdot(y, 0) = x + iy
\]
\end{rem}
\begin{thm}[Fundamental theorem of algebra]
Every non-constant, complex polynomial has a complex root. I.e. for $n \in \natn$, $\alpha_0, \cdots, \alpha_n \in \cmpln$, $\alpha_n \ne 0$ there is some $x \in \cmpln$ such that
\[
\sum_{i = 0}^n \alpha_i x^i = \alpha_0 + \alpha_1 x + \alpha_2 x^2 + \cdots + \alpha_n x^n = 0
\]
\end{thm}
\begin{proof}
Not here.
\end{proof}
\end{document}

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\documentclass[../../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\section{Sequences and Limits}
\begin{defi}
Let $M$ be a set (usually $M$ is $\realn$ or $\cmpln$). A sequence in $M$ is a mapping from $\natn$ to $M$. The notation is $(x_n)_{n \in \natn} \subset M$ or $(x_n) \subset M$. $x_n$ is called element of the sequence at $n$.
\end{defi}
\newpage
\begin{eg}
Some real sequences are
\begin{itemize}
\item $x_n = \frac{1}{n} ~~~\left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \cdots\right)$
\item $x_n = \sum_{k=1}^n k ~~~\left(1, 3, 6, 10, 15, \cdots\right)$
\item $x_n =$ "smallest prime factor of $n$" $~~~(*, 2, 3, 2, 5, 2, 7, 2, 3, 2, \cdots)$
\end{itemize}
\end{eg}
\begin{defi}[Convergence]
Let $(x_n) \subset \realn$ be a sequence, and $x \in \realn$. Then
\[
(x_n) \text{ converges to } x \iff \forall \epsilon > 0 ~\exists N \in \natn: ~~|x_n - x| < \epsilon ~~\forall n \ge N
\]
A complex sequence $(z_n) \subset \cmpln$ converges to $z \in \cmpln$ if the real and imaginary parts of $(z_n)$ converge to the real and imaginary parts of $z$. $x$ (or $z$) is called the limit of the sequence. Common notation:
\noindent\begin{minipage}{0.3\textwidth}
\[
x_n \longrightarrow x
\]
\end{minipage}
\begin{minipage}{0.3\textwidth}
\[
x_n \convinf x
\]
\end{minipage}
\begin{minipage}{0.3\textwidth}
\[
\lim_{n \rightarrow \infty} x_n = x
\]
\end{minipage}
\noindent If a sequence converges to $0$ it is called a null sequence.
\end{defi}
\begin{eg}\leavevmode
\begin{enumerate}[(i)]
\item $x \in \realn$, $x_n = x$ (constant sequence). This sequence converges to $x$. To show this, let $\epsilon > 0$. Then for $N = 1$:
\[
|x_n - x| = |x - x| = 0 < \epsilon
\]
\item $x_n = \frac{1}{n}$ is a null sequence. Let $\epsilon > 0$. By the Archimedean property:
\[
\exists N \in \natn: ~~\frac{1}{\epsilon} < N
\]
Then for $n \ge N$:
\[
|x_n - 0| = |x_n| = \frac{1}{n} \le \frac{1}{N} < \epsilon
\]
\item The sequence
\[
x_n =
\begin{cases}
1 &, n \text{ even} \\
-1 &, n \text{ odd}
\end{cases}
\]
does not converge.
\end{enumerate}
\end{eg}
\begin{rem}
A property holds for almost every (a.e.) $n \in \natn$ if it doesn't hold for only finitely many $n$. (e.g. $n < 10$ is true for a.e. $n \in \natn$)
\end{rem}
\begin{thm}
A sequence $(x_n) \subset \realn$ (or $\cmpln$) has at most one limit.
\end{thm}
\begin{proof}
Propose that $x, \tilde{x}$ are different limits of $(x_n)$. Without loss of generality (w.l.o.g.) we can write $x < \tilde{x}$. Now define $\epsilon = \frac{1}{2}(\tilde{x} - x) > 0$.
\begin{align}
x_n \longrightarrow x &\iff \exists N_1: ~~x_n \in (x - \epsilon, x + \epsilon) = \left(x - \epsilon, \frac{x + \tilde{x}}{2}\right) \\
x_n \longrightarrow \tilde{x} &\iff \exists N_2: ~~x_n \in (\tilde{x} - \epsilon, \tilde{x} + \epsilon) = \left(\frac{x + \tilde{x}}{2}, x + \epsilon\right)
\end{align}
Since these intervals are disjoint, the proposition led to a contradiction.
\end{proof}
\begin{thm}
Let $(x_n) \subset \realn$ (or $\cmpln$) be sequence with limit $x \in \realn$. Then for $m \in \natn$
\[
\lim_{n \rightarrow \infty} x_{n+m} = x
\]
\end{thm}
\begin{proof}
\reader
\end{proof}
\begin{defi}
The sequence $(x_n) \subset \realn$ is bounded above if $\{x_n \setvert n \in \natn \}$ is bounded above. A number $K \in \realn$ is an upper bound if $\forall n \in \natn: ~x_n \le K$.
\end{defi}
\begin{thm}
Every convergent sequence is bounded.
\end{thm}
\begin{proof}
Let $(x_n) \subset \realn$ converge to $x \in \realn$. For $\epsilon = 1$ we trivially know that
\begin{equation}
\exists N \in \natn ~\forall n \ge N: ~~|x_n - x| < \epsilon = 1
\end{equation}
Let
\begin{equation}
K = \max \{x_1, x_2, \cdots, x_N, |x| + 1\}
\end{equation}
Then
\begin{equation}
|x_n| \le K ~~\forall n \in \natn
\end{equation}
This is trivial for $n \le N$. For $n > N$ we can use the triangle inequality:
\begin{equation}
|x_n| = |(x_n - x) + x| \le |x_n - x| + |x| \le |x| + 1
\end{equation}
\end{proof}
\begin{thm}\label{215}
If $(x_n) \subset \realn$ bounded and $(y_n) \subset \realn$ null sequence, then $(x_n) \cdot (y_n)$ is also a null sequence.
\end{thm}
\begin{proof}
If $(x_n)$ is bounded, this means that $\exists K \in (0, \infty)$ such that
\begin{equation}
|x_n| \le K ~~\forall n \in \natn
\end{equation}
Since $(y_n)$ is a null sequence we know that
\begin{equation}
\forall \epsilon > 0 ~\exists N \in \natn ~\forall n \ge N: ~~|y_n| < \epsilon
\end{equation}
Now let $\epsilon > 0$, then $\exists N \in \natn$ such that
\begin{equation}
\forall n \ge N: ~~|y_n| < \frac{\epsilon}{K}
\end{equation}
\begin{equation}
|x_n \cdot y_n| = |x_n||y_n| \le K \frac{\epsilon}{K} = \epsilon
\end{equation}
Therefore $(x_n)(y_n)$ is a null sequence.
\end{proof}
\begin{thm}[Squeeze theorem]
Let $(x_n), (y_n), (z_n) \subset \realn$ be sequences such that
\[
x_n \le y_n \le z_n
\]
for a.e. $n \in \natn$, and let $x_n \rightarrow x$, $z_n \rightarrow x$. Then
\[
\lim_{n \rightarrow \infty} y_n = x
\]
\end{thm}
\begin{proof}
Let $\epsilon > 0$. Then $\exists N_1, N_2, N_3 \in \natn$ such that
\begin{align}
&\forall n \ge N_1: ~~x_n \le y_n \le z_n \\
&\forall n \ge N_2: ~~|x_n - x| < \epsilon \\
&\forall n \ge N_3: ~~|z_n - x| < \epsilon
\end{align}
Choose $N = \max \{N_1, N_2, N_3\}$. Then
\begin{equation}
\forall n \ge N: ~~-\epsilon < x_n - x \le y_n - x \le z_n - x < \epsilon
\end{equation}
Therefore $|y_n - x| < \epsilon$
\end{proof}
\begin{eg}
$\forall n \in \natn: ~~n \le n^2$ (why?).
\[
\implies 0 \le \frac{1}{n^2} \le \frac{1}{n} \implies \lim_{n \rightarrow \infty} \frac{1}{n^2} = 0
\]
\end{eg}
\begin{thm}
Let $(x_n), (y_n) \subset \realn$ and $x_n \rightarrow x$, $y_n \rightarrow y$. Then $x \le y$.
\end{thm}
\begin{proof}
\reader
\end{proof}
\begin{rem}
If $x_n < y_n ~~\forall n \in \natn$, then $x=y$ can still be true.
\end{rem}
\begin{lem}\label{220}
Let $(x_n) \in \realn$ and $x \in \realn$.
\[
(x_n) \longrightarrow x \iff (|x_n - x|) \text{ is null sequence}
\]
Especially:
\[
(x_n) \text{ null sequence} \iff |x_n| \text{ null sequence}
\]
\end{lem}
\begin{proof}
\begin{equation}
||x_n - x| - 0| = |x_n - x|
\end{equation}
\end{proof}
\begin{thm}\label{thm:lims}
Let $(x_n), (x_n) \subset \realn$ (or $\cmpln$) with $x_n \rightarrow x$, $y_n \rightarrow y$ ($x, y \in \realn$). Then all of the following are true:
\begin{enumerate}[(i)]
\item
\[
\limn x_n + y_n = x + y = \limn x_n + \limn y_n
\]
\item
\[
\limn x_n y_n = xy = \limn x_n \cdot \limn y_n
\]
\item If $y \ne 0$:
\[
\limn \frac{x_n}{y_n} = \frac{x}{y} = \frac{\limn x_n}{\limn y_n}
\]
\end{enumerate}
\end{thm}
\begin{proof}\leavevmode
\begin{enumerate}[(i)]
\item Let $\epsilon > 0$. Then $\exists N_1, N_2 \in \natn$ such that
\begin{align}
&\forall n \ge N_1: ~~|x_n - x| < \frac{\epsilon}{2} \\
&\forall n \ge N_2: ~~|y_n - y| < \frac{\epsilon}{2}
\end{align}
Now choose $N = \max \{N_1, N_2\}$. Then $\forall n \ge N$:
\begin{equation}
\begin{split}
|x_n + y_n - (x+y)| &= |(x_n - x) + (y_n - y)| \\
&\le |x_n - x| + |y_n - y| \\
&< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{split}
\end{equation}
\begin{equation}
\implies x_n + y_n \longrightarrow x + y
\end{equation}
\item
\begin{equation}
\begin{split}
0 \le |x_ny_n - xy| &= |(x_ny_n - x_ny) + (x_ny - xy)| \\
&\le |x_n(y_n - y)| + |(x_n - x)y| \\
&= |x_n||y_n - y| + |x_n - x||y| \longrightarrow 0
\end{split}
\end{equation}
Therefore $|x_ny_n - xy|$ is a null sequence and
\begin{equation}
x_ny_n \longrightarrow xy
\end{equation}
\item Now we need to show that if $y \ne 0$ then $\frac{1}{y_n} \rightarrow \frac{1}{y}$. We know that $|y| > 0$. So $\exists N \in \natn$ such that
\begin{equation}
\forall n \ge N: ~~|y_n - y| < \frac{|y|}{2}
\end{equation}
This implies that
\begin{equation}
\forall n \ge N: ~~0 < \frac{|y|}{2} \le |y_n|
\end{equation}
From this we now know that $\frac{1}{y_n}$ is defined and bounded
\begin{equation}
\left|\frac{1}{y_n}\right| = \frac{1}{|y_n|} \le \frac{2}{|y|}
\end{equation}
So finally
\begin{equation}
\begin{split}
\left| \frac{1}{y_n} - \frac{1}{y} \right| = \left| \frac{1}{y_n} \left(1 - y_n \frac{1}{y}\right) \right| = \left| \frac{1}{y_n} \right| \left| 1 - y_n \frac{1}{y} \right| \longrightarrow 0
\end{split}
\end{equation}
And therefore
\begin{equation}
\begin{split}
y_n \longrightarrow y \implies &\frac{y_n}{y} \longrightarrow 1 \\
\implbl{\cref{215}} &\left|1 - \frac{y_n}{y}\right| \text{ is a null sequence} \\
\implbl{\cref{220}} &\frac{1}{y_n} \longrightarrow \frac{1}{y}
\end{split}
\end{equation}
\end{enumerate}
\end{proof}
\begin{cor}\label{cor:polynomial}
Let $k \in \natn$, $a_0, \cdots, a_k, b_0, \cdots, b_k \in \realn$ and $b_k \ne 0$. Then
\[
\limn \frac{a_0 + a_1n + a_2n^2 + \cdots + a_{k-1}n^{k-1} + a_kn^k}{b_0 + b_1n + b_2n^2 + \cdots + b_{k-1}n^{k-1} + b_kn^k} = \frac{a_k}{b_k}
\]
\end{cor}
\begin{proof}
Multiply the numerator and the denominator with $\frac{1}{n^k}$
\begin{equation}
\frac{\frac{a_0}{n^k} + \frac{a_1}{n^{k-1}} + \frac{a_2}{n^{k-2}} + \cdots + \frac{a_{k-1}}{n} + a_k}{\frac{b_0}{n^k} + \frac{b_1}{n^{k-1}} + \frac{b_2}{n^{k-2}} + \cdots + \frac{b_{k-1}}{n} + b_k} \stackrel[n \rightarrow \infty]{}{\longrightarrow} 0
\end{equation}
\end{proof}
\begin{eg}
Let $x \in (-1, 1)$. Then $\limn x^n = 0$
\end{eg}
\begin{proof}
For $x = 0$ this is trivial. For $x \ne 0$ it follows that $|x| \in (0, 1)$ and $\frac{1}{|x|} \in (1, \infty)$. Choose $s = \frac{1}{|x|} - 1 > 0$ and apply the Bernoulli inequality (Theorem \autoref{thm:bernoulli}).
\begin{equation}
(1 + s)^n \ge 1 + n \cdot s
\end{equation}
\begin{equation}
0 \le |x|^n = \left(\frac{1}{1+s}\right)^n = \frac{1}{(1+s)^n} \le \frac{1}{1 + n\cdot s} = \frac{1 + n \cdot 0}{1 + n \cdot s} \stackrel{\autoref{cor:polynomial}}{\longrightarrow} 0
\end{equation}
The squeeze theorem now tells us that $|x^n| = |x|^n \rightarrow 0$ and therefore $x^n \rightarrow 0$.
\end{proof}
\begin{defi}
A sequence $(x_n) \subset \realn$ is called monotonic increasing (decreasing) if $x_{n+1} \ge x_n$ ($x_{n+1} \le x_n$) $\forall n \in \natn$.
\end{defi}
\begin{thm}[Monotone convergence theorem]\label{thm:monotone}
Let $\rseqdef{x}$ be a monotonic increasing (or decreasing) sequence that is bounded above (or below). Then $\seq{x}$ converges.
\end{thm}
\begin{proof}
Let $\seq{x}$ be monotonic increasing and bounded above. Define
\begin{equation}
x = \sup \underbrace{\{x_n \setvert n \in \natn \}}_A
\end{equation}
Now let $\epsilon > 0$, then $x - \epsilon$ is not an upper bound of $A$, this means $\exists N \in \natn$ such that $x_N > x - \epsilon$. The monotony of $\seq{x}$ implies that
\begin{equation}
\forall n \ge N: ~~x_n > x - \epsilon
\end{equation}
So therefore
\begin{equation}
x - \epsilon < x_n < x + \epsilon \implies |x_n - x| < \epsilon
\end{equation}
\end{proof}
\begin{rem}
\begin{align*}
\seq{x} \text{ is monotonic increasing} &\iff \frac{x_{n+1}}{x_n} \ge 1 ~~\forall n \in \natn \\
\seq{x} \text{ is monotonic decreasing} &\iff \frac{x_{n+1}}{x_n} \le 1 ~~\forall n \in \natn
\end{align*}
\end{rem}
\begin{eg}
Consider the following sequence
\begin{align*}
x_1 &= 1 \\
x_{n+1} &= \frac{1}{2}\left(x_n + \frac{a}{x_n}\right), ~~a \in [0, \infty)
\end{align*}
Notice that $0 < x_n ~~\forall n \in \natn$. For $n \in \natn$ one can show that
\[
\begin{split}
x_{n+1}^2 = \frac{1}{4} \left(x_n^2 + 2a + \frac{a^2}{x_n^2} \right) &= \frac{1}{4} \left(x_n^2 - 2a + \frac{a^2}{x_n^2} \right) + a \\
&= \frac{1}{4} \left(x_n - \frac{a}{x_n} \right)^2 + a \ge a
\end{split}
\]
So $x_n^2 \ge a ~~\forall n \ge 2$, and therefore $\frac{a}{x_n} \le x_n$. Finally
\[
x_{n+1} = \frac{1}{2}\left(x_n + \frac{a}{x_n}\right) \le \frac{1}{2}\left(x_n + x_n\right) = x_n ~~\forall n \ge 2
\]
This proves that $\seq{x}$ is monotonic decreasing and bounded below.
\end{eg}
\begin{thm}[Square root]
This theorem doubles as the definition of the square root. Let $a \in [0, \infty)$. Then $\exists! x \in [0, \infty)$ such that $x^2 = a$. Such an $x$ is called the square root of $a$, and is notated as $x = \sqrt{a}$.
\end{thm}
\begin{proof}
First we want to prove the uniqueness of such an $x$. Assume that $x^2 = y^2 = a$ with $x, y \in [0, \infty)$. Then $0 = x^2 - y^2 = (x-y)(x+y)$.
\begin{align}
&\implies x + y = 0 \implies x = y = 0 \\
&\implies x - y = 0 \implies x = y
\end{align}
Now to prove the existence, review the previous example.
\begin{equation}
x_n \longrightarrow x \text{ for some } x \in [0, \infty)
\end{equation}
By using the recursive definition we can write
\begin{equation}
2x_n \cdot x_{n+1} = x_n^2 + a \longrightarrow x^2 + a
\end{equation}
\begin{equation}
\implies 2x^2 = x^2 + a \implies x^2 = a
\end{equation}
\end{proof}
\begin{rem}
Analogously $\exists! x \in [0, \infty) ~\forall a \in [0, \infty)$ such that $x^n = a$. (Notation: $\sqrt[n]{a}$ or $x = a^{\frac{1}{n}}$). We will also introduce the power rules for rational exponents. Let $x, y \in \realn$, $u, v \in \ratn$.
\noindent\begin{minipage}[t]{.33\linewidth}
\[
(x \cdot y)^u = x^u y^u
\]
\end{minipage}
\begin{minipage}[t]{.33\linewidth}
\[
x^u \cdot x^v = x^{u+v}
\]
\end{minipage}
\begin{minipage}[t]{.33\linewidth}
\[
(x^u)^v = x^{u \cdot v}
\]
\end{minipage}
\end{rem}
\begin{thm}
Let $x, y \in \realn$, $n \in \natn$. Then
\[
0 \le x < y \implies \sqrt[n]{x} < \sqrt[n]{y}
\]
Let $n, m \in \natn$, $n < m$, $x \in (1, \infty)$, $y \in (0, 1)$. Then
\noindent\begin{minipage}{.5\linewidth}
\[
\sqrt[n]{x} > \sqrt[m]{x}
\]
\end{minipage}
\begin{minipage}{.5\linewidth}
\[
\sqrt[n]{y} < \sqrt[m]{y}
\]
\end{minipage}
\end{thm}
\begin{proof}
\reader
\end{proof}
\begin{thm}
Let $a \in (0, \infty)$. Then
\noindent\begin{minipage}{.5\linewidth}
\[
\limn \sqrt[n]{n} = 1
\]
\end{minipage}
\begin{minipage}{.5\linewidth}
\[
\limn \sqrt[n]{a} = 1
\]
\end{minipage}
\end{thm}
\begin{proof}
Let $\epsilon > 0$. Then
\begin{equation}
\frac{n}{(n + \epsilon)^n} \convinf 0
\end{equation}
This means that
\begin{equation}
\exists N \in \natn ~\forall n \ge N: ~~\frac{n}{(n + \epsilon)^n} < 1
\end{equation}
Therefore
\begin{equation}
n < (1 + \epsilon)^n \implies 1 - \epsilon < 1 \le \sqrt[n]{n} < 1 + \epsilon \iff \left| \sqrt[n]{n} - 1 \right| < \epsilon
\end{equation}
This proves the first statement. The second statement is trivially true for $a = 1$, so let $a > 1$. Then $\exists n \in \natn$ such that $a < n$:
\begin{align}
&\implies 1 < \sqrt[n]{a} < \sqrt[n]{n} \conv{} 1 \\
&\implbl{Squeeze} \sqrt[n]{a} \convinf 1
\end{align}
Now let $a < 1$. Then $\frac{1}{a} < 1$
\begin{equation}
\limn \sqrt[n]{a} = \limn \frac{1}{\sqrt[n]{\frac{1}{a}}} \convinf \frac{1}{1} = 1
\end{equation}
\end{proof}
\begin{defi}
Let $z \in \cmpln$, $x, y \in \realn$ such that $z = x + iy$.
\[
|z| := \sqrt{z\bar{z}} = \sqrt{x^2 + y^2}
\]
\end{defi}
\begin{thm}
Let $u, v \in \cmpln$. Then
\begin{align*}
|u \cdot v| &= |u||v| & \left| \frac{1}{u} \right| &= \frac{1}{|u|} & |u + v| &\le |u| + |v|
\end{align*}
\end{thm}
\begin{proof}
\begin{equation}
|uv| = \sqrt{uv \cdot \bar{uv}} = \sqrt{u\bar{u} \cdot v\bar{v}} = \sqrt{u\bar{u}} \cdot \sqrt{v\bar{v}} = |u||v|
\end{equation}
\begin{equation}
\left| \frac{1}{u} \right| |u| = \left| \frac{1}{u} u \right| = |1| \implies \left| \frac{1}{u} \right| = \frac{1}{|u|}
\end{equation}
For the final statement, remember that complex numbers can be represented as $z = x + iy$, and then
\begin{align}
&\Re(z) \le |\Re(z)| \le |z| \\
&\Im(z) \le |\Im(z)| \le |z|
\end{align}
So therefore
\begin{equation}
\begin{split}
|u + v|^2 &= (u + v) \cdot (\bar{u} + \bar{v}) \\
&= u\bar{u} + v\bar{u} + u\bar{v} + v\bar{v} \\
&= |u|^2 + 2\Re(\bar{u}v) + |v|^2 \\
&\le |u|^2 + 2|\bar{u} v| + |v|^2 \\
&= |u|^2 + 2|u||v| + |v|^2 \\
&= (|u| + |v|)^2
\end{split}
\end{equation}
\end{proof}
\begin{lem}\label{lem:cmplxnull}
Let $\cseqdef{z}$, $z \in \cmpln$.
\[
\seq{z} \conv{} z \iff (|z_n - z|) \text{ null sequence}
\]
\end{lem}
\begin{proof}
Let $x_n = \Re(z_n)$ and $y_n = \Im(z_n)$. Then $x = \Re(z)$ and $y = \Im(z)$. First we prove the "$\impliedby$" direction. Let $(|z_n - z|)$ be a null sequence.
\begin{equation}
0 \le |x_n| - |x| = |\Re(z_n - z)| \le |z_n - z| \conv{} 0
\end{equation}
Analogously, this holds for $y_n$ and $y$. We know that $(|x_n - x|)$ is a null sequence if $x_n \conv{} x$ (same for $y_n$ and $y$), therefore
\begin{equation}
\implies z_n \conv{} z
\end{equation}
To prove the "$\implies$" direction we use the triangle inequality:
\begin{equation}
\begin{split}
0 \le |z_n - z| &= |(x_n - x) + i(y_n - y)| \\
&\le |x_n - x| + \underbrace{|i(y_n - y)|}_{|y_n - y|} \conv{} 0
\end{split}
\end{equation}
By the squeeze theorem, $|z_n - z|$ is a null sequence.
\end{proof}
\begin{rem}
\Cref{lem:cmplxnull} allows us to generalize \Cref{thm:lims} and \Cref{cor:polynomial} for complex sequences.
\end{rem}
\begin{defi}[Cauchy sequence]
A sequence $\rseqdef{x}$ (or $\cmpln$) is called Cauchy sequence if
\[
\forall \epsilon > 0 ~\exists N \in \natn ~\forall n, m \ge N: ~~|x_n - x_m| < \epsilon
\]
\end{defi}
\begin{thm}[Cauchy convergence test]
A sequence $\rseqdef{x}$ (or $\cmpln$) converges if and only if it is a Cauchy sequence.
\end{thm}
\begin{proof}
Firstly, let $\seq{x}$ converge to $x$, and let $\epsilon > 0$. Then
\begin{equation}
\exists N \in \natn ~\forall n \ge N: ~~|x_n - x| < \frac{\epsilon}{2}
\end{equation}
So therefore $\forall n, m \ge N$:
\begin{equation}
|x_n - x_m| = |x_n -x + x - x_m| \le |x_n - x| + |x - x_m| < \epsilon
\end{equation}
This proves the "$\implies$" direction of the theorem. To prove the inverse let $\seq{x}$ be a Cauchy sequence. That means
\begin{equation}
\exists N \in \natn ~\forall n, m \ge N: ~~|x_n - x_m| \le 1
\end{equation}
\begin{equation}
\begin{split}
\implies |x_n| = |x_n - x_N + x_N| &\le |x_n - x_N| + |x_N| \\
&\le |x_N| + 1 ~~\forall n \ge N
\end{split}
\end{equation}
We will now introduce the two auxiliary sequences
\begin{align}
y_n &= \sup \{x_k \setvert k \ge n \} & z_n &= \inf \{x_k \setvert k \ge n \}
\end{align}
$\seq{y}$ and $\seq{z}$ are bounded, and for $\tilde{n} \le n$
\begin{equation}
\{x_k \setvert k \ge \tilde{n} \} \supset \{x_k \setvert k \ge n \}
\end{equation}
\begin{align}
&\implies y_n = \sup \{ x_k | k \ge n \} \le \sup \{ x_k | k \ge \tilde{n} \} = y_{\tilde{n}} \\
&\implies \seq{x} \text{ monotonic decreasing and therefore converging to } y
\end{align}
Analogously, this holds true for $\seq{z}$ as well. Trivially,
\begin{equation}
z_n \le x_n \le y_n
\end{equation}
If $y = z$, then $\seq{x}$ converges according to the squeeze theorem. Assume $z < y$. Choose $\epsilon > \frac{y - z}{2} > 0$. If $N$ is big enough, then
\begin{align}
\sup \{ x_k \setvert k \ge N \} &= y_N > y - \epsilon \\
\inf \{ x_k \setvert k \ge N \} &= z_N < z + \epsilon
\end{align}
So for every $N \in \natn$, we know that
\begin{align}
\exists k \ge N&: ~~x_k > y - 2\epsilon \\
\exists l \ge N&: ~~x_l < z + 2\epsilon
\end{align}
For these elements the following holds
\begin{equation}
|x_k - x_l| \ge \epsilon = \frac{y - z}{2}
\end{equation}
This is a contradiction to our assumption that $\seq{x}$ is a Cauchy sequence, so $y = z$ and therefore $\seq{x}$ converges.
\end{proof}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item $x_n = (-1)^n$. For this sequence the following holds
\[
\forall n \in \natn: ~~|x_n - x_{n+1}| = 2
\]
So this sequence isn't a Cauchy sequence-
\item It is NOT enough to show that $|x_n - x_{n+1}|$ tends to $0$! Example: $\seq{x} = \sqrt{n}$
\[
\begin{split}
\sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} \\
&= \frac{\cancel{n} + 1 - \cancel{n}}{\sqrt{n+1} + \sqrt{n}} \\
&= \frac{1}{\sqrt{n+1} + \sqrt{n}} \convinf 0
\end{split}
\]
However $(\sqrt{n})$ doesn't converge.
\item We introduce the following
\begin{align*}
\text{Limes superior}& & \limsupn x_n &= \limn \sup \{ x_k \setvert k \ge n \} \\
\text{Limes inferior}& & \liminfn x_n &= \limn \inf \{ x_k \setvert k \ge n \}
\end{align*}
$\limsupn x_n \ge \liminfn x_n$ always holds, and if $\seq{x}$ converges then
\[
x_n \convinf x \iff \limsupn x_n = \liminfn x_n
\]
\end{enumerate}
\end{rem}
\begin{defi}
A sequence $\rseqdef{x}$ is said to be properly divergent to $\infty$ if
\[
\forall k \in (0, \infty) ~\exists N \in \natn ~\forall n \ge N: ~~x_n > k
\]
We notate this as
\[
\limn x_n = \infty
\]
\end{defi}
\begin{thm}
Let $\rseqdef{x}$ be a sequence that diverges properly to $\infty$. Then
\[
\limn \frac{1}{x_n} = 0
\]
Conversely, if $\seq{y} \subset (0, \infty)$ is a null sequence, then
\[
\limz \frac{1}{y_n} = \infty
\]
\end{thm}
\begin{proof}
Let $\epsilon > 0$. By condition
\begin{equation}
\exists N \in \natn ~\forall n \ge N: ~~|x_n| > \frac{1}{\epsilon} ~~\left( \iff \frac{1}{|x_n|} < \epsilon \right)
\end{equation}
Therefore $\frac{1}{x_n}$ is a null sequence. The second part of the proof is left as an exercise for the reader.
\end{proof}
\begin{rem}[Rules for computing]
In this remark we will introduce some basic "rules" for working with infinities. These rules are exclusive to this topic, and are in no way universal! This should become obvious with our first two rules:
\begin{align*}
\frac{1}{\pm\infty} &= 0 & \frac{1}{0} &= \infty
\end{align*}
Obviously, division by $0$ is still a taboo, however it works in this case since we are working with limits, and not with absolutes. Let $a \in \realn$, $b \in (0, \infty)$, $c \in (1, \infty)$, $d \in (0, 1)$. The remaining rules are:
\begin{align*}
a + \infty &= \infty & a - \infty &= -\infty \\
\infty + \infty &= \infty & -\infty - \infty &= -\infty \\
b \cdot \infty &= \infty & b \cdot (-\infty) &= -\infty \\
\infty \cdot \infty &= \infty & \infty \cdot (-\infty) &= -\infty \\
c^{\infty} &= \infty & c^{-\infty} &= 0 \\
d^{\infty} &= 0 & d^{-\infty} &= \infty
\end{align*}
There are no general rules for the following:
\begin{align*}
&\infty - \infty & &\frac{\infty}{\infty} & &0 \cdot \infty & &1^{\infty}
\end{align*}
\end{rem}
\begin{thm}
Let $\rseqdef{x}$ be a sequence converging to $x$, and let $\nseqdef{k}$ be a sequence such that
\[
\limn k_n = \infty
\]
Then
\[
\limn x_{k_n} = x
\]
\end{thm}
\begin{proof}
Let $\epsilon > 0$. Then
\begin{equation}
\exists N \in \natn ~\forall n \ge N: ~~|x_n - x| < \epsilon
\end{equation}
Furthermore
\begin{equation}
\exists \tilde{N} \in \natn ~\forall n \ge \tilde{N}: ~~k_n > N
\end{equation}
Therefore
\begin{equation}
\forall n \ge \tilde{N}: ~~|x_{k_n} - x| < \epsilon
\end{equation}
\end{proof}
\begin{eg}
Consider the following sequence
\[
x_n = \frac{n^{2n} + 2n^n}{n^{3n} - n^n}
\]
This can be rewritten as
\[
\frac{n^{2n} + 2n^n}{n^{3n} - n^n} = \frac{(n^n)^2 + 2(n^n)}{(n^n)^3 - (n^n)}
\]
Introduce the subsequence $k_n = n^n$:
\[
\limk\frac{k^2 + 2k}{k^3 - k} = 0 \implies \limn\frac{n^{2n} + 2n^n}{n^{3n} - n^n} = 0
\]
\end{eg}
\end{document}

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\documentclass[../../script.tex]{subfiles}
\begin{document}
\section{Sets and Functions}
\begin{defi}
A set is an imaginary "container" for mathematical objects. If $A$ is a set we write
\begin{itemize}
\item $x \in A$ for "$x$ is an element of $A$"
\item $x \notin A$ for $\neg x \in A$
\end{itemize}
There are some specific types of sets
\begin{enumerate}[(i)]
\item $\varnothing$ is the empty set which contains no elements. Formally: $\exists x \forall y ~y\notin x$
\item Finite sets: $\left\{1, 3, 7, 20\right\}$
\item Let $\Phi(x)$ be a statement and $A$ a set. Then $\left\{x \in A \,\vert\, \Phi(x)\right\}$ is the set of all elements from $A$ such that $\Phi(x)$ holds.
\end{enumerate}
There are relation operators between sets. Let $A, B$ be sets
\begin{enumerate}[(i)]
\item $A \subset B$ means "$A$ is a subset of $B$".
\item $A = B$ means "$A$ and $B$ are the same"
\end{enumerate}
Each element can appear only once in a set, and there is no specific ordering to these elements. This means that $\{1, 3, 3, 7\} = \{3, 1, 7\}$. There are also operators between sets
\begin{enumerate}[(i)]
\item $A \cup B$ is the union of $A$ and $B$.
\[
x \in A \cup B \iff x \in A \vee x \in B
\]
\item $A \cap B$ is the intersection of $A$ and $B$.
\[
x \in A \cap B \iff x \in A \wedge x \in B
\]
This can be expanded to more than two sets ($A \cup B \cup C$). We can also use the following notation. Let $A$ be a set of sets. Then
\[
\bigcup_{C \in A} C
\]
is the union of all sets contained in $A$.
\item $A \setminus B$ is the difference of $A$ and $B$.
\[
x \in A \setminus B \iff x \in A \wedge x \notin B
\]
\item The power set of a set $A$ is the set of all subsets of $A$. Example:
\[
\mathcal{P}(\{1, 2\}) = \{\varnothing, \{1\}, \{2\}, \{1, 2\}\}
\]
\end{enumerate}
\end{defi}
\begin{thm}
Let $A, B, C$ be sets. Then
\begin{align*}
A \setminus (B \cup C) &= (A \setminus B) \cap (A \setminus C) \\
A \setminus (B \cap C) &= (A \setminus B) \cup (A \setminus C) \\
A \cup (B \cap C) &= (A \cup B) \cap (A \cup C) \\
A \cap (B \cup C) &= (A \cap B) \cup (A \cap C)
\end{align*}
\end{thm}
\begin{proof}
Let $A, B, C$ be sets.
\begin{equation}
\begin{split}
x \in A \cap (B \cup C) &\iff x \in A \wedge x \in B \cup C \\
&\iff x \in A \wedge (x \in B \vee x \in C) \\
&\iff (x \in A \wedge x \in B) \vee (x \in A \wedge x \in C) \\
&\iff x \in A \cap B \vee x\ in A \cap C \\
&\iff x \in (A \cap B) \cup (A \cap C)
\end{split}
\end{equation}
The other equations are left as an exercise to the reader.
\end{proof}
\begin{defi}
Let $A, B$ be sets. For $x \in A$, $y \in B$ we call $(x, y)$ the ordered pair from $x, y$. The Cartesian product is defined as
\[
A \times B = \left\{(x, y) \,\vert\, x \in A \wedge y \in B\right\}
\]
\end{defi}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item $(x, y)$ is NOT equivalent to $\{x, y\}$. The former is an ordered pair, the latter a set. It is important to note that
\[
(x, y) = (a, b) \iff x = a \wedge y = b
\]
\item This can be extended to triplets, quadruplets, ...
\[
A \times B \times C = \left\{(x, y, z) \,\vert\, x \in A \wedge y \in B \wedge z \in C \right\}
\]
We use the notation $A \times A = A^2$
\item For $\realn^2$ ($\realn$ are the real numbers) we can view $(x, y)$ as coordinates of a point in the plane.
\end{enumerate}
\end{rem}
\begin{defi}
Let $A$, $B$ be sets. A mapping $f$ from $A$ to $B$ assigns each $x \in A$ exactly one element $f(x) \in B$. $A$ is called the domain and $B$ the codomain.
\begin{figure}[h]
\centering
\begin{tikzpicture}[ele/.style={fill=black,circle,minimum width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner sep=-2pt}]
\node[ele] (a1) at (0,4) {};
\node[ele] (a2) at (0,3) {};
\node[ele] (a3) at (0,2) {};
\node[ele] (a4) at (0,1) {};
\node[ele] (b1) at (4,4) {};
\node[ele] (b2) at (4,3) {};
\node[ele] (b3) at (4,2) {};
\node[ele] (b4) at (4,1) {};
\node[draw,fit= (a1) (a2) (a3) (a4),minimum width=2cm, label=below:$A$] {} ;
\node[draw,fit= (b1) (b2) (b3) (b4),minimum width=2cm, label=below:$B$] {} ;
\draw[->,thick,shorten <=2pt,shorten >=2pt] (a1) -- (b4);
\draw[->,thick,shorten <=2pt,shorten >=2] (a2) -- (b3);
\draw[->,thick,shorten <=2pt,shorten >=2] (a3) -- (b1);
\draw[->,thick,shorten <=2pt,shorten >=2] (a4) -- (b4);
\end{tikzpicture}
\caption{A mapping $f: A \rightarrow B$}
\label{fig:mapping}
\end{figure}
As shown in figure \ref{fig:mapping}, every element from $A$ is assigned exactly one element from $B$, but not every element from $B$ must be assigned to an element from $A$, and elements from $B$ can be assigned more than one element from $A$. The notation for such mappings is
\[
f: A \longrightarrow B
\]
A mapping that has numbers ($\natn$, $\realn$, $\cdots$) as the codomain is called a function.
\end{defi}
\newpage
\begin{eg}\leavevmode
\begin{enumerate}[(i)]
\item
\begin{align*}
f: \natn &\longrightarrow \natn \\
n &\longmapsto 2n + 1
\end{align*}
\item
\begin{align*}
f: \realn &\longrightarrow \realn \\
x &\longmapsto
\begin{cases}
0 & x \text{ rational} \\
1 & x \text{ irrational}
\end{cases}
\end{align*}
\item Addition on $\natn$
\[
f: \natn \times \natn \longrightarrow \natn
\]
Instead of $f(x, y)$ we typically write $x + y$ for addition.
\item The identity mapping is defined as
\begin{align*}
\idf_A: A &\longrightarrow A \\
x &\longmapsto x
\end{align*}
\end{enumerate}
\end{eg}
\begin{rem}[Mappings as sets]\leavevmode
\begin{enumerate}[(i)]
\item A mapping $f: A \rightarrow B$ corresponds to a subset of $F = A \times B$, such that
\begin{align*}
&\forall x \in A ~\forall y, z \in B ~~(x, y) \in F \wedge (x, z) \in F \implies y = z \\
&\forall x \in A ~\exists y \in B ~~(x, y) \in F
\end{align*}
\item Simply writing "Let the function $f(x) = x^2$..." is NOT mathematically rigorous.
\item
\[
f \text{ is a mapping from } A \text{ to } B \iff f(x) \text{ is a value in } B
\]
\item
\[
f, g: A \longrightarrow B \text{ are the same mapping} \iff \forall x \in A ~~f(x) = g(x)
\]
\end{enumerate}
\end{rem}
\begin{defi}
We call $f: A \rightarrow B$
\begin{itemize}
\item injective if $\forall x, \tilde{x} \in A ~~f(x) = f(\tilde{x}) \implies x = \tilde{x}$
\item surjective if $\forall y \in B, \exists x \in A ~~f(x) = y$
\item bijective if $f$ is injective and surjective
\end{itemize}
\begin{figure}[h]
\centering
\begin{subfigure}[b]{0.45\textwidth}
\begin{tikzpicture}[ele/.style={fill=black,circle,minimum width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner sep=-2pt}]
\node[ele] (a1) at (0,4) {};
\node[ele] (a2) at (0,3) {};
\node[ele] (a3) at (0,2) {};
\node[ele] (a4) at (0,1) {};
\node[ele] (b1) at (4,4) {};
\node[ele] (b2) at (4,3.25) {};
\node[ele] (b3) at (4,2.5) {};
\node[ele] (b4) at (4,1.75) {};
\node[ele] (b5) at (4, 1) {};
\node[draw,fit= (a1) (a2) (a3) (a4),minimum width=2cm, label=below:$A$] {} ;
\node[draw,fit= (b1) (b2) (b3) (b4) (b5),minimum width=2cm, label=below:$B$] {} ;
\draw[->,thick,shorten <=2pt,shorten >=2pt] (a1) -- (b4);
\draw[->,thick,shorten <=2pt,shorten >=2] (a2) -- (b3);
\draw[->,thick,shorten <=2pt,shorten >=2] (a3) -- (b1);
\draw[->,thick,shorten <=2pt,shorten >=2] (a4) -- (b5);
\end{tikzpicture}
\caption{Injective mapping. There is at most one arrow per point in $B$}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\centering
\begin{tikzpicture}[ele/.style={fill=black,circle,minimum width=.8pt,inner sep=1pt},every fit/.style={ellipse,draw,inner sep=-2pt}]
\node[ele] (b1) at (4,4) {};
\node[ele] (b2) at (4,3) {};
\node[ele] (b3) at (4,2) {};
\node[ele] (b4) at (4,1) {};
\node[ele] (a1) at (0,4) {};
\node[ele] (a2) at (0,3.25) {};
\node[ele] (a3) at (0,2.5) {};
\node[ele] (a4) at (0,1.75) {};
\node[ele] (a5) at (0, 1) {};
\node[draw,fit= (a1) (a2) (a3) (a4) (a5),minimum width=2cm, label=below:$A$] {} ;
\node[draw,fit= (b1) (b2) (b3) (b4),minimum width=2cm, label=below:$B$] {} ;
\draw[->,thick,shorten <=2pt,shorten >=2pt] (a1) -- (b4);
\draw[->,thick,shorten <=2pt,shorten >=2] (a2) -- (b3);
\draw[->,thick,shorten <=2pt,shorten >=2] (a3) -- (b1);
\draw[->,thick,shorten <=2pt,shorten >=2] (a4) -- (b5);
\draw[->,thick,shorten <=2pt,shorten >=2] (a5) -- (b2);
\end{tikzpicture}
\caption{Surjective mapping. There is at least one arrow per point in $B$}
\end{subfigure}
\caption{Visualizations of injective and surjective mappings}
\end{figure}
\end{defi}
\begin{eg}\leavevmode
\begin{enumerate}[(i)]
\item
\begin{align*}
f: \natn &\longrightarrow \natn \\
n &\longmapsto n^2
\end{align*}
is not surjective (e.g. $n^2 \ne 3$), but injective.
\item
\begin{align*}
f: \intn &\longrightarrow \natn \\
n &\longmapsto n^2
\end{align*}
is neither surjective nor injective.
\item
\begin{align*}
f: \natn &\longrightarrow \natn \\
n &\longmapsto
\begin{cases}
\frac{n}{2} & n \text{even} \\
\frac{n+1}{2} & n \text{odd}
\end{cases}
\end{align*}
is surjective but not injective.
\end{enumerate}
\end{eg}
\begin{defi}[Function compositing]
Let $A, ~B, ~C$ be sets, and let $f: A \rightarrow B, ~g: B \rightarrow C$. Then the composition of $f$ and $g$ is the mapping
\begin{align*}
g \circ f : A &\longrightarrow C \\
x &\longmapsto g(f(x))
\end{align*}
\end{defi}
\begin{rem}
Compositing is associative (why?), but not commutative. For example let
\noindent\begin{minipage}{.5\linewidth}
\begin{align*}
f: \natn &\longrightarrow \natn \\
n &\longmapsto 2n
\end{align*}
\end{minipage}
\begin{minipage}{.5\linewidth}
\begin{align*}
g: \natn &\longrightarrow \natn \\
n &\longmapsto n + 3
\end{align*}
\end{minipage}
Then
\begin{align*}
&f \circ g (n) = 2(n + 3) = 2n + 6 \\
&g \circ f (n) = 2n + 3
\end{align*}
\end{rem}
\begin{thm}
Let $f: A \rightarrow B$ be a bijective mapping. Then there exists a mapping $\inv{f}: B \rightarrow A$ such that $f \circ \inv{f} = \emph{\idf}_B$ and $\inv{f} \circ f = \emph{\idf}_A$. $\inv{f}$ is called the inverse function of $f$.
\end{thm}
\begin{proof}
Let $y \in B$ and $f$ bijective. That means $\exists x \in A$ such that $f(x) = y$. Due to $f$ being injective, this $x$ must be unique, since if $\exists \tilde{x} \in A$ s.t. $f(\tilde{x}) = f(x) = y$, then $x = \tilde{x}$. We define $f(x) = y$ and $\inv{f}(y) = x$, therefore
\begin{equation}
f \circ \inv{f}(y) = f(\inv{f}(y)) = f(x) = y = \idf_B(y) \implies f \circ \inv{f} = \idf_B
\end{equation}
and equivalently
\begin{equation}
\inv{f} \circ f(x) = \idf_A(x) \implies \inv{f} \circ f = \idf_A
\end{equation}
\end{proof}
\end{document}

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\documentclass[../../script.tex]{subfiles}
% !TEX root = ../../script.tex
\begin{document}
\section{Vector Spaces}
We introduce the new field $\field$ which will stand for any field. It can be either $\realn$, $\cmpln$ or any other set that fulfils the field axioms.
\begin{defi}
A vector space is a set $V$ with the operations
\noindent\begin{minipage}[t]{.5\linewidth}
\[
\text{Addition}
\]
\[
\begin{split}
+: V \times V &\longrightarrow V \\
(x, y) &\longmapsto x + y
\end{split}
\]
\end{minipage}
\begin{minipage}[t]{.5\linewidth}
\[
\text{Scalar Multiplication}
\]
\[
\begin{split}
\cdot: \field \times V &\longrightarrow V \\
(\alpha, y) &\longmapsto \alpha x
\end{split}
\]
\end{minipage}
We require the following conditions for these operations
\begin{enumerate}[(i)]
\item $\exists 0 \in V ~\forall x \in V: ~~x + 0 = x$
\item $\forall x \in V ~\exists (-x) \in V: ~~x + (-x) = 0$
\item $\forall x, y \in V: ~~x + y = y + x$
\item $\forall x, y, z \in V: ~~(x + y) + z = x + (y + z)$
\item $\forall \alpha \in \field ~\forall x, y \in V: ~~\alpha (x + y) = \alpha x + \alpha y$
\item $\forall \alpha, \beta \in \field ~\forall x \in V: ~~(\alpha + \beta)x = \alpha x + \beta x$
\item $\forall \alpha, \beta \in \field ~\forall x \in V: ~~(\alpha\beta )x = \alpha(\beta x)$
\item $\forall x \in V: ~~1 \cdot x = x$
\end{enumerate}
Elements from $V$ are called vectors, elements from $\field$ are called scalars.
\end{defi}
\begin{rem}
We now have two different addition operations that are denoted the same way:
\begin{enumerate}[(i)]
\item $+: V \times V \rightarrow V$
\item $+: \field \times \field \rightarrow \field$
\end{enumerate}
Analogously there are two neutral elements and two multiplication operations.
\end{rem}
\begin{eg}\leavevmode
\begin{enumerate}[(i)]
\item $\field$ is already a vector space
\item $V = \field^2$. In the case that $\field = \realn$ this vector space is the two-dimensional Euclidean space. The neutral element is $(0, 0)$, and the inverse is $(\chi_1, \chi_2) \rightarrow (-\chi_1, -\chi_2)$. This can be extended to $\field^n$.
\item $\field$-valued sequences:
\[
V = \set[\chi \in \field ~~\forall n \in \natn]{\seq{\chi}_{n \in \natn}}
\]
\item Let $M$ be a set. Then the set of all $\field$-valued functions on $M$ is a vector space
\[
V = \set[f: M \rightarrow \field]{f}
\]
\end{enumerate}
\end{eg}
\begin{defi}
Let $V$ be a vector space, let $x, x_1, \cdots, x_n \in V$ and let $M \subset V$.
\begin{enumerate}[(i)]
\item $x$ is said to be a linear combination of $x_1, \cdots, x_n$ if $\exists \alpha_1, \cdots, \alpha_n \in \field$ such that
\[
x = \sum_{k=1}^n \alpha_k x_k
\]
\item The set of all linear combinations of elements from $M$ is called the \textit{span}, or the \textit{linear hull} of $M$
\[
\spn M := \set[n \in \natn, ~\alpha_1, \cdots, \alpha_n \in \field, ~x_1, \cdots, x_n \in V]{\sum_{k=1}^n \alpha_k x_k}
\]
\item $M$ (or the elements of $M$) are said to be linearly independent if $\forall \alpha_1, \cdots, \alpha_n \in \field, ~x_1, \cdots, x_n \in V$
\[
\series[n]{k} \alpha_k x_k = 0 \implies \alpha_1 = \alpha_2 = \cdots = \alpha_n = 0
\]
\item $M$ is said to be a generator (of $V$) if
\[
\spn M = V
\]
\item $M$ is said to be a basis of $V$ if it is a generator and linearly independent.
\item $V$ is said to be finite-dimensional if there is a finite generator.
\end{enumerate}
\end{defi}
\begin{eg}\leavevmode
\begin{enumerate}[(i)]
\item For $V = \realn^2$ consider the vectors $x=(1, 0)$, $y=(1,1)$. These vectors are linearly independent, since
\[
\alpha x + \beta y = \alpha(1, 0) + \beta(1, 1) = (0, 0) \implies \alpha + \beta = 0 \wedge \beta = 0
\]
So therefore $\alpha = \beta = 0$. We can show that $\spn\{x, y\} = \realn^2$ because
\[
(\alpha, \beta) = (\alpha - \beta)x + \beta y
\]
So $\set{x, y}$ is a generator, hence $\realn^2$ is finite-dimensional.
\item For $V = \realn^3$ consider $x=(1, -1, 2)$, $y=(2, -1, 0)$, $z=(4, -3, 3)$. These vectors are linearly dependent because
\[
2x + y - z = (0, 0, 0)
\]
\item Let $V = \set[f:\realn\rightarrow\realn]{f}$. Consider the vectors
\[
\begin{split}
f_n: \realn &\longrightarrow \realn \\
x &\longmapsto x^n
\end{split}
\]
The $f_0, f_1, \cdots, f_n, \cdots$ are linearly independent, because
\[
0 = \series_{k=0}^n \alpha_k f_k = \series_{k=0}^n \alpha_k x^k
\]
implies $\alpha_0 = \alpha_1 = \cdots = \alpha_n = 0$. The span of the $f_k$ is the set of all polynomials of $(\le n)$-th degree. The function $x \mapsto (x-1)^3$ is a linear combination of $f_0, \cdots, f_3$:
\[
(x-1)^3 = x^3 - 3x^2 + 3x - 1
\]
\end{enumerate}
\end{eg}
\begin{rem}
Let $V$ be a vector space, $y \in V$ a linear combination of $y_1, \cdots, y_n$, and each of those a linear combination of $x_1, \cdots, x_n$. I.e.
\[
\exists \alpha_1, \cdots, \alpha_n \in \field: ~~y = \series[n]{k} \alpha_k y_k
\]
and
\[
\exists \beta_{k,l} \in \field: ~~y_k = \series[n]{l} \beta_{k,l} x_l
\]
Then
\[
y = \series[n]{k} \alpha_k y_k = \series[n]{k}\alpha_k\series[n]{l}\beta_{k, l} x_l = \series[n]{l}\underbrace{\left(\series[n]{k}\alpha_k\beta_{k,l}\right)}_{\in \field} x_l
\]
So therefore
\[
\spn(\spn(M)) = \spn(M)
\]
\end{rem}
\begin{thm}
Let $V$ be a finite-dimensional vector space, and let $x_1, \cdots, x_n \in V$. Then the following are equivalent
\begin{enumerate}[(i)]
\item $x_1, \cdots, x_n$ is a basis.
\item $x_1, \cdots, x_n$ is a minimal generator (Minimal means that no subset is a generator).
\item $x_1, \cdots, x_n$ is a maximal linearly independent system (Maximal means that $x_1, \cdots, x_n, y$ is not linearly independent).
\item $\forall x \in V$ there exists a unique $\alpha_1, \cdots, \alpha_n \in \field$
\[
x = \series[n]{k} \alpha_k x_k
\]
\end{enumerate}
\end{thm}
\begin{proof}
First we prove "(i) $\implies$ (ii)". Let $x_1, \cdots, x_n$ be a basis of $V$. By definition $x_1, \cdots, x_n$ is a generator. Assume that $x_2, \cdots, x_n$ is still a generator, then
\begin{equation}
\exists \alpha_2, \cdots, \alpha_n \in \field: ~~x_1 = \series[n]{k} \alpha_k x_k
\end{equation}
However this contradicts the linear independence of the basis. Next, to prove "(ii) $\implies$ (iii)" let $x_1, \cdots, x_n$ be a minimal generator. Let $\alpha_1, \cdots, \alpha_n \in \field$ such that
\begin{equation}
0 = \series[n]{k} \alpha_k x_k
\end{equation}
Assume that one coefficient is $\ne 0$ (w.l.o.g. $\alpha_1 = 0$). Then
\begin{equation}
x_1 = \sum_{k=2}^n -\frac{\alpha_k}{\alpha_1} x_k
\end{equation}
$x_1, \cdots, x_n$ is a generator, i.e. for $x \in V$
\begin{equation}
\exists \beta_1, \cdots, \beta_n \in \field: ~~x = \series[n]{k} \beta_k x_k = \sum_{k=2}^n\left(\beta_k - \frac{\alpha_k}{\alpha_1}\right)x_k
\end{equation}
But this implies that $x_2, \cdots, x_n$ is a generator. That contradicts the assumption that $x_1, \cdots, x_n$ was minimal.
\begin{equation}
\implies \alpha_1 = \alpha_2 = \cdots = \alpha_n = 0
\end{equation}
Now let $y \in V$. Then
\begin{equation}
\exists \gamma_1, \cdots, \gamma_n \in \field: ~~y = \series[n]{k} \gamma_k x_k
\end{equation}
So $x_1, \cdots, x_n, y$ is linearly dependent, and therefore $x_1, \cdots, x_n$ is maximal. To prove "(iii) $\implies$ (iv)" let $x_1, \cdots, x_n$ be a maximal linearly independent system. If $y \in V$, then
\begin{equation}
\exists \alpha_1, \cdots, \alpha_k, \beta \in \field: ~~\series[n]{k} \alpha_k x_k + \beta y = 0
\end{equation}
Assume $\beta = 0$, then consequently
\begin{equation}
x_1, \cdots, x_n \text{ linearly independent} \implies \alpha_1 = \alpha_2 = \cdots = \alpha_n = 0
\end{equation}
This is a contradiction, so therefore $\beta \ne 0$:
\begin{equation}
y = \series[n]{k} -\frac{\alpha_k}{\beta} x_k
\end{equation}
The uniqueness of these coefficients are left as an exercise for the reader. Finally, to finish the proof we need to show "(iv) $\implies$ (i)". By definition
\begin{equation}
V = \spn\set{x_1, \cdots, x_n}
\end{equation}
Hence, $\set{x_1, \cdots, x_n}$ is a generator. In case
\begin{equation}
0 = \series[n]{k} \alpha_k x_k
\end{equation}
holds, then $\alpha_1 = \cdots = \alpha_n = 0$ follows from the uniqueness.
\end{proof}
\begin{cor}
Every finite-dimensional vector space has a basis.
\end{cor}
\begin{proof}
By condition, there is a generator $x_1, \cdots, x_n$. Either this generator is minimal (then it would be a basis), or we remove elements until it is minimal.
\end{proof}
\begin{lem}\label{lem:steinitz}
Let $V$ be a vector space and $x_1, \cdots, x_k \in V$ a linearly independent set of elements. Let $y \in V$, then
\[
x_1, \cdots, x_k, y \text{ linearly independent} \iff y \notin \spn\set{x_1, \cdots, x_k}
\]
\end{lem}
\begin{proof}
To prove "$\impliedby$", assume $y \ne \spn\set{x_1,\cdots,x_k}$. Therefore $x_1, \cdots, x_k, y$ must be linearly independent. To see this, consider
\begin{equation}
0 = \series[n]{k} \alpha_k x_k + \beta y ~~\alpha_1, \cdots, \alpha_n \in \field
\end{equation}
Then $\beta = 0$, otherwise we could solve the above for $y$, and that would contradict our assumption. The argument works in the other direction as well.
\end{proof}
\begin{thm}[Steinitz exchange lemma]
Let $V$ be a finite-dimensional vector space. If $x_1, \cdots, x_m$ is a generator and $y_1, \cdots, y_n$ a linear independent set of vectors, then $n \le m$. In case $x_1, \cdots, x_m$ and $y_1, \cdots, y_n$ are both bases, then $n=m$.
\end{thm}
\begin{hproof}
Let $K \in \set{0, \cdots, \min\set{m, n} - 1}$ and let
\begin{equation}
x_1, \cdots, x_K, y_{K+1}, \cdots, y_n
\end{equation}
be linearly independent. Assume that
\begin{equation}
x_{K+1}, \cdots, x_m \in \spn\set{x_1, \cdots, x_k, y_{K+2}, \cdots, y_n}
\end{equation}
Then
\begin{equation}
y_{K+1} \in \spn\set{x_1, \cdots, x_m} \subset \spn\set{x_1, \cdots, x_K, y_{K+2}, \cdots, y_m}
\end{equation}
This contradicts with the linear independence of $x_1, \cdots, x_K, y_{K+2}, \cdots y_n$. Furthermore,
\begin{equation}
\exists x_i \in V: ~~x_i \notin \spn\set{x_1, \cdots, x_K, y_{K+ 2}, \cdots, y_n}
\end{equation}
W.l.o.g. $x:i = x_{K+1}$. By \Cref{lem:steinitz}, $x_1, \cdots, x_{K+1}, y_{K+2}, \cdots y_n$ is linearly independent. We can now sequentially replace $y_i$ with $x_i$ without losing the linear independence. Assume $n > m$, then this process leads to a linear independent system $x_1, \cdots, x_m, y_{m+1}, \cdots, y_n$. But since $x_1, \cdots, x_m$ is a generator, $y_{m+1}$ is a linear combination of $x_1, \cdots, x_m$. If $x_1, \cdots, x_m$ and $y_1, \cdots, y_n$ are both bases, then we cannot change the roles and therefore $m = n$.
\end{hproof}
\begin{defi}
The amount of elements in a basis is said to be the dimension of $V$, and is denoted as $\dim V$ .
\end{defi}
\begin{eg}\leavevmode
\begin{enumerate}[(i)]
\item Let $V = \realn^n$ (or $\cmpln^n$). Define
\[
e_k = (0, 0, \cdots, 0, \underset{\substack{\uparrow\\\mathrlap{\text{\hspace{-1.5em}k-th position}}}}{1}, 0, \cdots, 0)
\]
Then $e_1, \cdots, e_n$ is a basis, in fact, it is the standard basis of $\realn^n$ ($\cmpln^n$).
\item Let $V$ be the vector space of polynomials
\[
V = \set[n \in \natn, ~\alpha_1, \cdots, \alpha_n \in \realn, ~~f(x) = \sum_{k=1}^n \alpha_k x^k ~~\forall x \in \realn]{f:\realn \longrightarrow \realn}
\]
This space has the basis
\[
\set[n \in \natn_0]{x \longmapsto x^n}
\]
\end{enumerate}
\end{eg}
\begin{cor}
In an $n$-dimensional vector space, every generator has at least $n$ elements, and every linearly independent system has at most $n$ elements.
\end{cor}
\begin{proof}
Let $M \subset \spn\set{x_1, \cdots, x_n}$. Then
\begin{equation}
V = \spn M \subset \spn{x_1, \cdots, x_n}
\end{equation}
Hence, $x_1, \cdots, x_n$ is a generator. On the other hand, assume
\begin{equation}
\exists y \in M \setminus \spn\set{x_1, \cdots, x_n}
\end{equation}
Then $x_1, \cdots, x_n, y$ is linearly independent (\Cref{lem:steinitz}), and we can sequentially add elements from $M$ until $x_1, \cdots, x_n, y_{n+1}, \cdots, y_{n+m}$ is a generator.
\end{proof}
\begin{defi}[Vector subspace]
Let $V$ be a vector space. A non-empty set $W \subset V$ is called a vector subspace if
\[
\forall x, y \in W ~\forall \alpha \in \field: ~~x + \alpha y \in W
\]
\end{defi}
\begin{eg}
Consider
\[
W = \set[chi \in \realn]{(\chi, \chi) \in \realn^2}
\]
This is a subspace, because
\[
(\chi, \chi) + \alpha(\eta, \eta) = (\chi + \alpha\eta, \chi + \alpha\eta)
\]
However,
\[
A = \set[\chi^2 + \eta^2 = 1]{(\chi, \eta) \in \realn^2}
\]
is not a subspace, because $(1, 0), (0, 1) \in A$, but $(1, 1) \notin A$.
\end{eg}
\begin{rem}\leavevmode
\begin{enumerate}[(i)]
\item Every subspace $W \subset V$ contains the $0$ and the inverse elements.
\item Let $W \subset V$ be a subspace. Then
\[
\forall x_1, \cdots, x_n \in W, ~\alpha_1, \cdots, \alpha_n \in \field: ~~\series[n]{k} \alpha_k x_k \in W
\]
Furthermore, $M \subset W \implies \spn M \subset W$.
\item $M \subset V$ is a subspace if and only of $\spn M = M$.
\item Let $I$ be an index set, and $W_i \subset V$ subspaces. Then
\[
\bigcap_{i \in I} W_i
\]
is also a subspace
\item The previous doesn't hold for unions.
\item Let $M \subset V$:
\[
\spn M = \bigcap_{W \supset M \text{ subspace of } V} W
\]
\end{enumerate}
\end{rem}
\end{document}

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\documentclass[11pt]{report}
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\crefname{thm}{Thm.}{Thms.}
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\begin{document}
\title{Mathemacs for Physicists}
\author{https://www.github.com/Lauchmelder23}
\maketitle
\tableofcontents
\subfile{chapters/FaN.tex}
\subfile{chapters/real_analysis_1.tex}
\subfile{chapters/linear_algebra.tex}
\end{document}