# Matej Balog

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## Analysis I: Sequences and Series

Chapters:

### 1 Field axioms

Proposition Let $a \in \mathbb{R}$. Then $-(-a) = a$.

Proof \begin{align} (-a) + (-(-a)) = 0 & \text{ by A4} \\ (-a) + a = 0 & \text{ by A4 and A1} \\ a = (-(-a)) & \text{ by uniqueness of additive inverses} \end{align}

Proposition For any $a \in \mathbb{R}$: $a \cdot 0 = 0$.

Proof \begin{align} a \cdot (0 + 0) = a \cdot 0 & \text{ by A3} \\ a \cdot 0 + a \cdot 0 = (a \cdot 0) + 0 & \text{ by D and A3} \\ a \cdot 0 = 0 & \text{ subtracting } a \cdot 0 \text{ (by A1 - A4)} \end{align}

### 2 Order axioms

We will assume the existence of a subset $\mathbb{P}$ of $\mathbb{R}$ called the positive numbers which satisfies the following three axioms:

1. given $a, b \in \mathbb{P}$, then also $a + b \in \mathbb{P}$
2. given $a, b \in \mathbb{P}$, then also $ab \in \mathbb{P}$
3. (trichotomy axiom) for any $a \in \mathbb{R}$, precisely one of $a \in \mathbb{P}$, $a = 0$, $-a \in \mathbb{P}$ holds

Proposition $1 \in \mathbb{P}$

Proof $1 = 0$ is impossible by $Z$.
Also if $-1 \in \mathbb{P}$, then $(-1)(-1) = -(-1) = 1 \in \mathbb{P}$ by $P2$, but that contradicts the trichotomy axiom.
The only remaining possibility is $1 \in \mathbb{P}$.

### 3 The Completness Axiom

Proposition (Approximation Property) Let $B$ be a non-empty, bounded above subset of $\mathbb{R}$. Let $b = \sup{B}$ and $\varepsilon > 0$. Then there exists $x \in B$ with $\sup{B} - \varepsilon < x \leq \sup{B}$.

Theorem (Archimedean Property) $\mathbb{N}$ is not bounded above.

### 4 Complex Numbers

This chapter was made into the separate course Introduction to Complex Numbers.