# Functional analysis/Set theory
## Introduction

In this chapter some standard results are collected from the set theory, which are to be used in the further sequence of lecture contents. In particular, the Hahn-Banach-theorem, which is actually already a result from the linear algebra, is introduced. The evidence for these theorems can be found in the books/Wikibooks Topology and Lineare Algebra.

## Axiom of Choice

The Axiom of choice is a axiom of the Zermelo-Fraenkel-set. It was formulated for the first time by Ernst Zermelo 1904. The Axiom of choice states that for every set X=⋃$_{i∈I}${S$_i$}{\textstyle X=\bigcup _{i\in I}\{S_{i}\}} as a union of non-empty sets S$_i${\textstyle S_{i}} a function for selection of an element exists. The Function $F$ selects an element $s_{i}$ from each of these non-empty set S$_i${\textstyle S_{i}}.
    F:X→Y{\textstyle F:X\rightarrow Y} with F(S$_i$)=s$_i$∈S$_i${\textstyle F(S_{i})=s_{i}\in S_{i}} with Y:=⋃$_{i∈I}$S$_i${\textstyle Y:=\bigcup _{i\in I}S_{i}}.

### Domain and Range

Please note that the following two sets are different:
* (M1) X=⋃$_{i∈I}${S$_i$}{\textstyle X=\displaystyle \bigcup _{i\in I}\, \{S_{i}\}}
* (M2) Y=⋃$_{i∈I}$S$_i${\textstyle Y=\displaystyle \bigcup _{i\in I}S_{i}}

### Example - Domain and Range

With the sample quantities S$_1$:={K, Z}{\textstyle S_{1}:=\{K, Z\}}, S$_2$:={1, 2, 3, 4, 5, 6}{\textstyle S_{2}:=\{1, 2, 3, 4, 5, 6\}}, S$_3$:={1, 2, ..., 32}{\textstyle S_{3}:=\{1, 2, ..., 32\}} applies:
* (M1) X=⋃$_{i∈I}${S$_i$}={S$_1$, S$_2$, S$_3$}{\textstyle X=\displaystyle \bigcup _{i\in I}\, \{S_{i}\}=\{S_{1}, S_{2}, S_{3}\}}_, i.e. X{\textstyle X} is a set of sets containing 3 elements.
* (M2) Y=⋃$_{i∈I}$S$_i$={K, Z, 1, ...., 32}{\textstyle Y=\displaystyle \bigcup _{i\in I}S_{i}=\{K, Z, 1, ...., 32\}}, Y{\textstyle Y} is an union of sets containing 34 elements.

### Finite Sets

For finite sets, the property can be derived from other axioms. Therefore, the selection axiom is only interesting for infinite sets.

### Definition: Choice Function

Be X{\textstyle X} a set of non-empty sets. Then $F$ an choice function applies to S{\textstyle S}
    ∀S∈X:F(S)∈S.{\textstyle \forall S\in X:F(S)\in S.}
F{\textstyle F} selects exactly one element from every set S{\textstyle S} in X{\textstyle X}.

### Axiom of Choice

The axiom of choice is then:
    For any set X{\textstyle X} of non-empty sets there is a choice function F{\textstyle F}.

### Example:

Be X={{0, 2}, {1, 2, 5, 7}, {4}}{\textstyle X=\{\{0, 2\}, \{1, 2, 5, 7\}, \{4\}\}} on X{\textstyle X}
    $F(\{0,2\})=2;\quad F(\{1,2,5,7\})=2;\quad F(\{4\})=4$
defined function F{\textstyle F} is a choice function for X{\textstyle X}.

### Choice as an element in the product space

The lecture also addresses the vector space of the sequences. The product space of sets $S_{i}$ can be used to represent the selection of a tupel $(s_{1},\ldots .,s_{n})$, e.g.. With F({0, 2})=2;F({1, 2, 5, 7})=2;F({4})=4{\textstyle F(\{0, 2\})=2;\quad F(\{1, 2, 5, 7\})=2;\quad F(\{4\})=4}_ and the index set I={1, 2, 3}{\textstyle I=\{1, 2, 3\}} you can written the result of selection in the following way:
    $(s_{1},s_{2},s_{3})=(2,2,4)\in \displaystyle \prod _{k\in I}S_{k}=S_{1}\times S_{2}\times S_{3}$

### Alternative formulations

* The power set of any non-empty set has a choice function (Zermelo 1904).
* Given any set X, if the empty set is not an element of X and the elements X$_i${\textstyle X_{i}} of X{\textstyle X} are pairwise disjoint, then there exists a set C such that its intersection with any of the elements X$_i${\textstyle X_{i}} of X{\textstyle X} contains exactly one element.
* Let $I$ arbitray non-empty index set and $(S_{i})_{i\in I}$ a family of non-empty sets $S_{i}$. It exists a function $F$ with the domain $I$, that maps every index $i\in I$ to a single element of $S_{i}$: $F(i)\in S_{i}$.

### Existence of choice function without axiom

In the following cases, a choice function exists even without the requirement of a valid axiom of choice:
* For a finite quantity X={S$_1$, …, S$_n$}{\textstyle X=\{S_{1}, \ldots , S_{n}\}} of non-empty set, it is trivial to specify a choice function: You select any particular element $s_{i}$ from any set $S_{i}$. You don't need the axiom of choice for this. A formal proof would use Induction over the size of the finite set.
* It is also possible to define a choice function for subsets $S_{i}$ of non-empty of the natural numbers: Due to the fact that all sets have a lover bound in the countable set, the smallest element is selected from each subset is chosen.
* Similarly, an explicit choice function (without the use of the axiom of choice) can be defined for a set of real numbers by selecting element with the smallest absolute value from each set $S_{i}\subset \mathbb {R}$. If there are two options $s\in \mathbb {R}$ and $s\in \mathbb {R}$ the positive value will be selected.
* Even for sets of intervals of real numbers, a choice function can be defined as the center of lower bound (center or the upper bound) of the interval as the selected element from each interval $S_{i}$.

### Existence of selection function with required Axiom of Choice

For the following cases, the selection axiom is required to obtain the existence of a choice function:
* It is not possible to prove the existence of a choice function F{\textstyle F} for a general countable set of sets $S_{i}$ that contain just two elements ZF] (not ZFC, i.e. ZF is without the axiom of choice).
* The same applies, e.g., to the existence of a choice function for the set of all non-empty subsets of real numbers.
This leads to the question whether theorems for which the axiom of choice is usually required (e.g. Hahn-Banach theorem) can be proven without the axiom of choice and the main conclusions of the theorem are still valid.

## Zorn's Lemma

Suppose (X, ≺){\textstyle \left(X, \prec \right)} is a partially ordered set that has the property that every chain $(s_{n})_{n\in \mathbb {N} }$ with $s_{n}\prec s_{n+1}$ in $X$ has an upper bound in $X$. Then the set $X$ contains at least one maximal element.

## Vector space

Be K{\textstyle \mathbb {K} } a field and V=(V, +){\textstyle V=(V, +)} a commutative group. $V$ is called V{\textstyle V} a K{\textstyle \mathbb {K} }-vector space when an function is
    ⋅:K×V→V{\textstyle \cdot :\mathbb {K} \times V\rightarrow V} with (λ, v)↦λ⋅v{\textstyle (\lambda , v)\mapsto \lambda \cdot v}
is defined which fulfills the following properties with $\lambda ,\mu \in \mathbb {K}$ and $v,w\in V$
* (ES) 1⋅v=v{\textstyle 1\cdot v=v} (Scalar Multiplication with the neutral element in K{\textstyle \mathbb {K} })
* (AS) λ⋅(μ⋅v)=(λ⋅μ)⋅v.{\textstyle \lambda \cdot (\mu \cdot v)=(\lambda \cdot \mu )\cdot v.}_ (associative scalar multiplication)
* (DV) λ⋅(v+w)=λ⋅v+λ⋅w.{\textstyle \lambda \cdot (v+w)=\lambda \cdot v+\lambda \cdot w.}_ (vectors distributiv)
* (DS) (λ+μ)⋅v=λ⋅v+μ⋅v.{\textstyle (\lambda +\mu )\cdot v=\lambda \cdot v+\mu \cdot v.}_ (Skalare distributiv)

## Learning Task

* Consider the space of all continuous functions C([a, b], R){\textstyle {\mathcal {C}}([a, b], \mathbb {R} )} from an interval [a, b]{\textstyle [a, b]} to R{\textstyle \mathbb {R} }. Define a partial order on C([a, b], R){\textstyle {\mathcal {C}}([a, b], \mathbb {R} )}.
* Define a scalar multiplication and an addition on the vector space V:=C([a, b], R){\textstyle V:={\mathcal {C}}([a, b], \mathbb {R} )}! Is there alternative definitions for addition ⊕{\textstyle \oplus } and multiplication ⊙{\textstyle \odot } with a scalar on $V$, which fulfill the properties of a vector space mentioned above?
* How can we define a distance between two continuous functions $f$ and $g$ with an integral V{\textstyle V}? (Preparation for the topology and norms on a space of functions)