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 $X=\bigcup _{i\in I}\{S_i\}$ as a union of non-empty sets $S_i$ a function for selection of an element exists. The Function $F$ selects an element $s_i$ from each of these non-empty set $S_i$.

$F:X\to Y$ with $F(S_i)=s_i\in S_i$ with $Y:=\bigcup _{i\in I}{S}_i$.

Domain and Range

Please note that the following two sets are different:

• (M1) $X=\bigcup _{i\in I}\{S_i\}$  
• (M2) $Y=\bigcup _{i\in I}{S}_i$  

Example - Domain and Range

With the sample quantities $S_1:=\{K,Z\}$, $S_2:=\{1,2,3,4,5,6\}$, $S_3:=\{1,2,...,32\}$ applies:

• (M1) $X=\bigcup _{i\in I}\{S_i\}=\{S_1,S_2,S_3\}$_, i.e. $X$ is a set of sets containing 3 elements.
• (M2) $Y=\bigcup _{i\in I}{S}_i=\{K,Z,1,....,32\}$, $Y$ 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 $X$ a set of non-empty sets. Then $F$ an choice function applies to $S$

$\mathrm{\forall }S\in X:F(S)\in S.$

$F$ selects exactly one element from every set $S$ in $X$.

Axiom of Choice

The axiom of choice is then:

For any set $X$ of non-empty sets there is a choice function $F$.

Example:

Be $X=\{\{0,2\},\{1,2,5,7\},\{4\}\}$ on $X$

$$F(\{0,2\})=2;F(\{1,2,5,7\})=2;F(\{4\})=4$$

defined function $F$ is a choice function for $X$.

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 $(s_1,\dots .,s_n)$, e.g.. With $F(\{0,2\})=2;F(\{1,2,5,7\})=2;F(\{4\})=4$_ and the index set $I=\{1,2,3\}$ you can written the result of selection in the following way:

$$(s_1,s_2,s_3)=(2,2,4)\in \prod _{k\in I}{S}_k=S_1\times S_2\times S_3$$

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 $X_i$ of $X$ are pairwise disjoint, then there exists a set C such that its intersection with any of the elements $X_i$ of $X$ 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 $X=\{S_1,\dots ,S_n\}$ 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 $S_i\subset ℝ$. If there are two options $s\in ℝ$ and $s\in ℝ$ 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 $F$ 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 $(X,\prec )$ 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 $𝕂$ a [[w:en:Field (mathematics) |field]] and $V=(V,+)$ a commutative group. $V$ is called $V$ a $𝕂$-vector space when an function is

$\cdot :𝕂\times V\to V$ with $(\lambda ,v)\mapsto \lambda \cdot v$

is defined which fulfills the following properties with $\lambda ,\mu \in 𝕂$ and $v,w\in V$

• (ES) $1\cdot v=v$ (Scalar Multiplication with the neutral element in $𝕂$)
• (AS) $\lambda \cdot (\mu \cdot v)=(\lambda \cdot \mu )\cdot v.$_ (associative scalar multiplication)
• (DV) $\lambda \cdot (v+w)=\lambda \cdot v+\lambda \cdot w.$_ (vectors distributiv)
• (DS) $(\lambda +\mu )\cdot v=\lambda \cdot v+\mu \cdot v.$_ (Skalare distributiv)

Learning Task

• Consider the space of all continuous functions $𝒞([a,b],ℝ)$ from an interval $[a,b]$ to $ℝ$. Define a partial order on $𝒞([a,b],ℝ)$.
• Define a scalar multiplication and an addition on the vector space $V:=𝒞([a,b],ℝ)$! Is there alternative definitions for addition $\oplus$ and multiplication $\odot$ 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 $V$? (Preparation for the topology and norms on a space of functions)

See also

• Lemma von Zorn (engl.)
• Vector space - definition
• Wikibook: Functional Analysis
• Continuous Function
• Partial ordered sets

Literature

• Thomas Jech: The Axiom of Choice. North Holland, 1973, ISBN 0-7204-2275-2.
• Paul Howard, Jean E. Rubin: Consequences of the Axiom of Choice. American Mathematical Society, 1998, ISBN 0-8218-0977-6.
• Per Martin-Löf: 100 years of Zermelo’s axiom of choice: what was the problem with it? (PDF-File; 257 KB])

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