Axioms of set theory

The $\in$-relation §

Set theory is built on the postulate that there is a fundamental binary relation denoted $\in$ .

We don’t define explicitly sets and the relation $\in$, we will have nine axioms concerning ∈ and sets, and it is only in terms of these nine axioms that ∈ and sets are defined at all.

We can define the common relations:

• $x\notin y=\neg (x\in y)$
• $x\subseteq y:=\forall a:(a\in x⟹a\in y)$
• $x=y:=(x\subseteq y)\wedge (y\subseteq x)$ $x\subset y:=(x\subseteq y)\wedge \neg (x=y)$

The axioms §

The totality of all these nine axioms are called ZFC set theory, which is a shorthand for Zermelo-Fraenkel set theory with the axiom of Choice

1. Axiom on the $\in$-relation _The expression $x\in y$ is a proposition if, and only if, both $x$ and $y$ are sets:

$$\forall x:\forall y:(x\in y)\vee \neg (x\in y)$$

This prevents tue Russell’s paradox, in fact from this definition the russell set is not a set.

2. Axiom on the existence of an empty set There exists a set that contains no elements:

$$\exists y:\forall x\notin y$$

3. Axiom on pair sets Let $x$ and $y$ be sets. Then there exists a set that contains as its elements precisley $x$ and $y$:

$$\forall x:\forall y:\exists m:\forall u:(u\in m⟺(u=x\vee u=y))$$

The pair set $m$ is denoted by $\{x,y\}$. From this follows the existence of the singleton set $\{x\}:=\{x,x\}$.

4. Axiom on union sets Let $x$ be a set. Then there exists a setr whose elements are precisley the elements of the elements of $x$:

$$\forall x:\exists u:\forall y:(y\in u⟺\exists s:(y\in s\wedge s\in x))$$

The set $u$ is denoted by $\cup x$. We need this axioms to build sets with more than 2 elments.

We can now define the intersection and the relative complement of sets. Let $x$ be a set. We define the intersection of $x$ by:

$$\cap x:=\{a\in \cup x|\forall b\in x:a\in b\}$$

We can also define the difference of two sets $u\subseteq m$:

$$m\setminus u:=\{x\in m|x\notin u\}$$

5. Axiom of replacement Let $R$ be a functional relation and $m$ a set. Then the image of $m$ under $R$, denoteb by $i{m}_R(m)$ is a set. A realtion $R$ is functional if:

$$\forall x:\exists !y:R(x,y)$$

Hence the name. The principle of restricted comprension (given a set $m$ and a predicate $P(x)$ then $\{y\in m|P(y)\}$ is a set) is a consequence of the replacement axioms. Without the costrain of the set $m$ is known as the principle of universal comprension, wich Russell proved to be inconsistent.

6. Axiom ont the existence of power sets Let $m$ be a set. Then there exists a set, denoted by $P(m)$, whose elements are precisely the subsets of $m$:

$$\forall x:\exists y:\forall a:(a\in y⟺a\subseteq x)$$

7. Axiom of infinity _There exists a set that contains the empty set and, together with every other element $y$, it also contains the set $\{y\}$ as an element:

$$\exists x:\varnothing \in x\wedge \forall y:(y\in x⟹\{y\}\in x)$$

Consider one of this sets $x$. It follows that:

$$x=\{\varnothing ,\{\varnothing \},\{\{\varnothing \}\},\dots \}$$

We can define:

$$0:=\varnothing ,1:=\{\varnothing \},2:=\{\{\varnothing \}\},\dots$$

Accordin to our axioms the set $\mathbb{N}:=x$ is a set. We then can define $\mathbb{R}:=P(\mathbb{N})$.

There is a modern variant, insted of $\{y\}\in x$ we impose $y\cup \{y\}\in x$.

8. Axiom of choice Let $x$ be a set whose elements are non-empty and mutually disjoint. Then there exists a set $y$ which contains exactly one element of each element of $x$.

$$\forall x:P(x)⟹\exists y:\forall a\in x:\exists |b\in a:a\in y,$$

where

$$P(x)⟺(\exists a:a\in x)\wedge (\forall a:\forall b:(a\in b\wedge b\in x)⟹\cap \{a,b\}=\varnothing )$$

The axiom of choice is independent of the other 8 axioms, which means that one could have set theory with or without the axiom of choice. However, standard mathematics uses the axiom of choice and hence so will we. There is a number of theorems that can only be proved by using the axiom of choice. Amongst these we have:

• every vector space has a basis;
• there exists a complete system of representatives of an equivalence relation;

9. Axiom of foundation . Every non-empty set $x$ contains an element $y$ that has none of its elements in common with $x$:

$$\forall x:(\exists a:a\in x)⟹\exists y\in x:\cap \{x,y\}=\varnothing$$

An immediate consequence of this axiom is that there is no set that contains itself as an element.