# Definition:Class/Zermelo-Fraenkel

## Definition

Denote with $\textrm{ZF}$ the language of set theory endowed with the Zermelo-Fraenkel axioms.

A **class** in $\textrm{ZF}$ is a formal vehicle capturing the intuitive notion of a class, namely a collection of all sets such that a particular condition $P$ holds.

In $\textrm{ZF}$, **classes** are written using class builder notation:

- $\set {x : \map P x}$

where $\map P x$ is a statement containing $x$ as a free variable.

More formally, a **class** $\set {x: \map P x}$ serves to define the following definitional abbreviations involving the membership symbol:

\(\displaystyle y \in \set {x: \map P x}\) | \(\quad \text{for} \quad\) | \(\displaystyle \map P y\) | |||||||||||

\(\displaystyle \set {x: \map P x} \in y\) | \(\quad \text{for} \quad\) | \(\displaystyle \exists z \in y: \forall x: \paren {x \in z \iff \map P x}\) | |||||||||||

\(\displaystyle \set {x: \map P x} \in \set {y: \map Q y}\) | \(\quad \text{for} \quad\) | \(\displaystyle \exists z: \paren {\map Q z \land \forall x: \paren {x \in z \iff \map P x} }\) |

where:

- $x, y ,z$ are variables of $\textrm{ZF}$
- $P, Q$ are propositional functions.

Through these "rules", every statement involving $\set {x: \map P x}$ can be reduced to a simpler statement involving only the basic language of set theory.

This definition "overloads" the membership symbol $\in$ since its operands could now be either sets or **classes**.

That such does not lead to ambiguity is proved on Class Membership Extension of Set Membership.

### Class Variables

In deriving general results about $\textrm{ZF}$ which mention **classes**, it is often convenient to have **class variables**, which denote an arbitrary **class**.

By convention, these variables are taken on $\mathsf{Pr} \infty \mathsf{fWiki}$ to be the (start of) the capital Latin alphabet, i.e. $A, B, C$, and so on.

## Caution

Unlike in von Neumann-Bernays-GĂ¶del set theory, it is prohibited to quantify over **classes**.

That is, expressions like:

- $\forall A: \map P A$

are ill-defined; admitting them without further consideration would lead us to Russell's Paradox.