Axioms of ZF
ZF Axioms#
There are Eight axioms.
- Extensionality. If \(X\) and \(Y\) have the same elements \(X=Y\).
- Pairing. There is a set \(\{a,b\}\) for any \(a\) and \(b\).
- Schema of Separation (for every \(P\) an axiom). Given property \(P\) with parameter \(p\), for any \(X\) and \(p\) there is a set \(Y=\{u\in X: P(u,p)\}\) that contains all those \(u\in X\) that have property \(P\).
- Union. For any \(X\) there is a set \(Y=\cup X\) the union of all elements of \(X\).
- Power Set. There is a set \(Y = P(X)\) for any \(X\), the set of all subsets of \(X\).
- Infinity. There is an infinite set.
This is formulated with $\exists S (\emptyset \in S \wedge (\exists x\in S) x \cup \{x\}\in S)$; i.e. there is an inductive set. - Schema of Replacement. If a class \(F\) is a function, then for any \(X\) there is a set \(Y=F(X)=\{F(x):x\in X\}\).
- Regularity. Every nonempty set has an \(\in\)-minimal element.
Possibly we can add choice:
- Choice. Every family of nonempty sets has a choice function.
We used the informal notion of classes for practical reasons, since they can be replaced with the notion of formulas.
Russell's Paradox#
Originally, in naive set theory, people accepted, without acknowledging, the following axiom:
Axiom Schema of Comprehension. Given a property \(P\), there is a set \(Y=\{x:P(x)\}\).
Which led to Russell's Paradox: \(S = \{X:X\not\in X\}\). Problem: is it \(S\in S\) or \(S\not\in S\)? Abandoning the Schema of Comprehension, we get Schema of Separation. But it turned out that Separation Axioms are too weak to develop set theory with its usual operations and constructions. For example, they are not sufficient to prove that the union \(X\cup Y\) of two sets exists, or to define the notion of a real number.