Category:Axioms/Zermelo-Fraenkel Axioms

From ProofWiki
Jump to navigation Jump to search

This category contains axioms related to Zermelo-Fraenkel Axioms.
Related results can be found in Category:Zermelo-Fraenkel Axioms.


The Axiom of Extension

Let $A$ and $B$ be sets.

The axiom of extension states that:

$A$ and $B$ are equal

if and only if:

they contain the same elements.


That is, if and only if:

every element of $A$ is also an element of $B$

and:

every element of $B$ is also an element of $A$.


This can be formulated as follows:

$\forall x: \paren {x \in A \iff x \in B} \iff A = B$


The Axiom of the Empty Set

$\exists x: \forall y \in x: y \ne y$


The Axiom of Pairing

For any two sets, there exists a set to which only those two sets are elements:

$\forall a: \forall b: \exists c: \forall z: \paren {z = a \lor z = b \iff z \in c}$


The Axiom of Specification

For any well-formed formula $\map P y$, we introduce the axiom:

$\forall z: \exists x: \forall y: \paren {y \in x \iff \paren {y \in z \land \map P y} }$

where each of $x$, $y$ and $z$ range over arbitrary sets.


The Axiom of Unions

For every set of sets $A$, there exists a set $x$ (the union set) that contains all and only those elements that belong to at least one of the sets in the $A$:

$\forall A: \exists x: \forall y: \paren {y \in x \iff \exists z: \paren {z \in A \land y \in z} }$


The Axiom of Powers

For every set, there exists a set of sets whose elements are all the subsets of the given set.

$\forall x: \exists y: \paren {\forall z: \paren {z \in y \iff \paren {w \in z \implies w \in x} } }$


The Axiom of Infinity

There exists a set containing:

$(1): \quad$ a set with no elements
$(2): \quad$ the successor of each of its elements.

That is:

$\exists x: \paren {\paren {\exists y: y \in x \land \forall z: \neg \paren {z \in y} } \land \forall u: u \in x \implies u^+ \in x}$


The Axiom of Replacement

For any function $f$ and subset $S$ of the domain of $f$, there is a set containing the image $\map f S$.

More formally, let us express this as follows:


Let $\map P {x, z}$ be a propositional function, which determines a function.

That is, we have:

$\forall x: \exists ! y : \map P {x, y}$.

Then we state as an axiom:

$\forall A: \exists B: \forall y: \paren {y \in B \iff \exists x \in A : \map P {x,y}}$


The Axiom of Foundation

For all non-empty sets, there is an element of the set that shares no element with the set.

That is:

$\forall S: \paren {\paren {\exists x: x \in S} \implies \exists y \in S: \forall z \in S: \neg \paren {z \in y} }$

The antecedent states that $S$ is not empty.