# Equivalence of Formulations of Axiom of Unions/Proof

## Theorem

In the context of class theory, the following formulations of the Axiom of Unions are equivalent:

### Formulation 1

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 $A$:

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

### Formulation 2

Let $x$ be a set (of sets).

Then its union $\bigcup x$ is also a set.

## Proof

It is assumed throughout that the axiom of extensionality and the axiom of specification both hold.

### Formulation $1$ implies Formulation $2$

Let formulation $1$ be axiomatic:

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 $A$:

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

Thus it is posited that for a given set of sets $A$ the union of $A$ exists:

$x := \bigcup A := \set {y: \exists z \in A: y \in z}$

and that this is a set.

Formulation $2$ asserts that given the existence of $\bigcup A$, it is axiomatic that $\bigcup A$ is itself a set.

Hence it follows that the truth of formulation $2$ follows from acceptance of the truth of formulation $1$.

From Class Union Exists and is Unique, which depends on:

the axiom of extensionality
the axiom of specification

$\bigcup A$ is unique for a given $A$.

$\Box$

### Formulation $2$ implies Formulation $1$

Let formulation $2$ be axiomatic:

Let $x$ be a set (of sets).

Then its union $\bigcup x$ is also a set.

Let $A$ be a set of sets.

Let us create the class:

$x := \bigcup A := \set {y: \exists z \in A: y \in z}$

From Class Union Exists and is Unique, which depends on:

the axiom of extensionality
the axiom of specification

$\bigcup A$ exists and is unique for a given $A$.

We have asserted the truth of formulation $2$.

That is, $\bigcup A$ is a set.

As $A$ is arbitrary, it follows that $\bigcup A$ exists and is unique for all sets of sets $A$.

That is, the truth of formulation $1$ follows from acceptance of the truth of formulation $2$.

$\blacksquare$
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