0

In the Subset Axiom:

Let $\phi(x)$ be a formula. For every set $A$ there exists a set $S$ that consists of all the elements $x \in A$ such that $\phi(x)$ holds.

the formula $\phi(x)$ contains only one free variable ($\phi(x)$ could contain other bounded variables and parameters). Consider the following statement where $\phi$ is now a formula of two free variables $\phi(x, y)$.

Let $\phi(x, y)$ be a formula. For every set $A$ there exists a set $S$ that consists of all the elements $x \in A \text{ and } y \in A$ such that $\phi(x, y)$ holds.

Will the axiom and the above statement be compatible with each other? By compatible I mean no contradiction arises when both are true.

Tran Khanh
  • 546

1 Answers1

0

The statement can be shown to be true based on the Subset Axiom. We can write the collection $S$ in the statement in set notation as follows:

$S=\{x\in A, y \in A: \phi(x, y)\}$

To show that $S$ is a set, observe that:

$ \begin{align} \{x\in A, y \in A: \phi(x, y)\}&=\{x: (\exists{y}\in A)\phi(x, y)\} \bigcup \{y: (\exists{x}\in A)\phi(x, y)\}\\ &=\{\exists{y}\in A: \phi^{'}(x)\} \bigcup \{y\in A: \phi^{''}(y)\}\\&(\text{where }\phi^{'}(x)=(\exists{y}\in A)\phi(x, y)\text{ and }\phi^{''}(y)=(\exists{x}\in A)\phi(x, y))\\&=(\exists{x}\in A)\phi(x, y) \end{align} $

Since both $\{x \in A: \phi^{'}(x)\}$ and $\{y \in A: \phi^{''}(y)\}$ are sets according to the Subset Axiom, and the union of them is also a set. So, $S$ is a set.

Tran Khanh
  • 546