# Thread: Correct use of Axiom Schema of Comprehension

1. ## Correct use of Axiom Schema of Comprehension

Hello all

I'm going through the book "introduction to set theory" by Hrbace. on page 11 Hrbacek introduces another way on defining a set that builds on the 'Axiom Schema or Comprehension' method
$\{ x \in A | P(x) \}$

I'll post what he writes to explain his method... but to summarise he explains that
$\{ x | P(x) \}$ can be used as long as $P(x)$ implies $x \in A$

Now when he first posted this I was unaware as the point of this. I started 'googling' axiom scheme of comprehension and came across these three posts...

Russell's paradox: how is it resolved?

From here I learnt that the ASoC was developed by ZFC to get around this Russel Paradox. That is $\{ x | P(x) \}$ where $P(x)=x \notin x$

Here is what he writes...

We conclude this section with another notational convention.
Let $P(x)$ be a property of $x$ (and, possibly, of other parameters).
If there is a set $A$ such that, for all $x$, $P(x)$ implies $x \in A$. then $\{x \in A | P(X)\}$ exists, and, moreover, does not depend on A. That means that if $A'$ is another set such that for all $x$, P(x) implies $x \in A'$. THen $\{ x \in A' | P(x) \} = \{ x \in A | P(x) \}$ Prove It
I don't follow his line of reasoning on why these sets are the same and why they exist and dont depend on their set A or A'.

Dylan

2. By the ASoC, $x\in\{y\in A\mid P(y)\}\leftrightarrow x\in A\land P(x)$. Since $\forall x\,(P(x)\to x\in A)$, we have $x\in A\land P(x)\leftrightarrow P(x)$. One can show similarly that $P(x)\leftrightarrow x\in\{y\in A'\mid P(y)\}$ provided that $\forall x\,(P(x)\to x\in A')$.

So, suppose that $\forall x\,(P(x)\to x\in A)$ for some set A. For this particular A, $\{y\in A\mid P(y)\}$ is a set by ASoC. Moreover, for any other set A' such that $\forall x\,(P(x)\to x\in A')$, the set $\{y\in A'\mid P(y)\}$ exists by ASoC and is equal to $\{y\in A\mid P(y)\}$ as shown above. So, if at least one set A exists such that $\forall x\,(P(x)\to x\in A)$, we can introduce the notation $\{y\mid P(y)\}$. It means: choose some set A such that $\forall x\,(P(x)\to x\in A)$ and form $\{y\in A\mid P(y)\}$. All such sets for various A's are equal, so the result is well-defined.

3. Thank you.