# Correct use of Axiom Schema of Comprehension

• Mar 19th 2011, 05:07 PM
drgonzo
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
$\displaystyle \{ x \in A | P(x) \}$

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

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 $\displaystyle \{ x | P(x) \}$ where $\displaystyle P(x)=x \notin x$

Here is what he writes...

Quote:

We conclude this section with another notational convention.
Let $\displaystyle P(x)$ be a property of $\displaystyle x$ (and, possibly, of other parameters).
If there is a set $\displaystyle A$ such that, for all $\displaystyle x$, $\displaystyle P(x)$ implies $\displaystyle x \in A$. then $\displaystyle \{x \in A | P(X)\}$ exists, and, moreover, does not depend on A. That means that if $\displaystyle A'$ is another set such that for all $\displaystyle x$, P(x) implies $\displaystyle x \in A'$. THen $\displaystyle \{ x \in A' | P(x) \} = \{ x \in A | P(x) \}$ 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
• Mar 20th 2011, 12:58 PM
emakarov
By the ASoC, $\displaystyle x\in\{y\in A\mid P(y)\}\leftrightarrow x\in A\land P(x)$. Since $\displaystyle \forall x\,(P(x)\to x\in A)$, we have $\displaystyle x\in A\land P(x)\leftrightarrow P(x)$. One can show similarly that $\displaystyle P(x)\leftrightarrow x\in\{y\in A'\mid P(y)\}$ provided that $\displaystyle \forall x\,(P(x)\to x\in A')$.

So, suppose that $\displaystyle \forall x\,(P(x)\to x\in A)$ for some set A. For this particular A, $\displaystyle \{y\in A\mid P(y)\}$ is a set by ASoC. Moreover, for any other set A' such that $\displaystyle \forall x\,(P(x)\to x\in A')$, the set $\displaystyle \{y\in A'\mid P(y)\}$ exists by ASoC and is equal to $\displaystyle \{y\in A\mid P(y)\}$ as shown above. So, if at least one set A exists such that $\displaystyle \forall x\,(P(x)\to x\in A)$, we can introduce the notation $\displaystyle \{y\mid P(y)\}$. It means: choose some set A such that $\displaystyle \forall x\,(P(x)\to x\in A)$ and form $\displaystyle \{y\in A\mid P(y)\}$. All such sets for various A's are equal, so the result is well-defined.
• Mar 20th 2011, 02:51 PM
drgonzo
Thank you.