Correct use of Axiom Schema of Comprehension

**drgonzo** · March 19th 2011, 05:07 PM

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?
ZFC and Russell's Paradox
[Russell's Paradox] Help with Notation

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

**emakarov** · March 20th 2011, 12:58 PM

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.

**drgonzo** · March 20th 2011, 02:51 PM

Thank you.

Thread: Correct use of Axiom Schema of Comprehension

LinkBack

Thread Tools

Display

Correct use of Axiom Schema of Comprehension

Similar Math Help Forum Discussions

Set Term Comprehension

[SOLVED] Probability of getting at least one correct answer, with 2 out of 5 correct options

comprehension -/sinx-cosx/=√1-sin2x

Bounded Comprehension Principle

Prove use of axiom