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Math Help - Correct use of Axiom Schema of Comprehension

  1. #1
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    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?
    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
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  2. #2
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    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.
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  3. #3
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    Thank you.
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