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Thread: Correct use of Axiom Schema of Comprehension

  1. #1
    Feb 2011

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

    Here is what he writes...

    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'.

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  2. #2
    MHF Contributor
    Oct 2009
    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.
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  3. #3
    Feb 2011
    Thank you.
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