In Jech's Introduction to Set Theory he writes

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 $A$. What is the meaning of this statement. specifically the does not depend on $A$ part.

Acc. to me such a case could happen if $A$ is not the parameter in the statement but from what I understand $x$ is not the parameter here because the statement says for all $x$.

I am confused can somebody clear this up

  • 1
    $\begingroup$ In this context, $\{x\in A|P(x)\}$ is the set of all $x$ such that $P(x)$. If $B$ is any other set such that $P(x)\Rightarrow x\in B$, then $\{x\in A:P(x)\}=\{x\in B:P(x)\}$. $\endgroup$ – Renan Maneli Mezabarba Mar 16 '17 at 16:12
  • $\begingroup$ but why does this not depend on $A$ $\endgroup$ – arrhhh Mar 16 '17 at 16:13
  • $\begingroup$ It only depends on the existence of some such $A$, not on the specific $A$. $\endgroup$ – Fabio Somenzi Mar 16 '17 at 16:19
  • $\begingroup$ Because as Maneli Mezbarba said, this works for any set $B$ with the required property $\endgroup$ – Max Mar 16 '17 at 16:20

In this context, $\{x\in A|P(x)\}$ is the set of all $x$ such that $P(x)$, that we may write as $\{x|P(x)\}$.

Indeed, one clearly has $\{x\in A|P(x)\}\subset \{x|P(X)\}$. On the other hand, if $P(x)$ holds, then your hypothesis implies that $x\in A$, hence $x\in A$ and $P(x)$, proving the reverse inclusion.

For instance, take $P(x)$ as "$x\subset y$", for a fixed set $y$. Then the axiom of power set says that there exists a set $A$ such that $x\subset y\Rightarrow x\in A$, and one defines $\wp(y)=\{x\in A:x\subset y\}$. However, any other set $B$ such that $x\subset y\Rightarrow x\in B$ yields $\wp(y)=\{x\in B:x\subset y\}$.

The requirement of a set $A$ such that $P(x)\Rightarrow x\in A$ is here to avoid "Russell's paradox" situations.

  • $\begingroup$ what I have understood from the example and please correct me if i am wrong is that - $A $ and $B$ are merely notation as by the power set axiom there will always be some set to which $x$ will belong whether I write that as $A$ or $B$ doesnt matter. $\endgroup$ – arrhhh Mar 16 '17 at 16:44
  • $\begingroup$ in the second paragraph you have proved is that $ \{x \in A|P(x)\} = \{x|P(x)\}$ by saying $A = B$ if $ A \subset B$ and $ B \subset A $ $\endgroup$ – arrhhh Mar 16 '17 at 16:48
  • $\begingroup$ For your first comment: if I understood what you wrote, then you're right, it doesn't matter if you take $A$ or $B$, as long as they satisfy the condition ``if $P(x)$ then $x$ is a member of it''. For your second comment: you are right, I used the axiom of extension. $\endgroup$ – Renan Maneli Mezabarba Mar 16 '17 at 19:16

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