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I have two questions regarding this issue.

  1. In addition to Axiom Schema of Replacement, Which axioms do we need to deduce Axiom of Separation?

  2. How to deduce Axiom of Separation from Axiom Schema of Replacement?

Axiom of Separation: $\forall w \forall x \exists y \forall z [z \in y \iff (z \in x \land \varphi(z, w, x ) )]$

Axiom Schema of Replacement:$\forall w \forall A [ (\forall x \in A \implies \exists ! y \varphi(x,y,w, A)) \implies (\exists B \forall x (x \in A \implies \exists y \in B \varphi(x,y,w, A)))]$

Many thanks!

Andrés E. Caicedo
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Akira
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1 Answers1

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Hint : consider $\psi(x,y, w, A) : x=y\land x\in A \land \phi(x, w, A)$

Maxime Ramzi
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  • Hi @Max, I can interpret the role of $x$ as $\psi(x,y, w, A)$ holds true for $x, y, w$. Can you interpret the role of A in $\psi(x,y, w, A)$? Is it redundant to include both $x$ and $A$ in $\psi(x,y, w, A)$, since $x$ is clearly in $A$. – Akira Dec 23 '17 at 02:52
  • Yeah of course I introduced some redundancies – Maxime Ramzi Dec 23 '17 at 09:24
  • But from https://math.stackexchange.com/questions/932781/definability-and-the-separation-and-replacement-axiom-schemata, and from https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory#6._Axiom_schema_of_replacement, they both include $\color{blue}{A}$ into $\varphi(x,y,\vec{w}, \color{blue}{A})$! Please clarify my doubt! – Akira Dec 23 '17 at 09:36
  • Sorry I thought you meant when I wrote $x\in A$. For the $A$ in the formula, it's surely not dedundant, but finding examples where it isn't may be tough. By the way I just took a look at your axiom of replacement and there's something wrong with it : it shouldn't be "$\forall x\in A, \exists ! y$ blabla" but "for all $x\in A$, there's at most one $y$ blabla" – Maxime Ramzi Dec 23 '17 at 09:42
  • So any statement of Axiom Schema of Replacement that uses $\exists ! y$ is wrong? As a result, the statements of Axiom Schema of Replacement in two links, which I have given above, are wrong two? Is there any symbols that stand for "there's at most one y"? – Akira Dec 23 '17 at 09:47
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    Well it's not wrong, it's another version; less useful in practice. Now that I think about it, with a slight adjustment you can still deduce the axiom schema of separation from it, but not as easily as what I did (there must be a case distinction). There's no designated "consensual" symbol. Sometimes people use $\exists^{\leq 1} y$ (which is an abbreviation, like $\exists !$) – Maxime Ramzi Dec 23 '17 at 09:59