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In $\S2.5$ the book writes $\{A:\varphi(A)\}$ is translated:$\{x:x\ is\ a\ set\&\varphi(x)\}$.But with this translation it seems impossible to prove theorem 86:$B\in\mathscr{P}A\leftrightarrow B\subseteq A$.

I can't prove sufficiency. Here's my attempt:

  1. By hypothesis $B\subseteq A$.
  2. By power set axiom there is a set, say B*, $\forall x(x\ is\ a\ set\rightarrow(x\in B^*\leftrightarrow x\subseteq A))$.
  3. By definition of power set and translation of $\{A:\varphi(A)\}$,$\mathscr{P}A=\{x:x\ is\ a\ set\&x\subseteq A\}$

And according to definition by abstraction, if I want to get $\forall x (x\in \mathscr{P}A\leftrightarrow x\ is\ a\ set\&x\subseteq A)$ I have to prove$\exists B\forall x(x\in B\leftrightarrow x\ is\ a\ set\&x\subseteq A)$. The only way seems to be using the formula $\forall x(x\ is\ a\ set\rightarrow(x\in B^*\leftrightarrow x\subseteq A))$, but this doesn't entail $\exists B\forall x(x\in B\leftrightarrow x\ is\ a\ set\&x\subseteq A)$ and I'm stucked here.

I suspect $\{A:\varphi(A)\}$ should be translated as$\{x:x\ is\ a\ set\rightarrow\varphi(x)\}$, that way, the proof will be simple.

joggingrat
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  • Suppes defines ${\mathbf{x}|\varphi(\mathbf{x})}=\mathbf{y}\leftrightarrow \forall \mathbf{x}(\mathbf{x}\in \mathbf{y}\leftrightarrow \varphi(x))\wedge \mathbf{y}\ is\ a\ set\vee(\mathbf{y}=0\wedge\neg\exists \mathbf{B}\forall \mathbf{x}(\mathbf{x}\in \mathbf{B}\leftrightarrow \varphi(x)))$, so I don't have $\varphi(B)\rightarrow B\in{x|\varphi(x)}$, take $\varphi (x)$ be $x=x$ will cause a contradiction. – joggingrat Sep 11 '24 at 14:50

1 Answers1

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The key property of the set-builder operator (aka: definition by abstraction) is: $B ∈ \{ x ∣ φ(x) \} ↔ φ(B)$.

Thus, from $B ⊆ A$ we get $B ∈ \{ x ∣ x⊆A \}$.

You can see page 20 for the use of capital variables: to say "A is..." means "if y is a set then ...".

Thus $B ⊆ A$ is "$\text {x is a set and y is a set } → ∀z(z∈x → z∈y)$".

Mauro ALLEGRANZA
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