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We have that any subset of P(A) is transitive.

And if {x} ∈ P(A) then {x} is a transitive set, then are these following steps true?

∴ x ∈ {x}
∴ x ⊆ {x}


if {x} ∈ P(A)
∴ x ∈ P(A)
since x ⊆ {x} ∈ P(A)

Is this true?


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up vote 2 down vote accepted

I'm not sure what you are trying to prove, however I do need to point out the following:

Claim: $x\subseteq\{x\}$ if and only $x=\varnothing$.

Proof. If $x=\varnothing$ then this is trivial. On the other hand if $x\subseteq\{x\}$ then either $x=\varnothing$ or for every $y\in x$, $y\in\{x\}$. From this follows that $y=x$ and therefore $x\in x$ which is a contradiction to the axiom of regularity. Therefore $x\subseteq\{x\}$ implies $x=\varnothing$.

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Yup, I was trying to prove that if all the subsets of A are transitive, then A itself is also transitive. I knew that the only power of a subset in which all of its element are transitive themselves is P={ø,{ø}} ; which is the power of A={ø}. The problem was that I didn't know for sure if x=ø. Thanks. – OFRBG Sep 26 '12 at 23:42
@Fiire: Note the triviality of what you wished to have shown, if all subsets of $A$ are transitive then in particular $A$ itself is transitive, since $A\subseteq A$. Indeed as you wrote, and I pointed out in my answer, this is true only for $\varnothing$ and $\{\varnothing\}$. – Asaf Karagila Sep 26 '12 at 23:46
Yes, I know it's trivial since any set is subset of itself, but I'm more on the side of proving things through unconventional and orthodox means. But, thanks s lot. – OFRBG Sep 27 '12 at 0:19

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