Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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}

then:

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

Is this true?

Thanks

share|improve this question

1 Answer 1

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 $x=\varnothing$.

share|improve this answer
    
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. –  Fiire 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. –  Fiire Sep 27 '12 at 0:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.