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
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
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 $x=\varnothing$.