About definition of transitive set (with sets or ur-elements)!!

--1) let $A$ a set, $A$ is transitive set if $\forall B \in A (B \subseteq A )$

--2) let $A$ a set, $A$ is transitive set if $\forall x \in A (x \subseteq A )$

I think that 2) is correct, in fact if $A:=\{\emptyset, \{\emptyset\},a,f\}$ then in case 1) $A$ is transitive set but in case 2) $A$ is not transitive set because $a,f \nsubseteq A$... is correct?

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The two definitions are exactly equivalent; they just use different names for the elements of $A$. – user61527 Aug 20 '13 at 13:47
@T.Bongers, I think that first definition is only for sets of sets.. or not? – mle Aug 20 '13 at 13:49
No, they just have different letters. – user61527 Aug 20 '13 at 13:50
In Zermelo-Fraenkel Set Theory, all objects are sets. – João Júnior Aug 20 '13 at 13:50
@JoãoJúnior, okok is true.. but if I am in ZF with ur-element? just out of curiosity... – mle Aug 20 '13 at 13:53

The two definitions are equivalent. They are also equivalent to this one:

A set $A$ is transitive if $\forall\heartsuit\in A(\heartsuit\subseteq A)$.

The only difference is what we're calling the variable ranging over $A$.

As for your particular example, it depends on what $a$ and $f$ are. In the case that $a=\bigl\{\emptyset,\{\emptyset\}\bigr\}$ and $f=a\cup\{a\},$ then $A$ is indeed transitive. It also works if $f=\bigl\{\emptyset,\{\emptyset\}\bigr\}$ and $a=f\cup\{f\}.$ It also works if $a$ and $f$ are elements of $$\Bigl\{\emptyset,\{\emptyset\},\bigl\{\emptyset,\{\emptyset\}\bigr\}\Bigr\}.$$ Otherwise, you're correct that $A$ is not transitive, regardless of what $a$ and $f$ might be (sets or ur-elements).

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okok.. thanks...!! :) ;) – mle Aug 20 '13 at 13:55
As a side observation, one can readily prove that if $A$ is a transitive set, then there are no ur-elements in $A$ (i.e.: any set with an ur-element in it is non-transitive). – Cameron Buie Aug 20 '13 at 14:00
I can edit the definition for this case "$A$ is transitive set if $A - \{B \in A|B \subseteq A\}=\emptyset$.. is correct? – mle Aug 20 '13 at 14:10
That is another perfectly fine definition, and works whether or not we have ur-elements in our theory. – Cameron Buie Aug 20 '13 at 14:11
okok thanks soo much!! :) – mle Aug 20 '13 at 14:16