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

---let $A$ a set, $A$ is inductive if $\emptyset \in A$ and $\forall x \in A (x^+ \in A)$

---let $A$ a set, $A$ is inductive if $\emptyset \in A$ and $\forall Y \in A (Y^+ \in A)$

an example of inductive set is $\{\{\},\{\}^+,\{\}^{+^+},...\{\}^{+^+{^{...{^+}}}},...\}$, but it is inductive set for first and second definition, but if I have $\{a,b,c,\{\},\{\}^+,\{\}^{+^+},...\{\}^{+^+{^{...{^+}}}},...\}$ with $a,b,c$ ur-elements then it is a inductive set for second definition and is not a inductive set for first definition...

therefore, in definition of inductive set $A$ must be a pure set, or not?

edit: the variable $x$ refers both ur-elements and sets, but the variable $Y$ refers only sets
The two definitions, as written, are $\alpha$-equivalent: all you've done is rename a bound variable (from $x$ to $Y$). Did you mean something else? [Edit: and the same goes for your other question.] – Clive Newstead Aug 20 '13 at 21:45
How is $x^+$ defined when $x$ is an ur-element? – Cameron Buie Aug 20 '13 at 21:47
@CameronBuie, in fact for first definition $\{a,b,c,\{\},\{\}^+,\{\}^{+^+},...\{\}^{+^+{^{...{^+}}}},...\}$ is not inductive set.. is not defined $a^+,b^+,c^+$ with $a,b,c$ ur elements :) – mle Aug 20 '13 at 21:49
If $x^+$ is not defined for urelements $x$, and yet the variable $x$ can refer to urelements, then the first formula is not even well-defined. [FWIW, it would have been good to specify your convention on variable-naming in the question!] – Clive Newstead Aug 20 '13 at 22:03
As far as I can see (or guess), you are working with a convention (not mentioned in the question but suggested in a comment) that capital letters are variables ranging over sets and lower-case letters range over everything, including urelements as well as sets. Your first "definition" uses the notation $x^+$, so it would make sense only if the + operation were defined for urelements as well as sets. But you said in another comment that it isn't defined for urelements. So I conclude that the first definition doesn't make sense. – Andreas Blass Aug 20 '13 at 22:16