What is the Inductive Property in Set Theory?

[Dragonfall]

A set X is *this* iff whenever x is a subset of X, then x is also an element of X. Note that this is not transitivity.

 

[Preno]

Are you sure such a set exists? Wouldn't you have an injection from the class of ordinals to X if it had this property, like 0->x,1->{x},... ?

 

[Dragonfall]

EDIT: Yes, you're right. Nevermind, this turned out to be power-set closure.

 

[0xDEADBEEF]

I think your set is empty.
1.) Let's discount subsets of one element, So if your set contains 0. Then it doesn't need to contain {0}, {{0}}, {{{0}}}, as well as {}, {{}}, {{{}}}...
2.) Now we can limit ourselves to subsets of order 2. We start with two elements a and b. This produces a chain of sets of type {a,b}, {a,{a,b}} {a,{a,{a,b}}} well map these onto \mathbb{N}
3.) We will now show that the cardinality of our set is undefined.
a) The cardinality of the set of all of an infinit set's subsets is one cardinality higher than that of the set.
b) 1 is part of the set and {1,2} is part of the set as well as {1,{2,3}} and {1,{2,{3,4}}} are part of the set; and we can go on producing chains of arbitrary elements with this brace pattern, so the set contains all subsets of \mathbb{N} (its powerset) but by the same argument it contains the powerset's powerset, and so on.

 

[Elucidus]

I believe the closest thing at the axiomatic level is "inductuve"

Since \mathbb{N} = \{ , \{\}, \{ , \{\}\}, \dots \}<br />

--Elucidus