Set Containing Itself: X and {X}

[Outlined]

A set is not allowed to has itself as a member: $$X \notin X$$. But I wonder if this is allowed: $$\{X\} \in X$$.

 

[CRGreathouse]

A set is not allowed to has itself as a member: $$X \notin X$$. But I wonder if this is allowed: $$\{X\} \in X$$.

Neither are allowed in ZF. But there are theories in which both are allowed -- for example, ZF without the Axiom of Foundation.

 

[disregardthat]

It's not allowed since the set {{X},X} have no disjoint element of itself.

 

[disregardthat]

Neither are allowed in ZF. But there are theories in which both are allowed -- for example, ZF without the Axiom of Foundation.

Is it possible to construct such a set in ZF without Axiom of Foundation?

 

[Hurkyl]

No -- such a construction would amount to a theorem of anti-foundation.

Specific axioms of anti-foundation may provide constructions, however.

 

[JeremyHorne]

The confusion here, I think, is in the designator "X", where this letters is trying to stand for both an element as well as a set. {X} in X means the element X as a set is found within the set, which violates the axiom of regularity, otherwise called "axiom of foundation", designed to prevent such expressions.