system
Given the following post, respond as if you're commenting on Reddit:
human
Is it possible to powerset a set infinitely many times?

and if so does that mean i have found a set so big that if i powerset it again it will still be the same set? because powersetting a set that have been powersetted infinitely many times one more time will not make it any bigger (because infinity+1=infinity duh!)
gpt
No such set exists.

Suppose that we were given a set *A* such that *X* = P(P(P(...P(A)...))) exists. In fact, let X*_0_* = X and let X*_1_* be the set such that P(X*_1_*) = X*_0_*. Generalize this: define a recursive sequence X*_n_* such that P(X*_n+1_*) = X*_n_*. So we are considering the sequence of sets:

> (X*_0_*, X*_1_*, ..., X*_n_*, ...)

Any set *S* is a subset of itself, hence *S* is always an element of P(S). This means that X*_n+1_* is an element of X*_n_*, for all *n*. Thus our sequence is an infinitely descending chain of sets. This is not allowed by the axiom of regularity (also called the axiom of foundation).

> and if so does that mean i have found a set so big that if i powerset it again it will still be the same set?

Even if you were allowed to make sense of X = P(P(P(...P(A)...))) and this set existed, the axioms would certainly not imply that P(X) = X. For one, the power set P(S) always has strictly larger cardinality than S, for any set S. ~~Secondly, if P(S) = S, then this means that any element *x* in S would also be a subset of *S*, by definition of power set. Thus the set *S* would have non-empty intersection with all of its (non-empty) elements, again contradictory to the axiom of regularity.~~ (There is an error in the crossed-out text. See a follow-up below.)