Is the isomorphism class of a fixed cardinality a set(not a proper class)? Or a fixed ordinality for that matter?
By "isomorphism" I mean just bijection for cardinals and order preserving bijection for ordinals, in the category of sets.



Is the isomorphism class of a fixed cardinality a set?
The isomorphism classes of the set {} and the ordinal 0 are sets, each with a single element.


let $\kappa$ be an non-zero ordinal and $\alpha$ # $\beta$ be the [natural sum](http://en.wikipedia.org/wiki/Ordinal_arithmetic#Natural_operations) of the ordinals $\alpha$ and $\beta$. Since the natural sum is the restriction of the addition of [surreal numbers](http://en.wikipedia.org/wiki/Surreal_numbers) to ordinals, and the surreal numbers are an [ordered field](http://en.wikipedia.org/wiki/Ordered_field), the natural sum is cancellative and strictly increasing. Therefore, for any ordinal $\gamma$, the set {$\gamma$#0, $\gamma$#1, ... $\gamma$#$\kappa$} of ordinals that are the natural sum of $\gamma$ and an ordinal less than $\kappa$ is order-isomorphic to $\kappa$. These sets are different for different values $\gamma$, so this provides a bijection between the ordinals and a subclass of the isomorphism class of $\kappa$. If the isomorphism class of $\kappa$ was a set, this subclass would be a set and therefore the ordinals would also form a set. The ordinals cannot form a set, therefore the isomorphism class of $\kappa$ does not form a set.


Assuming the axiom of choice, every set is well-orderable. Given a well-ordering of a set there is a bijection between the order-isomorphism class of that set and a subclass of it's isomorphism(bijection) class. if the set is non-empty, then since it's order-isomorphism class is not a set it's isomorphism class is not a set either.


This can also be proved without the axiom of choice. Suppose that a set x is bijectible
with {y}$\sqcup$ z, where $\sqcup$ means disjoint union, for two sets y and z. Then for any set w, x is bijectible with {w}$\sqcup$z , and since these sets are different for different values of w, this gives a bijection of the class of all sets with a subclass of the isomorphism class of x. Since the class all sets cannot be a set, the isomorphism class of x cannot be a set. It remains to be shown that any non-empty set x is bijectible with {y}$\sqcup$z for two sets y and z. to see this, simply choose one element v from x (this does not require the axiom of choice, because we are only making one choice of one element from one set) and write x as the union of {v} and the set of other elements of x. these two sets are disjoint, so the obvious surjection from their disjoint union to their union is a bijection. The union is x, so v and the set of other elements of x are the y and z we were looking for, respectively.


