When Can You State that a Given Class is a Set?

Question

Just wondering if the following is of any interest (I am an amateur in these areas, so this might be so much malarkey).

Let $\mathcal X$ be a class in ZFC set theory. Let $\mathbb \Phi:\mathcal X \to \mathcal X $ be a defined bijective correspondence with the property that for every $X \in \mathcal X$, $\mathbb \Phi(X) \ne X$. Then $\mathcal X$ is a set.

Question 1: Can the above statement even be formulated in ZFC?

Question 2: If if does make sense, is it true or could it be used in an axiomatc framework?

Answer 1

First, note that the principle you're considering is false: consider the class $\mathcal{X}$ of ordinals which are either limits or successors of limits. Then there is an obvious self-bijection of $\mathcal{X}$ with no fixed points - namely, for $\alpha\in\mathcal{X}$ we send $\alpha$ to $\alpha+1$ if $\alpha$ is a limit and we send $\alpha$ to the predecessor of $\alpha$ if $\alpha$ is not a limit - but $\mathcal{X}$ is a proper class since no unbounded class of ordinals is a set.

Another example which may be easier to think about at first: let $\mathcal{A}$ be the class of all sets of the form $x\times\{0\}$ and let $\mathcal{B}$ be the class of all sets of the form $x\times\{1\}$. Their union $\mathcal{C}:=\mathcal{A}\cup\mathcal{B}$ is clearly a proper class, but by swapping $0$ and $1$ we get a self-bijection of $\mathcal{C}$ with no fixed points.

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As to expressing it appropriately, there is a serious problem: you can't quantify over class functions in set theory (that is, you can't say "for some $\Phi$"). What you can do is express the principle as a scheme: for each formula $\Phi$ defining a class function and each formula $\chi$ defining a class, we can write a sentence which says "$\Phi$ defines a fixed-point-free self-bijection of the class defined by $\chi$." However, this is enough to express what you want (ignoring its falsity for the moment), so it's not a huge problem.

Answer 2

The claim is that if $\mathcal X$ can be bijected to itself with no fixed points, $\mathcal X$ is a set. The closest we can get to formulating this in ZFC is to introduce a unary predicate $\chi(x)$ meaning $x\in\mathcal X$ and a binary predicate $\phi(x,\,y)$ meaning $\Phi(x)=y$, and write (I hope I don't mess up the details!)$$(\forall x(\chi(x)\to\forall y,\,z(\phi(x,\,y)\land\phi(x,\,z)\to y=z\ne x))\land \exists y (\chi(y)\land \phi(y,\,x)))\to(\exists z\forall x(\chi(x)\leftrightarrow x\in z)).$$(Since ZFC's objects are all sets, "this is a set" means "there's a set with the members this has".) Note each choice of the predicates $\chi,\,\phi$ has its own version of this statement; we can't quantify over them in one statement, because the logic is first-order.

So that's question 1 answered. The conjecture isn't in general true, though. For example, let's biject the class of ordinals to itself according to the following rule: write an ordinal $\alpha$ as $\gamma+n$ with $n$ finite and $\gamma$ a limit ordinal; then $\Phi(\gamma+n)=\gamma+n+(-1)^n$. This doesn't have fixed points, but the ordinals don't form a set.

Answer 3

I don't know if it would be consistent with ZFC, but you can to the best of my knowledge formulate said condition in the language of sets (or in any other language), but it would require infinitely many axioms.

A class of sets is basically a family of sets given by a formula (possibly with parameters). You basically want to consider the following axiom for any pair or two formulas (possibly involving parameters) $\phi(x)$ and $\psi(x,y,\bar{z})$, the first one having $1$ free variable and the second one $2+n$.

For every $\bar{a}$ (array of $n$ parameters), if $\psi(x,y,\bar{a})$ defines a bijection $\mathcal{X}\rightarrow\mathcal{X}$ with no fixed points, where $\mathcal{X}$ is the class defined by $\phi(x)$, then $\mathcal{X}$ is a set.

You can do this for every pair of formulas to formulate your condition.