For which P can one claim the existence of the set of all sets that satisfy P?

Question

Given a property P, is there some rules that are sufficient or necessary to determine if there exists a set of all sets with property P?

Answer 1

In ZFC, the following are equivalent:

1. The class $\{x\mid P(x)\}$ is a set.
2. There is an ordinal $\alpha$ such that whenever $P(x)$, then $x$ has rank at most $\alpha$.
3. There is a set $B$ such that whenever $P(x)$, then $x\in B$.
4. There is no class function $F$ from the class $\{x\mid P(x)\}$ onto the ordinals.
5. There is an ordinal $\theta$ such that there is no function mapping $\{x\mid P(x)\}$ onto $\theta$.
6. There is a set $C$ such that there is no function mapping $\{x\mid P(x)\}$ onto $C$.
7. There is an ordinal $\theta$ that does not map injectively into $\{x\mid P(x)\}$.
8. There is a set $D$ that does not map injectively into $\{x\mid P(x)\}$.

Proof. (1 iff 2) If the class is a set, then it must be contained in some $V_\alpha$, and so every element will have rank at most $\alpha$. The converse is the Separation axiom.

(2 iff 3) Use that every $V_\alpha$ is a set, and every set is contained in some $V_\alpha$.

(1 implies 4) The ordinals are not a set, so this follows by the Replacement axiom.

(4 implies 2) Map each $x$ for which $P(x)$ holds to its rank.

(1 implies 5) For every set, there is an ordinal onto which it does not map, namely, the successor of its cardinality.

(5 iff 6) Every set is bijective with an ordinal.

(5 implies 7) If a class does not map surjectively onto $\theta$, then $\theta$ cannot map injectively into the class.

(7 iff 8) Every set is bijective with an ordinal.

(7 implies 2) If $\theta$ does not map injectively into $\{x\mid P(x)\}$, then that class cannot contain sets of arbitrarily large rank.

QED

Meanwhile, the following notions are strictly weaker in ZFC, if ZFC is consistent:

• There is a map from the ordinals onto $\{x \mid P(x)\}$.
• There is a bijection of $\{x\mid P(x)\}$ with the ordinals.
• There is a bijection of${x\mid P(x)}$ with $V$, the entire set-theoretic universe.

There reason that they are weaker in general is that it is relatively consistent with ZFC that there is no definable (from parameters) well-ordering of the universe. In such a model $V$, there is no class surjection or bijection from the ordinals to $V$, since this would provide the desired well-ordering, but $V$ is not a set. Similarly, in such a model, there is no bijection from the class of ordinals to the entire universe, but the class of ordinals is not a set. Such a model can be constructed using the forcing technique, by an Easton support iteration that adds a Cohen subset to unboundedly many regular cardinals.

Addendum. Let me add that there can be no purely syntactic characterization of the properties $P$ for which $\{x\mid P(x)\}$ is a set. The reason is that some properties determine sets in some models of ZFC, but not in others. So the question of whether this class is a set depends not just on the syntactic features of $P$, but on the properties of the universe in which the class is to be formed. An example of this is the property $P(x)\iff CH\wedge x=x$, which determines a set just in case $\neg CH$.

Answer 2

If the property $\mbox{P}$ is expressible by an equivalent first-order formula $\varphi(x)$ of set theory in the free variable $x$, an exact (i.e. sufficient and necessary) condition for the existence of some set $A$ of all sets $x$, which satisfy $\mbox{P}$ in $\mbox{ZFC}$, i.e. for

$\models_{ZFC} \exists A \forall x (x \in A \leftrightarrow \varphi(x))$,

is by some axiom of separation the condition

$\models_{ZFC} \exists B \forall x (\varphi(x) \rightarrow x \in B)$.

I.e. you need only by the completeness theorem of first-order logic to

1. express $\mbox{P}$ as an equivalent first-order formula $\varphi(x)$ of set theory,
2. find some suitable set $B$, and
3. prove $\vdash_{ZFC} \forall x (\varphi(x) \rightarrow x \in B)$ in your metatheory.