Proving a collection is a set in ZF

Question

Suppose $C$ is a set and consider

$$ F:= \{ x : x \cup C = C \} $$

Is it true that $F$ is a set? I say yes.

My attempt:

Define $G:= \{x \in P(C) : x \cup C = C \}$

where $P(C)$ is the power set of $C$. (It is a set by the axiom of the power set, since $C$ is a set).

By the axiom of specification the latter is a set.

Notice that:

$$ F = G $$

By extension $F$ and $G$ are equal.

Now some questions:

1. It is not explicitly stated, but is every $x$ in $F$ a set? I guess the answer is yes, because under ZF union makes sense only for sets (considering pairing and union axioms)

2. The axiom of extension tells us that, provided $A$ and $B$ are sets, they are equal iff their elements are equal. Then is it correct to say that $F$ and $G$ are equal? Technically I hadn't proved $F$ is a set.

Answer 1

Technically, in ZFC, everything is a set. When we express "let $\alpha$ be an ordinal" as "let $\alpha$ be an element of the class of ordinals" or "let $\alpha \in\operatorname{On}$", then that is a priori an abuse of notation because in ZFC proper, there is no object $\operatorname{On}$ that holds all ordinals; a set $\alpha$ can only have the property of being an ordinal, not being an element of the (non-set of) ordinals. Nevertheless, speaking of (proper) classes this way and allowing them only to the right of the $\in$-relation, makes semi-formal communication a lot easier. But strictly speaking, ZFC prevents us to even define $F$ as done, i.e., we are not "allowed" to use the class-builder notation to define an object of ZFC (=a set).

In fact, all valid notations are formally only introduced because some axiom of ZFC says that a set with certain properties exists and by Extension, this set is unique: By Axiom of Pairing, given sets $A$ and $B$, the exists a (unique) set that has precisely $A$ and $B$ as element; we introduce the notation $\{A,B\}$ for this. By Axiom of Union, given a set $A$, there exists a (unique) set that has as elements precisely the elements of elements of $A$; we introduce th notation $\bigcup A$ for this. Similarly, the notations $P(A)$ and $\{\,x\in A\mid \phi(A)\,\}$ and $\{\,f(x)\mid x\in A\,\}$ are motivated by the Axioms of Power Set, of Comprehension, of Replacement. By the way, other axioms of ZFC can also be written as postulating the existence of a set with specific properties, but that set is in general not unique so that we do not introduce a notation that might confusingly pretend uniqueness. Most prominently, the Axiom of Choice grants us a choice function; there is no the choice function. Likewise, the Axiom of Foundation grants us (for nonempty $A$) an element $x\in A$ such that $x\cap A=\emptyset$; but this is not unique and hence we cannot introduce a notation a la $\operatorname{FUND}(A)$ to denote it. Even the Axiom of Extension can be formulated in a way that it postulates the existence of a set with certain properties: Given sets $A$ and $B$, there exists a set $x$ such that $A=B\leftrightarrow(x\in A\leftrightarrow x\in B)$.

So to make a more formal sense of the problem statement, one should rather ask something like:

In ZFC and given a set $C$, does there exist a set $F$ such that $$\tag1 \forall x\colon (x\in F\leftrightarrow x\cup C=C)$$ is true?

And even in this reformulation, note that "$\forall x$" means "for all sets $x$" because everything is a set in ZFC.

And now we can proceed and show that $(1)$ holds true if (and only if) $F$ is the power set of $C$.

Answer 2

Everything in ZF set theory is a set, even elements. However when you try to define a set $S := \{x : P(x)\}$ it may be the case that $S$ it's not a set, but their elements are sets (set of all sets for example).

Then, It's usefull the notion of ''class'', if $P(x)$ is a property of $x$, then ``$\langle x : x\text{ is a set and }P(x)\rangle$'' is a class.

If $C$ is a set, we encounter the class

$$F = \langle x : x \text{ is a set and } x \cup C = C\rangle$$

and we ask if $F$ is a set. We could argue that by the axiom of union $F$ can be a set, but sadly, as far as I know, and what we read in this thread, the notion of class is not part of ZF set theory, so we can't prove that a class is a set unless we construct it/derive it from the theory alone. However, I think we can do the following.

We observe that

$$B \cup A = A \iff B \subset A \; \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (I)$$

for any sets $A,B$. Then you have a characterization for your class $F$:

$$F = \langle x : x \text{ is a set and } x \cup C = C\rangle \iff F = \langle x : x \text{ is a set and } x \subset C \rangle \quad\quad\quad (II)$$

Of course here we are out of ZF, we are relying only on the logic and the property $(I)$. Finally we observe that $(II)$ is indeed a set given by the power set axiom.

Is every $x$ in $F$ a set?

If you look it as a class, yes.

Is it correct to say that $F$ and $G$ are equal?

In ZF you can't say $F = G$ unless you prove that $F$ is indeed is a set. We have not argued that, we just saw that $F$ is a class that is equal to another class that happens to be a just the same as a set (the power set of C).