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Using the axioms of $ZF$, you ensure that from a set or multiple sets, you can also create a set. However, the question that arose in my mind is whether all subsets of this set that was created are also sets. In other words, is there a proof that every class $A$, where $A\subset C$and $C$ is a set, is also a set?

If A is defined by a formula $[ A = \{ x \in C : P(x) \} ]$, then by the axiom of schema, it becomes easy to infer that A is a set. But, we do not know if any subset of C can be defined by a formula.

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  • $\begingroup$ By definition, a set $A$ is a subset of $B$ iff for all $x$, $x\in A\implies x\in B$. Separation tells you certain things are sets, the definition tells you if they are subsets of a given set. $\endgroup$ Feb 23 at 5:06
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    $\begingroup$ There is no such thing as the "axiom of schema". $\endgroup$ Feb 23 at 5:07
  • $\begingroup$ Technically, when we stick to ZF only, we have no concept of any object that is not a set, but of course there's NBG and so on. $\endgroup$ Feb 23 at 5:18

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If $C$ is a class, $A$ is a set, and for all $x$, $x\in C\implies x\in A$, then $C$ is a set. But I do not follow your first paragraph...

Namely, let $D$ be defined by $$ D=\{x\in A\mid x\in C\}.$$

By the Axiom Schema of Separation, $D$ is a set (since $A$ is a set by assumption). I claim that $C=D$, and therefore $C$ is also a set.

Indeed, if $x\in D$ then $x\in C$. So $D\subseteq C$.

Conversely, if $x\in C$, then since $C\subseteq A$, and $x\in C$, then $x\in A$. Thus, $x\in A$ and $x\in C$, so $x\in D$. So $C\subseteq D$.

Therefore, $C=D$, and since $D$ is a set, then $C$ is a set.

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  • $\begingroup$ I didn't expect the proof to be this simple, thank you Mr Arturo. $\endgroup$ Feb 23 at 6:01
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Reading between the lines somewhat, it seems that the question you are really interested in, would be rather something like this: Suppose $(M,\varepsilon)$ is a model of set theory, $a\in M$ is a set and $C\subseteq M$ is some subset of the model with the property that each (actual) element of $C$ is an ($\varepsilon$-)element of $a$. Is $C$ then necessarily "realized" in $M$ as a subset of $a$, in that there is some $c\in M$ such that the $\in$-elements of $C$ are exactly the $\varepsilon$-elements of $c$?

The answer to this (interpretation of your) question is actually no, not necessarily: If ZFC is consistent, it has a countable model $M$. Hence, there are only countably many actual subsets of (what the model thinks is) $\mathbb{N}$ that are realized as sets in $M$.

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By definition, a class is a collection of sets given by a first order formula. If $A$ is given by the formula $\phi$ and $C$ is already a set, you can use the axiom (schema) of specification to prove that $A$ is a set. That's precisely what the axiom was invented for.

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    $\begingroup$ I do not disagree with you on this. When set A is defined by a formula, its proof becomes easy. However, we do not know if any subset of a set can be defined by a formula, and if not, I would appreciate your clarification. $\endgroup$ Feb 23 at 4:44
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If $A\subseteq B$, then $A\cap B = A$. The LHS is a set by separation and $B$ being a set, so the RHS is also set.

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