# Is every subset of a set also a set?

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.

• 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. Feb 23 at 5:06
• There is no such thing as the "axiom of schema". Feb 23 at 5:07
• 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. Feb 23 at 5:18

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.

• I didn't expect the proof to be this simple, thank you Mr Arturo. Feb 23 at 6:01

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 $$M$$.

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.

• 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. Feb 23 at 4:44

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.