A particular class which satisfies all ZF axioms except for the axiom of infinity

Suppose we have a non-empty transitive class $$\mathcal{C}$$, meaning that if $$x$$ is in the class, then all its elements are also in the class. Suppose also that $$\mathcal{C}$$ satisfies the axiom schema of specification and the following property: If a set $$x$$ is such that all its elements are elements of $$\mathcal{C}$$, then there exists an element $$y$$ in the class $$\mathcal{C}$$ such that $$x\subseteq y$$.

I have to show that $$\mathcal{C}$$ satisfies all ZF axioms except for the axiom of infinity. Also, I have to find an example of such a class $$\mathcal{C}$$ such that the axiom of infinity is not satisfied.

I could prove that such a class $$\mathcal{C}$$ satisfies all ZF axioms except for the axiom of infinity. It is not very hard. For example, for the axiom of union, I just take the usual union and show that it is contained in $$\mathcal{C}$$, so I should have that this union is contained in some element of $$\mathcal{C}$$, and then by specification the union is an element of $$\mathcal{C}$$. The other axioms are similar to prove.

So, it only remains to find a class $$\mathcal{C}$$ which does not satisfy the axiom of infinity. We have the functional relation $$V$$ inductively defined as $$V_{\alpha}=\bigcup_{\beta \in \alpha}\mathcal{P}\left (V_{\beta}\right )$$ for every ordinal $$\alpha$$, where $$V_{\emptyset}=\emptyset$$. It is known that if $$\alpha$$ is a limit ordinal, then $$V_{\alpha}$$ satisfies all ZFC axioms except (perhaps) for the axiom of infinity, which is satisfied if and only if $$\omega \in \alpha$$. Therefore, I have the candidate $$V_{\omega}$$, which satisfies all ZF axioms except for the axiom of infinity, which is not satisfied because $$\omega \not\in \omega$$. Of course, $$V_{\omega}$$ is transitive, but I could not prove that it satisfies that if a set $$x$$ is such that all its elements are elements of $$V_{\omega}$$, then there exists an element $$y\in V_{\omega}$$ such that $$x\subseteq y$$ (indeed, I think that assertion is false).

Which $$\mathcal{C}$$ would you take?

EDIT: Some of you were asking where this question came from. I took it from "Teoría axiomática de conjuntos: Una introducción" by Roberto Cignoli. It is written in spanish. This is a screenshot of the exercise:

• It seems there is something wrong with the problem since, as stated, it is false. On the basis of the other axioms, one easily check that $V_\omega\subseteq\mathcal C$, and therefore the requirement that infinity fails is incompatible with the requirement that subsets of $\mathcal C$ are subsets of members of $\mathcal C$. Commented Feb 20, 2019 at 22:34
• If I am not mistaken, not only $\cal C$ satisfy ZF(including infinity), even if you change "$\cal C$ satisfies the axiom schema of specification" into "$\cal C$ satisfies the axiom schema of specification to $Δ_0$ formulas" it still satisfy all of ZF
– Holo
Commented Feb 21, 2019 at 0:14
• It seems to me like someone wanted you to find a transitive model of ZF-Infinity, but instead made a mistake. Where did you find this question? Commented Feb 21, 2019 at 6:09
• $V_{\omega}$ is transitive and satisfies ZFC - Infty +($\neg$ Infty) but does not satisfy the last sentence of your 1st paragraph. No transitive model of ZF -Infty + ($\neg$ Infty) can satisfy that sentence. Commented Feb 21, 2019 at 7:02
• @solomeo I think you read it right and the question is just wrong. It is actually a common exercise to show that a transitive class is a model of all of ZF if and only if those two conditions hold. Commented Feb 21, 2019 at 17:32

I think this is true. Every element of $$V_\omega$$ is a member of $$V_n$$ for some $$n \in \omega$$ so every element of $$V_\omega$$ is finite. That means that if $$x \in V_\omega,~ \exists N \in \omega \text{ with } z \in V_N ~\forall z \in x$$. But then $$x \subset y= x \cup \{x\} \in V_{N+1} \subset V_\omega$$.
• Wasn't he asking for a proof that if all elements of $x$ are in $V_\omega$, then $\exists~y \in V_\omega$ with $x \subseteq y$? Commented Feb 21, 2019 at 2:43
• Robert, all elements of $V_\omega$ are in $V_\omega$. Commented Feb 21, 2019 at 3:11
• All elements of $\omega$ are in $V_\omega$, but that doesn't mean a set containing all of those elements is in $V_\omega$. Commented Feb 21, 2019 at 4:58