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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:

Specific exercise

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    $\begingroup$ 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 $\mathcal C$. $\endgroup$
    – Andrés E. Caicedo
    Commented Feb 20, 2019 at 22:34
  • $\begingroup$ 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 $\endgroup$
    – Holo
    Commented Feb 21, 2019 at 0:14
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    $\begingroup$ 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? $\endgroup$
    – Asaf Karagila ♦
    Commented Feb 21, 2019 at 6:09
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    $\begingroup$ $V_{\omega}$ 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. $\endgroup$
    – DanielWainfleet
    Commented Feb 21, 2019 at 7:02
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    $\begingroup$ @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. $\endgroup$
    – spaceisdarkgreen
    Commented Feb 21, 2019 at 17:32

1 Answer 1

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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 $x \subset y= x \cup \{x\} \in V_{N+1} \subset V_\omega$.

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    $\begingroup$ This does not address the question. $\endgroup$
    – Andrés E. Caicedo
    Commented Feb 21, 2019 at 2:18
  • $\begingroup$ Wasn't he asking for a proof that if all elements of $x$ are in $V_\omega$, then $\exists~y \in V_\omega$ with $x \subseteq y$? $\endgroup$
    – Robert Shore
    Commented Feb 21, 2019 at 2:43
  • $\begingroup$ Robert, all elements of $V_\omega$ are in $V_\omega$. $\endgroup$
    – Andrés E. Caicedo
    Commented Feb 21, 2019 at 3:11
  • $\begingroup$ All elements of $\omega$ are in $V_\omega$, but that doesn't mean a set containing all of those elements is in $V_\omega$. $\endgroup$
    – Robert Shore
    Commented Feb 21, 2019 at 4:58
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    $\begingroup$ No Robert, that's not it. $\endgroup$
    – Andrés E. Caicedo
    Commented Feb 21, 2019 at 5:56

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