Proving $V_{\kappa}$ is a model of ZFC for inaccessible $\kappa$

Prove that if $\kappa$ is an inaccessible cardinal, then $V_{\kappa}$ satisfies all the axioms of ZFC.

How is this done for the axiom of choice and for regularity?

• For Choice, show that the (usual) product of a $\{A_{\lambda}\}_{\lambda\in\Lambda}$, where $A_{\lambda},\Lambda\in V_{\kappa}$ and $A_{\lambda}\neq\emptyset$ is also in $V_{\kappa}$. As for regularity, it is trivial! – Arturo Magidin Dec 15 '10 at 20:45
• Depending on how choice is defined, there is an easier argument: If we use the statement "For any $X$ there is an $f:{\mathcal P}(X)\setminus\{\emptyset\}\to X$ such that $f(A)\in A$ for all $A$", then it is trivial that this holds in $V_\kappa$: If $X\in V_\kappa$, then $X\in V_\alpha$ for some smaller $\alpha<\kappa$, and (since choice holds in $V$), there is such an $f\in V_{\alpha+7}\subset V_\kappa$ (where 7 can be reduced but I felt lazy). – Andrés E. Caicedo Dec 15 '10 at 22:13
• @Andres: This is very good! @kittykat687: I really think a better title is needed for this question. – Asaf Karagila Dec 15 '10 at 22:24

Let's solve the general problem surrounding the question, with a few observations, each of them easy to see.

• Every $V_\alpha$ for any ordinal $\alpha$ satisfies Extensionality and Foundation, since all transitive sets satisfy Extensionality and Foundation.

• Every $V_\alpha$ satisfies Separation, for the simple reason that $A\subset B\in V_\alpha\implies A\in V_\alpha$.

• Every $V_\alpha$ satisfies Union, since $A\in V_\alpha\implies \bigcup A\in V_\alpha$.

• Every $V_\lambda$ for a limit ordinal $\lambda$ satisfies Pairing and Powerset, since the required set is added at the next stage below $\lambda$.

• Every $V_\alpha$ satisfies the Axiom of Choice (assuming this holds in $V$) in the choice-set version, since if ${\cal A}\in V_\alpha$ is a family of disjoint sets , then all choice sets $B\subset \bigcup {\cal A}$ selecting one element from each $A\in{\cal A}$ have the same or lower rank than ${\cal A}$, and so $B\in V_\alpha$.

• Every $V_\alpha$ with $\omega\lt\alpha$ satisfies Infinity, since $\omega\in V_\alpha$.

• The only remaining axiom is Replacement, and this is the only one that makes use of inaccessibility. But if $\kappa$ is inaccessible, then $V_\kappa$ satisfies Replacement, since if $A\in V_\kappa$ and $F:A\to V_\kappa$ is definable over $V_\kappa$, then $F''A$ has bounded rank below $\kappa$, since $|A|\lt\kappa$ and $\kappa$ is regular. Thus, $F''A\in V_\kappa$ as desired.

Finally, one can also consider the question about $H_\delta$, the sets of hereditary size less than $\delta$, and things are a bit nicer here in several ways.

• For any regular uncountable cardinal $\delta$, the set $H_\delta$ of sets having hereditary size less than $\delta$, satisfies $ZFC^-$, that is, all of ZFC except the Powerset axiom. One gets the easy axioms easily; Separation is easy since the subset also has small hereditary size; and Replacment follows from the fact that the union of fewer than $\delta$ many sets in $H_\delta$ still has size less than $\delta$ by the regularity of $\delta$.

In particular, this shows that ZFC proves that there are numerous transitive models of $ZFC^-$.

• If $\kappa$ is inaccessible, then $V_\kappa=H_\kappa$, and this satisfies full ZFC, since we get $ZFC^-$ in $H_\kappa$, and we get power set since $\kappa$ is a strong limit.
• could you give a bit more details on replacement for $H_\delta$ please? It is not so clear for me how to write the image of a class function as the union you talk about... – WrabbitW Nov 29 '17 at 20:52
• The point is that if $\delta$ is regular, then $H_\delta$ contains all its subsets of size less than $\delta$, since any such set will still have a small transitive closure. And then the point is every instance of replacement in $H_\delta$ amounts to the existence in $H_\delta$ of a particular such set, the one defined by whichever class function you have in mind. That is, if $F:A\to H_\delta$ is a class function, with $A\in H_\delta$, then $\{F(a)\mid a\in A\}$ is a subset of $H_\delta$ of size less than $\delta$. So it exists in $H_\delta$, verifying replacement. – JDH Nov 29 '17 at 21:18
• But then would'nt you also need regularity for separation and choice? As I proved (for choice for example) the choice function is a subset of a set in $H_\delta$ so it is also in $H_\delta$. – WrabbitW Nov 29 '17 at 22:13
• You don't need it for separation, since every subset of a small set is small. Similar for choice. – JDH Nov 29 '17 at 22:34
• Oh I see in those cases you are taking subset of sets in $H_\delta$ but for replacement you take a subset of $H_\delta$. – WrabbitW Nov 29 '17 at 22:36

The idea is that you need to remember that whenever ZFC requires you something to be a set, you want it to be an element of $V_\kappa$, as this is the model.

So to expand Arturo's comment, the axiom of choice says that when you have a set of non-empty sets the product is non-empty. Since the sets of $V_\kappa$ are elements of it in the world, you can take a product of them, from the inaccessibility of $\kappa$ you'd have that the rank of the product (in the big universe) is still less than $\kappa$ and thus the product remains within $V_\kappa$ as you'd like it to.

As for regularity, well... basically the relation $\in$ is the same as in the world, so it's well-founded and everything holds.