$V_k$ transitive model of ZFC when $k$ is inaccessible?

Question

Is $V_k$ transitive model of ZFC when $k$ is inaccessible? I know that $V_k$ is a model of ZFC, but not sure if it's transitive one. If it is, why is it?

Answer 1

Part of your question is answered in the following MSE thread: Proving $V_{\kappa}$ is a model of ZFC for inaccessible $\kappa$.

If $ \kappa $ is an inaccessible cardinal, then $ (V_{\kappa},\in|_{V_{\kappa} \times V_{\kappa}}) $ is a transitive model of $ \sf ZFC $, where $ \in $ is the actual set-membership relation.

Answer 2

Every set of the form $V_\alpha$ is transitive. To prove this one has to decide which definition of $V_\alpha$ one takes.

If one defines a rank function first, and $V_\alpha$ is the set of those whose rank is strictly smaller than $\alpha$ then this is obvious, as $y\in x$ implies $\operatorname{rank}(y)<\operatorname{rank}(x)$, so whenever $y\in x\in V_\alpha$ we immediately have $y\in V_\alpha$.

If one defines $V_\alpha$ by power sets and unions, then defines the rank as the least $\alpha$ such that $x\subseteq V_\alpha$, it is not hard to show the following two properties:

1. If $A$ is a transitive set then $\mathcal P(A)$ is a transitive set. This follows from the fact that $x\in\mathcal P(A)$ and $y\in x$ then $x\subseteq A$ and $y\in A$, and since $A$ is transitive $y\subseteq A$ and $y\in\mathcal P(A)$.

2. If $\{A_i\mid i\in I\}$ is an increasing $\subseteq$-chain of transitive sets, then $A=\bigcup A_i$ is transitive. To see this holds take $x\in A$ and $y\in x$ then $x\in A_i$ for some $i\in I$ and so $y\in A_i$ (it is assumed to be transitive), and so $y\in A$.

Now apply these to the creation rule of $V_\alpha$ and we immediately have that those sets are transitive.