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

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?

-
math.stackexchange.com/questions/310730/… - Are you the same user? You can request the accounts to be merged. –  Asaf Karagila Feb 23 '13 at 12:42
@AsafKaragila No... But I feel guilty of asking stupid questions. I should really read my textbooks carefully. Things get mixed up when you study bunch of things together.. –  Mark Zwazingker Feb 23 '13 at 12:50

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.

-

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.

-