Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.