Question: Define the cumulative hierarchy inductively as: $$V_0 = \emptyset$$ For each ordinal $\alpha$: $$V_{\alpha+1} = \mathcal{P}(V_\alpha)$$ ie the power set of $\alpha$ and for any nonzero limit ordinal $\lambda$: $$V_\lambda = \bigcup{\{V_\xi : \xi < \lambda\}}$$
If we use $V_\omega$ and $V_{\omega+1}$ as structures for Morse-Kelley, Do the Morse-Kelley axioms of Power Set and Infinity hold for $V_\omega$ and $V_{\omega+1}$?
My thoughts: Firstly, $\omega$ is a nonzero limit ordinal and $V_\omega$ is the power set of naturals.
The MK power set axiom says that the power set of a set is, itself, a set
The MK infinity axiom says that there exists a set $y : (\emptyset \in y) \land (x\in y \Rightarrow x\cup\{x\} \in y)$
$V_\omega$: The power set axiom holds because the power sets of any of the members of $V_\omega$ will themselves be elements of $V_\omega$ and thus sets.
The axiom of infinity holds because $\emptyset = 0$ and $x\cup\{x\} = x+1$ and we know that the successor of a natural number is a natural number and thus will be a member of $\omega$ and certainly a member of $V_\omega$
$V_{\omega+1}$: The power set axiom holds because we know that there are no sets with cardinality between $V_\omega$ & $V_{\omega+1}$ and so for all proper subsets of $V_{\omega+1}$, their power set will either be $V_{\omega+1}$ or a subset of $V_\omega$.
The axiom of infinity holds because, similar to above, $\alpha \subseteq V_\alpha$.
Is my reasoning correct or totally wrong? Also, despite my reasoning above, I am a little confused as I thought the fact the MK axioms are axioms means that it isn't a case of them being true or false.
Thanks
The axiom of infinity fails in $V_\omega$. Even though each of the natural numbers is itself in $V_\omega$, the set of all of them is not itself a member of $V_\omega$. It only shows up as an element of $V_{\omega+1}$.
In $V_{\omega+1}$ the MK axiom of infinity still fails. You do have $\omega$ as an element of $V_{\omega+1}$, but it is not a "set" (as implicitly demanded by the axiom), since there is no element of $V_{\omega+1}$ that contains $\omega$.
No, we don't know that. $V_\omega$ has cardinality $\aleph_0$ and $V_{\omega+1}$ has cardinality $2^{\aleph_0}$ -- and whether there are cardinalities between these or not is the continuum hypothesis. Fortunately this isn't really related to the power set axiom. It holds in the two structures, but not for the reason you claim.