# Does the MK axiom of infinity and the Power set axiom hold in the cumulative hierarchy?

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

• Fair enough. The question was asked as two separate parts: one whether the two axioms hold in $V_\omega$ and the second question said "now decide for $V_{\omega+1}$ – Andrew May 22 '16 at 2:08

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 $\omega$.
The power set axiom holds because we know that there are no sets with cardinality between $V_\omega$ and $V_{\omega+1}$.
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.
• Thanks so much for that. I thought the axiom of infinity only said that there exists such a set so even if $\omega \notin V_{\omega}$ there would still be such sets. Is this wrong? – Andrew May 22 '16 at 2:07
• @Andrew: Remember that the axioms are always claims about what is (or isn't) true inside the model. When the axiom of infinity says that such-and-such $y$ exists, it means that there is an element of the structure that qualifies as $y$ in the property the axiom states. So it doesn't matter that $\omega$ exists at the metalevel if it's not an element of the structure. – Henning Makholm May 22 '16 at 2:09
• Thanks again. Sorry to keep asking, your help raised for me another question: if $\omega \notin V_\omega \land \omega \in V_{\omega+1}$, does this mean that the power set of $\omega$ is not actually in $V_{\omega+1}$ and therefore the power set axiom fails in $V_{\omega+1}$? – Andrew May 22 '16 at 2:31
• @Andrew: Partially right. You're right that $\omega\in V_{\omega+1}$ and $\mathcal P(\omega)\notin V_{\omega+1}$. But this does not mean that the power set axiom fails -- because in the structure $V_{\omega+1}$, $\omega$ is a proper class rather than a set, and therefore the power set axiom is not supposed to apply to it there. – Henning Makholm May 22 '16 at 2:35