Assume that is consistent that there is a measurable cardinal.

Question: How many normal measures can a measurable cardinal carry?

Let $\mathrm{NM}(\kappa)$ denote the set of normal measures on $\kappa$. Trivially, $\#\mathrm{NM}(\kappa)\le 2^{2^\kappa}$.

One interesting case happens in a canonical inner model.

**Theorem (Kunen '71):**In $L[U]$, there is exactly one normal measure (on $\kappa$, the unique measurable cardinal).

We can also realize the other extreme:

**Theorem (Kunen-Paris '71):**There is a forcing extension in which $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

We will prove this one later today. In the middle, we have

**Theorem (Mitchell '74):**It is consistent relative to a measurable $\kappa$ of order $\delta$ that $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

And for one case, we can lower the large cardinal assumption used.

**Theorem (Apter-Cummings-Hamkins '07):**It is consistent relative to a measurable cardinal that that $\#\mathrm{NM}(\kappa)=\kappa^+$.

Finally, we can do it in all cases.

**Theorem (Friedman-Magidor '09):**Assume GCH. Suppose $\kappa$ is measurable and let $\mu\le \kappa^{++}$ be a cardinal. Then in a cofinality-preserving forcing extension, $\#\mathrm{NM}(\kappa)=\mu$.

The goal eventually will be to show the proof of this result.

**Lemma:**Suppose $j:V\rightarrow M$ is the ultrapower by a normal measure on $\kappa$. Let $G$ be $V$-generic for $\mathbb{P}$. Assume that in $V[G]$,

$$j_0:V[G]\rightarrow M[j_0(G)]$$

and

$$j_1:V[G]\rightarrow M[j_1(G)]$$

are elementary embeddings extending $j$. Then the following are equivalent:

1) $j_0=j_1$.

2) $j_0(G)=j_1(G)$.

3) The normal measure $U_0$ derived from $j_0$ is equal to the normal measure $U_1$ derived from $j_1$.

*Proof:*Exercise.

**Exercise (Levy-Solovay):**Show that every normal measure extends uniquely to a normal measure in any forcing extension by small forcing.

As promised, we will now prove the Kunen-Paris Theorem.

**Theorem (Kunen-Paris '71):**There is a forcing extension in which $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

*Proof*: By a preparation forcing if necessary, assume that $2^\kappa=\kappa^+$. Let $\mathbb{P}$ be the length $\kappa+1$ Easton support iteration that forces at cardinal stages $\gamma\le \kappa$ with $\mathbb{Q}_\gamma:=\mathrm{Add}(\gamma^+,1)$ (computed in the extension by $\mathbb{P}_\gamma$), trivial forcing at other stages. This is a standard way of forcing the GCH to hold below $\kappa$.

Let $G\ast H$ be $V$-generic for $\mathbb{P}=\mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa$. Let $j:V\rightarrow M$ be the ultrapower by a normal measure. Factor $j(\mathbb{P}_\kappa)\simeq \mathbb{P}_\kappa \ast \dot{\mathbb{P}}_{\kappa,j(\kappa)}$.

Since ${}^\kappa M[G]\subseteq M[G]$ in $V[G]$ (a name exercise which uses closure under $\kappa$-sequences of $M$ in $V$), we have that in $M[G]$, $\mathbb{P}_{\kappa,j(\kappa)}$ is $\le \kappa$-closed (it is the composition of increasingly closed posets starting with $\mathbb{Q}_\kappa=\mathrm{Add}(\kappa^+,1)$.

Furthermore, computing in $V[G]$, $\mathbb{P}_{\kappa,j(\kappa)}$ has at most $\kappa^+$ maximal antichains inside $M[G]$. since $|j(\kappa)|=\kappa^+$ by our cardinal arithmetic assumption, and $\mathbb{P}_{\kappa,j(\kappa)}$ is $j(\kappa)$-c.c. of size $j(\kappa)$ in $M[G]$. By enumerating the maximal antichains of $M[G]$ in order-type $\kappa^+$, we can meet them one by one, with the closure of the poset in $M[G]$, noticing that closure of the model $M[G]$ gives that the proper initial segments of this enumeration are in $M[G]$.

Since $j``G\subseteq G\ast G_{\kappa,j(\kappa)}$, we can lift the embedding to

$$j^+:V[G]\rightarrow M[j(G)].$$

Using the Lemma/Exercise above, different choices of $j(G)$ will give rise to different embeddings which will give different normal measures. By passing to finer antichains, we can assume in the enumeration $\langle A_\alpha:\alpha<\kappa^+\rangle$ that if $i<j$, then $A_j$ (strictly) refines $A_i$. Looking at how we built $G_{\kappa,j(\kappa)}$, we can form the tree of attempts to build the generic, noticing that at each level there are incompatible ways to extend the generic so far to meet the maximal antichain. This gives $2^{\kappa^+}=2^{2^\kappa}$ many different generics, and hence different normal measures. $\Box$

So it's not hard to force many normal measures. It is harder to force so that there are few normal measures.

We will need to use Hamkins's Gap Forcing Theorem which gives a sufficient condition for an ultrapower embedding in a generic extension to be the lift of a ground model embedding.

**Definition:**A forcing $\mathbb{P}$

*admits a closure point at $\delta$*if it factors as $\mathbb{P}\simeq \mathbb{Q}\ast \dot{\mathbb{R}}$ where $\mathbb{Q}$ is nontrivial, $|\mathbb{Q}|\le \delta$, and $\Vdash_{\mathbb{Q}} \mathbb{R} \textrm{ is }<\delta-\textrm{closed}$.

**Theorem (Hamkins '01, Gap Forcing Theorem):**If $V\subseteq V[G]$ admits a closure point at $\delta$ and $j:V[G]\rightarrow M[j(G)]$ is an ultrapower in $V[G]$ with $\mathrm{crit}(j)>\delta$, then $j\upharpoonright V:V\rightarrow M$ is a definable class in $V$.

Finally, we prove one more of the theorems in the introduction.

**Theorem (Apter-Cummings-Hamkins '07):**It is consistent relative to a measurable cardinal that that $\#\mathrm{NM}(\kappa)=\kappa^+$.

*Idea of the proof:*Start with at least $\kappa^+$ normal measures on $\kappa$ (e.g., by Kunen-Paris forcing). Force with $\mathrm{Col}(\kappa^+,2^{2^\kappa})$, and show that no new normal measures are added.

*Proof:*Again we assume $2^\kappa=\kappa^+$. Start with $\#\mathrm{NM}(\kappa)\ge \kappa^+$. Let $\mathbb{P}=\mathrm{Add}(\omega,1)\ast \dot{\mathrm{Col}}(\kappa^+, 2^{2^\kappa})$. The point of the Cohen forcing is to give $\mathbb{P}$ a closure point below $\kappa$. Suppose $c\ast G\subseteq \mathbb{P}$ is $V$-generic. Then every normal measure in $V$ generates a normal measure in $V[c]$ by Levy-Solovay, and these remain normal measures in $V[c][G]$ since $\mathrm{Col}(\kappa^+, 2^{2^\kappa})$ is $\le \kappa$-closed in $V[c]$. So there are at least $\kappa^+$ normal measures in $V[c][G]$.

To show the other inequality, suppose $U$ is a normal measure on $\kappa$ in $V[c][G]$. Let

$$j:V[c][G]\rightarrow M[c][j(G)]$$

be the ultrapower. By the gap forcing theorem, $j\upharpoonright V:V\rightarrow M$ is a definable class in $V$. So we can lift to $j\upharpoonright V[c]:V[c]\rightarrow M[c]$. We can now define $U$ in $V[c]$, as the derived measure from this embedding. $\Box$

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