The Axiom of Choice in the $\kappa$-Mantle

First, we show that all limit cardinals of $L$ are limit cardinals in $L[G]$. It is enough to prove that all double successors $\delta^{++}$ are preserved. This is obvious for $\delta\geq\kappa$ as $\mathbb{P}$ has size $\kappa$. For $\delta<\kappa$, $\mathbb{P}_{>\delta}$ is ${\leq}\delta^{++}$-closed so that all cardinals $\leq\delta^{++}$ are preserved in $L[G_{>\delta}]$. Furthermore, $\mathbb{P}_{<\delta}$ has size at most $\delta^{+}$ in $L[G_{>\delta}]$ by $\mathrm{GCH}$ in $L$. Hence $\delta^{++}$ is still a cardinal in $L[G]$.
Now we have to argue that all $\lambda\in I$ remain regular. Again, this is clear if $\lambda>\kappa$. On the other hand, assume $\delta\coloneqq\operatorname{cof}(\lambda)^{L[G]}<\lambda$. As $\mathbb{P}_{>\delta}$ is ${\leq}\delta$-closed, $\lambda$ is still regular in $L[G_{>\delta}]$. Hence, a witness to $\operatorname{cof}(\lambda)=\delta$ must be added in the extension of $L[G_{{>}\delta}]$ by $\mathbb{P}_{\leq\delta}$. But this forcing has size $<\lambda$ in $L[G_{{>}\delta}]$ and thus could not have added such a sequence. ∎

Suppose $S\subseteq\lambda$ is stationary, $\dot{C}$ is a $\mathbb{Q}$-name for a club in $\lambda$ and $p\in\mathbb{P}$. We will imitate the standard proof that a ${<}\kappa$-closed forcing preserves stationary sets. Let $\langle\leq_{\alpha}\mid\alpha<\lambda\rangle$ witness that $\mathbb{Q}$ satisfies $\mathrm{AA}(\lambda,\lambda)$.

Suppose toward a contradiction that $p\Vdash\dot{C}\cap\check{S}=\emptyset$. By the claim above, we can build sequences $\langle p_{\alpha}\mid\alpha<\lambda\rangle$ of conditions in $\mathbb{Q}$ and an increasing sequence $\langle\gamma_{\alpha}\mid\alpha<\lambda\rangle$ of ordinals below $\lambda$ so that

1. $(i)$

   $p_{0}=p$,

2. $(ii)$

   $p_{\beta}\leq_{\alpha}p_{\alpha}$ for all $\alpha\leq\beta<\lambda$ and

3. $(iii)$

   $p_{\alpha+1}\Vdash\check{\gamma}_{\alpha}\in\dot{C}$ for all $\alpha<\lambda$.

Let $D$ be the set of all limit points ${<}\lambda$ of $\{\gamma_{\alpha}\mid\alpha<\lambda\}$. For any $\alpha<\lambda$, we have

which shows that $D\cap S=\emptyset$, contradiction. ∎

For $\gamma<\kappa$ define $\leq_{\gamma}$ by $r\leq_{\gamma}q$ if $r\leq q$ and $r\upharpoonright\gamma=q\upharpoonright\gamma$ for $q,r\in\mathbb{P}$. We will only show that $(\mathrm{AA}.ii)$ holds. So let $p\in\mathbb{P}$, $\gamma<\kappa$ and $A\subseteq\mathbb{P}$ a maximal antichain. Let $\langle q_{\alpha}|\alpha<\delta\rangle$ be an enumeration of all conditions in $\mathbb{P}_{\leq gamma}$ below $p\upharpoonright\gamma+1$ with $\delta=|\mathbb{P}_{\leq\gamma}|$. We construct a $\leq_{\gamma}$-descending sequence $\langle p_{\alpha}|\alpha\leq\delta\rangle$ of conditions in $\mathbb{P}$ starting with $p_{0}=p$ as follows: If $\alpha\leq\delta$ then choose some $\leq_{\gamma}$-bound of $\langle p_{\beta}\mid\beta<\alpha\rangle$. This is possible as $\mathbb{P}_{>\gamma}$ is ${\leq}\delta$-closed, as the next forcing only appears at the next inaccessible. Moreover, if possible and $\alpha<\delta$ make sure that

is below a condition in $A$. This completes the construction. Set $q\coloneqq q_{\kappa}$, we will show that $q$ is compatible with at most $\delta$-many elements of $A$. Toward this goal, suppose $a\in A$ and $q$ is compatible with $a$. We may find some $\alpha<\kappa$ so that $a\upharpoonright\gamma+1=q_{\alpha}$. It follows that we must have succeeded in the construction of $p_{\alpha}$ with the additional demand that

is below a condition in $A$, but this can only be true for $a$. We have shown that for any $a\in A$ compatible with $q$ there is $\alpha<\delta$ with $q_{\alpha}^{\frown}q\upharpoonright(\gamma,\kappa)\leq a$ and note that no single $\alpha$ can witness this for more than one element of $A$. ∎

This follows immediately from Lemma 3.1, Lemma 3.6 and Proposition 3.4. ∎

Suppose $W$ is a $\kappa$-ground of $L[G]$. It is enough to find $\lambda\in I\cap\kappa$ such that $L[G_{>\lambda}]\subseteq W$. Clearly, $\mathbb{P}\in L\subseteq W$. As $\kappa$ is a limit of inaccessibles, we may take some $\lambda<\kappa$ inaccessible so that $W$ is a $\lambda$-ground. Thus $W\subseteq L[G]$ satisfies the $\lambda$-approximation property. We will show $G_{>\lambda}\in W$ (even $G_{\geq\lambda}\in W$). Find $\alpha$ with $\lambda=I_{\alpha}$, it is thus enough to show $M_{\geq\alpha}\in W$. To any $\lambda$-approximation $M_{\geq\alpha}\cap a$ of $M_{\geq\alpha}$ by $W$ corresponds some $a^{\prime}\subseteq\kappa\setminus\alpha\times\kappa$, $a^{\prime}\in W$ of size ${<}\lambda$ so that

We will show that all such restrictions of $M_{\geq\alpha}$ are in $W$. So let $a\in W$, $a\subseteq\kappa\setminus\alpha\times\kappa$, $|a|<\lambda$. As $0^{\#}$ does not exist in $W$, there is $b\in L$, $b\subseteq\kappa\setminus\alpha\times\kappa$ of size $<\lambda$ with $a\subseteq b$. For all $\alpha\leq\gamma<\kappa$, the set of $\beta<I_{\gamma}$ with $(\gamma,\beta)\in b$ is bounded in $I_{\gamma}$. As described earlier, we may think of conditions in $\mathbb{P}$ as partial $\kappa\times\kappa$ matrices. With this in mind, the conditions $p\in\mathbb{P}$ that contain information on the entry $e_{\gamma,\beta}$ for all $(\gamma,\beta)\in b$ form a dense set of $\mathbb{P}$. Thus $M\upharpoonright b=\langle e_{\gamma,\beta}\mid(\gamma,\beta)\in b\rangle$ is essentially a condition $p\in\mathbb{P}\subseteq W$ and hence $M\upharpoonright a=(M\upharpoonright b)\upharpoonright a\in W$. As $W\subseteq L[G]$ satisfies the $\lambda$-approximation property, we have $M_{\geq\alpha}\in W$.

∎

First suppose $a\subseteq\kappa$, $a\in\mathbb{M}_{\kappa}^{L[G]}$. If $\lambda<\kappa$ then $a\in L[G_{>\lambda}]$. As $\mathbb{P}_{>\lambda}$ is ${\leq}\lambda$-closed in $L$, $a\cap\lambda\in L$.
For the other direction assume $a\in L[G]$ is a fresh subset of $\kappa$ and assume $W$ is a $\kappa$-ground of $L[G]$. There is $\lambda<\kappa$ so that $W\subseteq L[G]$ satisfies the $\lambda$-approximation property. As $a$ is fresh, all the $\lambda$-approximations of $a$ in $W$ are in $W$. Thus $a\in W$. ∎

The map $\pi\colon\mathbb{P}\rightarrow\mathrm{Add}(\kappa,1)$ that maps $p\in\mathbb{P}$ to the information that $p$ has on $c_{\beta}$ is well-defined as $\mathbb{P}$ is a bounded support iteration of length $\kappa$. Clearly, $\pi$ is a projection. ∎

Note that any generic for $\mathrm{Add}(\kappa,1)^{L}$ is the characteristic function of a fresh subset of $\kappa$ so that $c_{\beta}\in\mathbb{M}_{\kappa}^{L[G]}$ for any $\beta<\kappa$. Of course, the sequence $\langle c_{\beta}|\beta<\kappa\rangle$ is not in $\mathbb{M}_{\kappa}^{L[G]}$, as one can compute the whole matrix $M$ (and thus the whole generic $G$) from this sequence. However, we can make this sequence fuzzy to result in an element of $\mathbb{M}_{\kappa}^{L[G]}$. Let $\sim$ be the equivalence relation of eventual coincidence on $({}^{\kappa}2)^{\mathbb{M}_{\kappa}^{L[G]}}$, i.e.

We call $\langle[c_{\beta}]_{\sim}|\beta<\kappa\rangle$ the fuzzy sequence.

Finally, we argue that $\mathbb{M}_{\kappa}^{L[G]}$ does not contain a choice sequence for the fuzzy sequence⁶⁶6That is, there is no sequence $\langle x_{\beta}\mid\beta<\kappa\rangle\in\mathbb{M}_{\kappa}^{L[G]}$ with $x_{\beta}\in[c_{\beta}]_{\sim}$ for all $\beta<\kappa$.. Heading toward a contradiction, let us assume that

is such a sequence. $L[G]$ knows about the sequence

where $\delta_{\beta}$ is the least $\delta$ with $x_{\beta}\upharpoonright(\kappa\setminus\delta)=c_{\beta}\upharpoonright(\kappa\setminus\delta)$. The set of $\lambda<\kappa$ that are closed under the map $\beta\longmapsto\delta_{\beta}$ is club in $\kappa$. As $\kappa$ is Mahlo in $L[G]$, there is an inaccessible $\alpha=I_{\alpha}<\kappa$ that is closed under $\beta\longmapsto\delta_{\beta}$. Now observe that

holds for all $\beta<I_{\alpha}$, so that $r_{\alpha}\in\mathbb{M}_{\kappa}^{L[G]}$. But this is impossible as clearly $r_{\alpha}$ is not fresh.

∎

First let us do it for $\mathbb{M}_{\mathsf{C}}$. Clearly, $\mathbb{M}_{\mathsf{C}}$ is closed under the Gödel operations. It is thus enough to show that $\mathbb{M}_{\mathsf{C}}\cap V_{\alpha}\in\mathbb{M}_{\mathsf{C}}$ for all $\alpha\in\mathrm{Ord}$. Let $W$ be any Cohen-ground. As Cohen-forcing is homogeneous, $\mathbb{M}_{\mathsf{C}}^{V}$ is a definable class in $W$. Hence, $\mathbb{M}_{\mathsf{C}}\cap V_{\alpha}=\mathbb{M}_{\mathsf{C}}\cap V_{\alpha}^{W}\in W$. As $W$ was arbitrary, this proves the claim.

Now onto $\mathbb{M}_{\omega_{1}}$. The above argument shows that all we need to do is show that $\mathbb{M}_{\omega_{1}}$ is a definable class in all associated grounds. So let $W$ be such a ground. There are two cases. First, assume that $\omega_{1}^{W}=\omega_{1}^{V}$. Then $W$ extends to $V$ via Cohen forcing, so $\mathbb{M}_{\omega_{1}}$ is definable in $W$. Next, suppose that $\omega_{1}^{W}<\omega_{1}^{V}$. This can only happen if $\omega_{1}^{V}$ is a successor cardinal in $W$, say $W\models\omega_{1}^{V}=\mu^{+}$. In this case, $W$ extends to $V$ via a forcing of $W$-size $\leq\mu$ and which collapses $\mu$ to be countable. It is well known that in this situation, $W$ extends to $V$ via $\mathrm{Col}(\omega,\mu)$, which is homogeneous as well, so once again, $\mathbb{M}_{\omega_{1}}$ is a definable class in $W$. ∎

This is a straightforward adaptation of the proof that Axiom A forcings preserve $\omega_{1}$. ∎

We define $\leq_{\alpha}$ independent of $\alpha<\kappa$ as the order $\leq^{\ast}$: Let $p\leq^{\ast}q$ iff $p\leq q$ and $p\upharpoonright\mathrm{supp}(q)=q$. The only nontrivial part is showing that for any antichain $A$ and any $p\in\mathbb{P}$ there is $q\leq^{\ast}p$ with

Let

for all $\alpha<\kappa$. We will proceed to find some $q$ with the desired property. For convenience, we may assume that $A$ is a maximal antichain. As $\kappa$ is Mahlo, there is a regular $\lambda<\kappa$ so that $p\mathbb{P}\upharpoonright\lambda$ and any $r\in\mathbb{P}\upharpoonright\lambda$ is compatible with some $a\in A\cap\mathbb{P}\upharpoonright{\lambda}$. As $V=L$, $\diamondsuit_{\lambda}$ holds. Thus there is a sequence $\vec{d}\coloneqq\langle d_{\alpha}\mid\alpha<\lambda\rangle$ with

1. $(\vec{d}.i)$

   $d_{\alpha}\in\mathbb{P}_{\leq\alpha}$ and

2. $(\vec{d}.ii)$

   for all $r\in\mathbb{P}_{\leq\lambda}$ there are stationarily many $\alpha<\lambda$ with $d_{\alpha}=r\upharpoonright\alpha$.

Construct a sequence

of conditions in $\mathbb{P}\upharpoonright\lambda$ with $q_{\alpha}\leq^{\ast}q_{\beta}$ for all $\alpha<\beta<\lambda$ as follows: Set $q_{0}=p$. If $q_{\beta}$ is defined for all $\beta<\alpha$, let first $q^{\prime}_{\alpha}=\bigcup_{\beta<\alpha}q_{\beta}$ and note that this is a condition. Let $\gamma_{\alpha}=\sup\mathrm{supp}(q_{\alpha}^{\prime})$. Now find $a\in A\cap\mathbb{P}\upharpoonright\lambda$ that is compatible with $d_{\gamma_{\alpha}}$ and let

Finally, set $q=\bigcup_{\alpha<\lambda}q_{\alpha}$. We have to show that $q$ is compatible with only a few elements of $A$, so suppose $b\in A$ is compatible with $q$. The properties of $\vec{d}$ guarantee that there is $\alpha<\lambda$ so that

1. $(\alpha.i)$

   $\gamma_{\alpha}=\alpha$ and

2. $(\alpha.ii)$

   $d_{\alpha}=b\upharpoonright\alpha$.

Hence in the construction of $q_{\alpha+1}$ we found some $a\in A\cap\mathbb{P}\upharpoonright\lambda$ compatible with $b\upharpoonright\alpha$ and have $q_{\alpha+1}\upharpoonright[\alpha,\lambda)\leq a\upharpoonright[\alpha,\lambda)$. If $a\neq b$, then $a\perp b$ and the incompatibility must lie in the interval $[\alpha,\lambda)$. But then $q_{\alpha+1}$ and $b$ are incompatible as well, contradiction. Thus $b=a$ and it follows that $q$ is compatible with at most $\lambda$-many elements of $A$. ∎

To see $(G.i)$, note that $\mathbb{P}$ collapses all cardinals ${<}\kappa$ to $\omega$, so $\omega_{1}^{L[G]}\geq\kappa$. As $\mathbb{P}$ satisfies $\mathrm{AA}(\kappa,\kappa)$, there is no surjection from $\omega$ onto $\kappa$ in $L[G]$.
Next, let us prove $(G.ii)$. Let $\dot{g}\in L$ be a name for $g$. In $L[G]$, find a decreasing sequence of conditions $\langle p_{n}\mid n<\omega\rangle$ in $G$ so that $p_{n}$ decides the value of $\dot{g}(\check{n})$ (from the perspective of $L$). Let $\alpha=\sup_{n<\omega}\sup\mathrm{supp}(p_{n})$. By $(G.i)$, $\alpha<\kappa$. But then $L[G_{\leq\alpha}]$ can compute the whole of $g$. ∎