# Destroying the Mahloness of $\kappa$ with a forcing of size $\kappa$ that is $\alpha$-distributive for all $\alpha<\kappa$

Exercise 21.4 of Jech's Set Theory says:

Let $$\kappa$$ be an inaccessible cardinal. There is a notion of forcing $$(P,<)$$ such that $$|P| = \kappa$$ and $$P$$ is $$\alpha$$-distributive for all $$\alpha < \kappa$$, and such that $$\kappa$$ is not a Mahlo cardinal in the generic extension.

The hint below says:

Forcing conditions are sets $$p \subseteq \kappa$$ such that $$|p \cap \gamma| < \gamma$$ for every regular $$\gamma \leq \kappa$$; $$p \leq q$$ if and only if $$p$$ is an end-extension of $$q$$, i.e., if $$q = p \cap \alpha$$ for some $$\alpha$$. To show that for any $$\alpha < \kappa$$, $$P$$ does not add any new $$\alpha$$-sequence, observe that for every $$p$$ there is a $$q \leq p$$ such that $$P_q = \{r \in P : r \leq q\}$$ is $$\alpha$$-closed.

While it was not difficult proving that the $$P$$ in the hint is $$\alpha$$-distributive, I'm not sure how we have $$\kappa$$ is not Mahlo in $$V[G]$$. My guess is that $$P$$ adds a stationary subset of $$\kappa$$ that does not intersect the regular cardinals below $$\kappa$$, so I tried the most straightforward approach of taking $$\bigcup G$$. Yet it does not seem to be club.

In $$V[G]$$, $$X := \bigcup G$$ is an unbounded subset of $$\kappa$$. Moreover, by the distributivity claim, $$\kappa$$ is still regular.

Consider the club $$\lim (X) = \{ \alpha : X \cap \alpha \text{ is unbounded in } \alpha \}$$ (This is always a club). If $$S$$ is stationary, where $$S$$ is the set of regular cardinals, then $$X \cap S \neq \emptyset$$. Say $$\lambda \in X \cap S$$. Then $$\lambda$$ is a regular cardinal and $$C \cap \lambda$$ is unbounded in $$\lambda$$. $$\lambda$$ also was a regular cardinal in $$V$$ (the property of being regular is downwards absolute). Then there must have been a condition $$p \in G$$, with $$X \cap \lambda \subseteq p$$. This is a contradiction.

Let me modify $$P$$ so that conditions are closed subsets of $$\kappa$$ (everything else is left unchanged). I'm afraid I don't see why $$P$$ as Jech has defined it destroys the Mahloness. If conditions are not closed, then the generic object is just stationary, not club, and we need a club set that misses the inaccessibles below $$\kappa$$ to show that $$\kappa$$ is not Mahlo in the extension.

Let $$\dot C$$ be a name for the generic club $$\bigcup \dot G$$.

Claim: $$\dot C$$ is forced to be club in $$\kappa$$.

Proof: Unbounded: Fix $$\alpha_0<\kappa$$ and put $$p_0:=\{\alpha_0\}$$. Given $$\alpha_n$$ and $$p_n$$, find $$\alpha_{n+1}> \max\{\sup(p_n),\alpha_n\}$$ and $$p_{n+1}\le p_n$$ such that $$\alpha_{n+1}\in p_{n+1}$$. Put $$\alpha_{\omega}=\sup_n \alpha_n$$ and $$p_{\omega}=\bigcup_n p_n\cup \{\alpha_{\omega}\}$$. It is easy to see that $$p_\omega$$ is a condition forcing $$\alpha_\omega$$ into the generic object.

Closed: Suppose we had $$p\in P$$ and $$\alpha<\kappa$$ such that $$p\Vdash \alpha\in\lim(\dot C)\setminus\dot C$$. We may assume, extending $$p$$ if necessary, that $$p$$ has a largest element. Note that we must have $$\alpha<\max(p)$$ (else we could just add $$\alpha$$ into $$p$$ and get an extension forcing $$\alpha$$ into the generic object). But then $$\alpha$$ must be a limit point of $$p$$. Suppose otherwise and fix $$G$$ which is $$P$$-generic over $$V$$ and contains $$p$$. Find $$q\le p$$ such that $$\alpha\in q$$. This is impossible. So $$\alpha\in\lim(p)\subseteq p$$.

Let $$G$$ be $$P$$-generic over $$V$$ and $$S=\{\gamma<\kappa:(\gamma\text{ is inaccessible})^V\}=\{\gamma<\kappa:(\gamma\text{ is inaccessible})^{V[G]}\}$$ (because $$P$$ adds no bounded subsets of $$\kappa$$). Suppose $$S\cap\lim({\dot C}_G)\neq\emptyset$$. Take some $$p\in G$$ and $$\gamma\in S$$ with $$\gamma\in \lim(p)$$. By definition, $$|p\cap \gamma|<\gamma$$, so $$p\cap \gamma$$ is bounded in $$\gamma$$, contradicting that $$\gamma$$ is a limit point of $$p$$.

This shows that $$S$$ misses a club, so $$S$$ is non-stationary in the extension. In particular, $$\kappa$$ is not Mahlo in $$V[G]$$. Note that, since $$P$$ adds no bounded subsets of $$\kappa$$, $$\kappa$$ remains inaccessible in the extension.

• Thank you. Can you elaborate on why the generic stationary set avoids the set of regular cardinals below $\kappa$? Regardless, it seems like we can change the definition of $P$ such that all conditions $p$ do not contain any regular cardinals and the proof passes perfectly (since $\alpha_\omega$ has cofinality $\omega$). Commented Mar 20, 2022 at 1:21
• @ClementYung Upon further reflection, this doesn't immediately kill Mahloness, because stationary sets can of course be disjoint. And it's clear that the generic need not be club, for instance the condition $\{\aleph_n:n\in\omega\}\cup\{\aleph_{\omega}+1\}$ forces that $\aleph_\omega$ is a limit point outside of the generic object. I'll need to think harder about why $\kappa$ isn't Mahlo. My apologies for the confussion! Commented Mar 20, 2022 at 3:34
• I don't see why you want to use closed conditions. $\bigcup G$ is unbounded in $\kappa$, so $\lim(\bigcup G)$ (all the limit points, not just those in $\bigcup G$) is club. Commented Mar 20, 2022 at 9:55
• And to add to what @Jonathan said, you can show that as the forcing was defined by Jech, none of the limit points is regular. Commented Mar 20, 2022 at 10:11