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I'm trying to see that every weakly inaccessible cardinal $\kappa$ in a ctm $M$ remains weakly inaccesible if we force with a forcing which preserves cofinalities (thus, preserves cardinalities). It's clear for me that this situation can't be reproduced for a $\kappa$ inaccesible because we can destroy it adding enough Cohen reals; namely, forcing with $\mathbb{P}=Fn(\kappa\times\omega,2)$.

My attempt to prove that we can preserve inaccessible cardinals is the following:

Let $\mathbb{P}\in M$ any forcing which preserves cofinalities (for example, the Cohen forcing which adds only one real) and $\kappa$ a cardinal number such that $(\kappa\;\text{is weakly inaccessible})^M$ then:

  • $(\kappa\;\text{is a cardinal number})^{M[G]}$ because $\mathbb{P}$ preserves cardinalities.
  • $(\kappa\;\text{is regular})^{M[G]}$ because $\mathbb{P}$ preserves cofinalities and $\aleph_0$ is an absolute.
  • $(\kappa\;\text{is a limit cardinal})^{M[G]}$ because given $\lambda<\kappa$ a cardinal number in $M[G]$ then $\lambda\in M$ (by transitivity) so $(\lambda\;\text{is a cardinal})^M$. Since $(\kappa\;\text{is weakly inaccessible})^M$ then $(\lambda^+<\kappa)^M$ and therefore $((\lambda^+)^M<\kappa)^{M[G]}$ once again because our forcing preserves cardinal numbers. So, If we were willing to prove that $(\lambda^+)^M=(\lambda^+)^{M[G]}$ then we will be done. In this regard, $\leq$ is obvious and $\geq$ holds because $((\lambda^+)^M>\lambda)^{M[G]}$ and $((\lambda^+)^M\;\text{is a cardinal})^{M[G]}$.

Could someone tell me if my attemp is in fact a right proof?

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Yes, your argument is essentially the correct one, although it's not clear why it is important that $\aleph_0$ is an absolute in the second argument.

The whole point is that if $\Bbb P$ preserves cofinalities, then it preserves the property "$\alpha$ is a cardinal" as well, since otherwise the least ground model cardinal which is not a cardinal will have its cofinality changed.

You can even look at this in a more general way:

Suppose that $M\subseteq N$ are two models of $\sf ZFC$ and $\operatorname{cf}^M(\alpha)=\operatorname{cf}^N(\alpha)$ for every $\alpha\in\mathrm{Ord}^M$, then being weakly inaccessible is absolute between $M$ and $N$.

You don't even need that $N$ was a generic extension.

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  • $\begingroup$ thanks for your answer and for that more general result. On the other hand, you're right: it's not necessary for the argument the absoluteness of $\aleph_0$. $\endgroup$ – Antoine Apr 15 '16 at 13:37

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