Forcing and violation of the $GCH$ at $\aleph_\omega$

Question

In page 295 of Kunen's Set theory the author asserts that if $M$ is a countable transitive model of the axiom of constructibility $V=L$ then no forcing extension of $M$ can satisfy the theory $$ZFC\,\wedge\,\forall\,n\in \omega\,[2^{\aleph_n}=\aleph_{n+1}]\,\wedge\,2^{\aleph_\omega}\geq\aleph_{\omega+2}$$ Could someone give any hint or explanation?

Answer 1

Here's a fun proof that $0^\sharp$ can't be added by forcing:

• Suppose $V$ is a model of ${\sf ZFC}$, and $\mathbb{P}$ is a forcing notion such that $\Vdash_\mathbb{P}\exists 0^\sharp$.

• Let $\nu$ be a name such that $\Vdash_\mathbb{P}\nu[G]=0^\sharp$.

• Now I claim that for each $n\in\omega$, either $\Vdash_\mathbb{P} n\in\nu[G]$ or $\Vdash_\mathbb{P} n\not\in \nu[G]$. Why? Well, suppose not. Then by an appropriate product forcing we get $G\times H$ which is $\mathbb{P}\times\mathbb{P}$-generic such that $\nu[G]\not=\nu[H]$.

• So what? Well, consider the extension $V[G\times H]$. Being $0^\sharp$ is an absolute property, so $V[G\times H]\models \nu[G]=0^\sharp=\nu[H]$ - contradiction.

• OK, so we have: $$(*) \quad \forall n[(\Vdash_\mathbb{P} n\in\nu[G])\vee(\Vdash_\mathbb{P} n\not\in \nu[G])].$$ So we may "compute $0^\sharp$ in $V$": let $X=\{i: \Vdash_\mathbb{P}i\in\nu[G]\}$, and note that $X\in V$ (by definability of forcing).

• Now we use the downwards absoluteness of being $0^\sharp$: since a generic extension satisfies "$X=0^\sharp$," $V$ must already satisfy that! So $V\models\exists 0^\sharp$.

So: if a generic extension of $V$ satisfies "$0^\sharp$ exists," then so does $V$.

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I'm blackboxing the claim that being $0^\sharp$ is absolute. This is by Shoenfield absoluteness, via the fact (provably in ${\sf ZFC}$) that if $0^\sharp$ exists, it's a $\Pi^1_2$ singleton.

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For some interesting further remarks on the non-genericity of $0^\sharp$, see http://www.math.sjsu.edu/~stanley/inv-gen-zerosharp.pdf.