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?
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2$\begingroup$ The statement implies an inner model with a measurable. You can't add measurable cardinals to $L$ via forcing. $\endgroup$– Asaf Karagila ♦May 4, 2016 at 14:49
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3$\begingroup$ Suppose there is such extension $M[G]$, then we would have that in $M[G]$ $0^\sharp$ exists, however in $L$ $0^\sharp$ does not exist, and so neither does in $M$, and you cannot add $0^\sharp$ to $M$ using forcing. $\endgroup$– Camilo Arosemena-SerratoMay 4, 2016 at 15:04
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$\begingroup$ Thank you both. But first, why that statements implies $0^\sharp$ holds in the generic extension and moreover, why is not possible to add $0^\sharp$ via forcing? Is there any reference to get in touch with that details? $\endgroup$– AntoineMay 4, 2016 at 19:15
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4$\begingroup$ This is an instance of a famous theorem of Jensen, the covering lemma for $L$ (the wikipedia page on this topic is surprisingly unhelpful and technical). Jech's set theory book discusses this result. For additional details, Devlin's book on constructibility is perhaps the most accessible reference. $\endgroup$– Andrés E. CaicedoMay 4, 2016 at 19:28
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$\begingroup$ @AndrésCaicedo I was under the impression that Devlin's book has a number of subtle errors, so I've stayed away from it - is his section on the covering lemma good? $\endgroup$– Noah SchweberJun 4, 2016 at 4:15
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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 $V$.
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.
For some interesting further remarks on the non-genericity of $0^\sharp$, see http://www.math.sjsu.edu/~stanley/inv-gen-zerosharp.pdf.
answered Jun 4, 2016 at 2:33
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$\begingroup$ Isn't $0^\#$ a $\Delta^1_3$ singleton? $\endgroup$– Asaf Karagila ♦Jun 4, 2016 at 4:03
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1$\begingroup$ @AsafKaragila No. $0^\sharp$ is a $\Delta^1_3$ real, but the set $\{0^\sharp\}$ is $\Pi^1_2$ (similarly to how, say, the true theory of first-order arithmetic $T$ has Turing degree $0^{(\omega)}$ but $\{T\}$ is $\Pi^0_2$). See Lemma 25.30 in Jech's big book. $\endgroup$ Jun 4, 2016 at 4:13