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The following is a result which I know as a weak form of Jensen's coding lemma$^*$ (first published in the book "Coding the universe"; also see http://www.jstor.org/stable/2273986):

For any class of ordinals $A\subseteq ON$ with $V=L[A]$, there is a class forcing $\mathcal{P}$ such that $\Vdash_\mathcal{P}V=L[a], a\subseteq\omega$.

(The coding theorem says more than this, but I doubt that the claim above has a substantially easier proof.)

My question is:

  • Can the class forcing $\mathcal{P}$ can assumed to be a set forcing, in the case when $A$ is a set?

I suspect this is true; however, the proof of the coding lemma is too difficult for me to untangle easily, so I'm not sure this is true.


$^*$ I've heard this called "Jensen's Coding Theorem," but the book is by Jensen and Beller & Welch; is this indeed due to Jensen, or all three of them?

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    $\begingroup$ Your displayed statement seems to be missing some assertion about $A$, since you say "for any class...$A$," but then don't mention $A$ again. $\endgroup$ Jul 12, 2014 at 0:25
  • $\begingroup$ Did you mean to say that $V=L[A]$ initially, and then you want the forcing to have $V[G]=L[a]$? $\endgroup$ Jul 12, 2014 at 1:05
  • $\begingroup$ Quite right, fixed. $\endgroup$ Jul 12, 2014 at 1:08
  • $\begingroup$ In that case, it is the last paragraph of my answer that is relevant, giving a positive answer. $\endgroup$ Jul 12, 2014 at 1:16
  • $\begingroup$ As far as I know, the proof of coding theorem goes back to 1975, which is due to Jensen. The book simplifies the proof, and gives some applications of the theorem. I may mention that the title of Beller's thesis is "Applications of Jensen's Coding Technique". $\endgroup$ Jul 12, 2014 at 2:51

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According to how I understand your question, the answer to your question is no. If $V$ is not of the form $L[x]$, where $x$ is a set, then it is impossible to have a set-forcing extension $V[G]$ of the form $L[a]$, where $a$ is a set. If $V[G]=L[a]$, where $a$ is a set and $G\subset P\in V$ is $V$-generic, then fix a name $\dot a$ for $a$, and consider $L[P][\dot a]$ (one should use well-ordered versions of these names to have a ZFC model). So $G$ is also $L[P][\dot a]$-generic, and $L[P][\dot a][G]$ has $a$ and hence includes $L[a]$ and is contained in $L[a]$, and so $L[P][\dot a][G]=L[a]=V[G]$. In particular, this means $$L[P][\dot a]\subset V\subset V[G]=L[P][\dot a][G]=L[a],$$ and so by the intermediate model theorem, $V$ must be a set-forcing extension of $L[p][\dot a]$, and so $V=L[P][\dot a][H]$ for some $H$ generic for a subforcing of $P$. In particular, this means that $V$ has the form $L[x]$ for the set $x=(P,\dot a,H)$, contrary to assumption.

It is easy to make models $V$ not of the form $L[x]$, simply by performing any of the usual class forcing, such as the usual Easton forcing or the canonical forcing of the GCH.

But I think that perhaps you meant to say that $V=L[A]$ as the hypothesis about $A$. In this case, if $A$ is a set, then of course we can make $V[G]=L[a]$ for $a\subset\omega$, simply by collapsing $\sup A$ to $\omega$ and then coding both $A$ and the collapsing function with a real. So under that interpretation of the question, the answer would be positive.

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    $\begingroup$ About your latter footnote question, I am not sure exactly how the credit parcels out, and I am hoping someone, perhaps Philip himself, will show up and explain it. $\endgroup$ Jul 12, 2014 at 0:38
  • $\begingroup$ OK, this was easier than I thought. Thanks! $\endgroup$ Jul 12, 2014 at 1:08
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    $\begingroup$ If $V=L[A]\models GCH,$ where $A$ is a set, then we can code the universe by a real using a set forcing, even without collapsing cardinals. If we allow collapsing cardinals, then the proof is easy as shown by Prof. Hamkins, but if we require the preservation of cardinals, then the proof becomes very complicated, in general. $\endgroup$ Jul 12, 2014 at 2:53

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