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In his answer to the MathOverflow question Gödel's Constructible Universe in Infinitary Logics, Prof. Hamkins gives a very interesting answer and proof to user46667's second question:

(2) What is $L_\infty$?

Here is Prof Hamkins' answer, and the first two sentences to his proof:

Theorem. $L_\infty$ is the entire set-theoretic universe $V$.

Proof. I claim that every set will arise in the construction process.... In infinitary logic, there are far more than only countably many formulas, and one can cook up a formula to define a specific set, by using the formulas that define its elements...."

My question is twofold:

First, what is the sense of "every" in the sentence containing "every set will arise in the construction process"? Since 'every' is a quantification term, its sense can be construed in one of two (and possibly more) ways:

  1. where 'every' is defined over a given universe of discourse, i.e. a model of set theory; and
  2. the unrestricted sense, where 'every' is defined over all possible sets.

By definition of $L_\infty$ and its associated infinitary language, the sense of 'every' as used in the proof seems to be (at least as I understand it) the unrestricted sense (the rest of the proof seems to confirm this also).

Second, if $L_\infty=V$, then how does one construe $G$ in the forcing extension $V[G]$, where $G$ is usually construed to exist 'outside' $V$?

Please help me out here so I can better understand the proof.

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You are making it way more complicated than it is. We work internally to a fixed universe of $\sf ZFC$, called $V$.

If we use $\cal L_{\infty,\infty}$ which allows any set-many quantifiers and disjunctions, then we can explicitly define every set of ordinals in the universe. It follows that every set of ordinals from $V$ is in $L_\infty$. Two models of $\sf ZFC$ with the same sets of ordinals (and in particular, the same ordinals) are equal. Therefore $V=L_\infty$.

Your second question has nothing to do with the first actually. Generic live outside the universe, in a larger universe. What we can conclude from $L_\infty=V$ is that $V$ is not a non-trivial generic extension of $L_\infty$.

If this bothers you, perhaps first you should ask yourself, where does the generic $G$ comes from if we assume $V=L$?

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    $\begingroup$ Or, perhaps more to the point: where does the generic $G$ come from if we assume $V = V$? (And why would we assume otherwise?) $\endgroup$ – user642796 Feb 15 '14 at 12:40
  • $\begingroup$ @Asaf: I guess the question one needs to ask is: can a generic set G be defined by some $\mathcal L_{\infty}{\infty}$ formula? If so, then by definition of $mathcal L_{\infty}{\infty}$ one should be able to "cook up" a formula adding an unlmited number of such generic sets to V. However, since one assumes that V=L$_infty$, one must conclude that these generic sets are already in V seemingly contradicting the fact that, say, V=L and CH are independent of ZFC. Why is it impossible to define a generic set G by a $\mathcal L_{\infty}{\infty}$ formula? $\endgroup$ – Thomas Benjamin Feb 24 '14 at 9:24
  • $\begingroup$ @Thomas: Why would it be possible to define a generic set explicitly by a formula? Formulas can define things inside the universe. Generic sets are not inside the universe. $\endgroup$ – Asaf Karagila Feb 24 '14 at 10:08
  • $\begingroup$ @Arthur: Thanks for editing my question, as I am not getting quite the hang of using mathjax. Would you be so kind as to give me the mathjax (LaTex) codes for infinite conjunction/disjunction with index set, and the codes for writing the symbols for infinitary languages? It would be much appreciated. $\endgroup$ – Thomas Benjamin Feb 24 '14 at 18:12
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    $\begingroup$ @Thomas: I have no idea what you're trying to ask. $\endgroup$ – Asaf Karagila Jul 25 '14 at 8:44

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