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:
- where 'every' is defined over a given universe of discourse, i.e. a model of set theory; and
- 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.
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 $V=L$?