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I just saw an interesting video from Hugh Woodin about Ultimate $L$. In it, he says one of the reasons $L$ is so interesting is because it not only settles many natural set theory questions, but is also "immune" to Cohen's forcing technique. In what sense can we make this precise?

I get the basic idea that we can't force into existence extra constructible sets that aren't actually constructible, but I'd like to get precise on what this means:

  • Are we saying that if we have $ZFC$ in the metatheory, and we have some model of $ZFC + V=L$, any forcing extension of it will no longer satisfy $V=L$, so that there is some absoluteness property involved?
  • Or are we saying something else involving how forcing works if the metatheory is $ZFC + V=L$?
  • Or both?
  • And how many of these results are only true for countable transitive models?

Also, any good references to learn about this stuff would be much appreciated.

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If we start with a model of $V=L$, then no (non trivial) generic extension will satisfy $V=L$.

So, to that extent, if you have adopted this axiom as "true", nothing can be changed by forcing. So, continuum function, diamonds, squares, trees, etc. must all be fixed already.

So the only changes we can make involve adding ordinals. Which, if you're a Platonist and take the universe to be complete, would be impossible.

Woodin points out that $L$ does not really accommodate large cardinal axioms above $0^\#$, which is the goal of the Ultimate-$L$ programme: have an axiom like $V=L$ which is compatible with very large cardinal axioms.

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  • $\begingroup$ Thanks, that is very interesting. Do we get something similar for $HOD$? $\endgroup$
    – Mike Battaglia
    Commented Jul 4 at 16:49
  • $\begingroup$ Oh my, no. $\rm HOD$ is the second order logic analogue of $L$, and that gets very finicky. You can make any model $\rm HOD$ of a class generic extension, it can accommodate all known large cardinal axioms, etc. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 4 at 18:38
  • $\begingroup$ Hm, that's kind of surprising. Well, it is interesting that $V=L$ proves that there is no forcing extension of any model of itself (and the universe, I guess?). How strong is the axiom "there is no forcing extension of any model of this theory?" or perhaps "there is no forcing extension of the universe?" Is there some way to formalize these things directly and add to ZFC? $\endgroup$
    – Mike Battaglia
    Commented Jul 4 at 20:28
  • $\begingroup$ What do you mean by "how strong is the axiom"? It follows from $V=L$, and we can get it with large cardinals, but that's not a particularly well studied axiom... $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 4 at 22:25
  • $\begingroup$ Oh, I was hoping it was well studied enough that someone would be able to say what follows from it. What is the correct way to phrase this property? "There is no forcing extension of the universe?" $\endgroup$
    – Mike Battaglia
    Commented Jul 4 at 22:30

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