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Sacks forcing allows us to build a model $V[G]$, such that there is no "intermediate model" between $V$ and $V[G]$, meaning if $V \subseteq W \subseteq V[G]$ is a model of ZFC then either $W = V$ or $W = V[G]$.

My question is:

  1. Whether we know how to force a model to have exactly $1$ intermediate model?
  2. Assuming the answer is Yes, do we know how to extend this result arbitrarily long? Meaning, do we know how to force a model where all the inner models (i.e standard, transitive classes that contain all the ordinals) are well ordered and have a $1-1$ correspondence with the ordinals?

This would necessate to start with $V_0 = L$, and then possibly take $V_1 = V_0[G]$ (where this would be a regular Sacks forcing) and somehow continue to create $V_\alpha$ for all $\alpha \in ORD$, without 'accidentally' creating any more inner models.

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Yes, there is a whole literature on this kind of thing. One of the main methods is to perform iterations and products of Sacks forcing, so as to realize a given (set-sized) partial order of inner models.

Marcia Groszek has been very active in this area. For example, you could begin with the following.

Although the methods are extremely flexible for achieving set-sized orders for the structure of inner models, and in particular, any ordinal can be realized. But to have actually exactly Ord many inner models, you cannot use Sacks forcing alone, since there would be too many reals. So I am less sure about question 2.

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    $\begingroup$ I'd imagine that for the second question, under some assumption, an iteration of generalized Sacks forcings might work. Assuming that they are minimal (at least under some mitigating assumptions, e.g. a proper class of inaccessible cardinals, and add a Sacks subset to each one, iteratively). $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 16, 2016 at 23:16
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    $\begingroup$ Yes, I have had similar thoughts. $\endgroup$
    – Joel David Hamkins
    Commented Oct 17, 2016 at 0:02
  • $\begingroup$ Very interesting! Thank you for the swift answer :). $\endgroup$
    – Alon Navon
    Commented Oct 17, 2016 at 22:44
  • $\begingroup$ The question seems to be related to "degrees of constructibility of sets". $\endgroup$
    – Mohammad Golshani
    Commented Oct 18, 2016 at 9:56

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