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Definition (1): A‎n‎‎ ‎inner ‎model ‎of ‎‎$‎ZFC‎$ ‎is a‎ ‎tarnsitive proper class ‎model ‎‎of $‎‎ZFC$ ‎which ‎contains ‎all ‎ordinal numbers. ‎Informally ‎we ‎denote ‎the ‎collection ‎of ‎all ‎inner ‎models ‎of ‎‎$‎ZFC‎$ ‎by ‎‎$‎‎Inn(ZFC)$.‎ ‎

Now ‎if we ‎‎consider the partial order (reflexive and transitive) $‎\langle ‎Inn(ZFC),\subseteq ‎\rangle‎‎‎‎$ ‎with ‎end ‎points ‎‎$‎L‎$ ‎and ‎‎$‎‎V$ ‎then ‎there ‎are ‎some ‎natural ‎questions ‎about ‎other properties of this partial order:‎ ‎

Question (1): ‎Is‎ $‎\langle ‎Inn(ZFC),\subseteq ‎\rangle‎$ ‎linear? ‎In the other words‎, ‎is ‎the ‎following ‎statement ‎true?‎ ‎

‎‎$‎‎‎\forall ‎M,N\in Inn(ZFC)‎~~~M\subseteq N~~\vee~~N\subseteq M$‎ ‎

Question (2): ‎Is‎ $‎\langle ‎Inn(ZFC),\subseteq ‎\rangle‎$ ‎dense?‎ ‎In the other words‎, ‎is ‎the ‎following ‎statement ‎true?‎ ‎‎

$‎‎‎\forall ‎M,N\in Inn(ZFC)‎~~~(M\subsetneq N \Longrightarrow ‎\exists ‎P\in Inn(ZFC)~~~M\subsetneq P \subsetneq N‎)$‎

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    $\begingroup$ It is consistent that the answer to Question 1 is no. This would be the case, for example, if $V$ was obtained by adding two mutually generic Cohen reals $c_1,c_2$ to $L$. Then $L[c_1]$ and $L[c_2]$ are incomparable. On the other hand, if $V=L$, the answer is obviously yes. $\endgroup$
    – Miha Habič
    Sep 21, 2013 at 12:25
  • $\begingroup$ Ok. What about question (2)? $\endgroup$
    – user36136
    Sep 21, 2013 at 12:55

2 Answers 2

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Your question is related to the concept of degrees of constructibility, where we say that $c\equiv_L d$ if and only if $L[c]=L[d]$. When $c$ is well-ordered in $L[c]$, then this will be an inner model of ZFC, and so the structure theory of the degrees of construtibility are a part of your informal treatment of inner models.

The inner models can be linear, for example, if you add a Sacks real generically over $L$, then there are precisely two inner models, $L$ and $L[s]$, because of the minimality property of the Sacks real. This also shows that the inner models may not be dense. Indeed, if one adds a Sacks real over any $V$, then there are no models between $V$ and $V[s]$, which is exactly the Sacks minimality property. Meanwhile, as Miha pointed out, it is easy to make it non-linear, and in fact many other patterns are possible.

Theorem. If $c$ is an $L$-generic Cohen real, then the inner models of $L[c]$ are densely ordered by inclusion, but not linearly ordered.

Proof. Every inner model $M$ with $L\subset M\subset L[c]$ is a forcing extension by a subalgebra of Cohen forcing, and all such subalgebras are themselves isomorphic to Cohen forcing. So the inner models are just of the form $L[d]$ for a Cohen real $d$, and the quotient forcing $L[d]\subset L[c]$ is still countably-dense and hence also isomorphic to Cohen forcing. But if we add a Cohen real $r$ over any model $V$, then we can let $r_0$ be the even digits of $r$, and find $V\subsetneq V[r_0]\subsetneq V[r]$. So we get density precisely because all the forcing extensions arise by Cohen forcing. Namely, if $L[d_0]\subsetneq L[d_1]$, then $L[d_1]=L[d_0][e]$ for some Cohen real $e$, and we may take every-other digit of $e$ to find an intermediate model between $L[d_0]$ and $L[d_1]$. The order is not linear, since we may split $c$ into even and odd parts, and find mutually generic reals $c_0$ and $c_1$, whose extensions $L[c_0]$ nad $L[c_1]$ are incomparable, yet have meet $L$ and join $L[c]$. QED

In answer to the related question What can the degrees of constructibility be?, I posted the following:

The article Initial segments of the degrees of constructibility by Marcia Groszek and Richard Shore (Israel Journal of Mathematics June 1988, Volume 63, Issue 2, pp 149-177) shows that

Any constructible, constructibly countable, (dual) algebraic lattice is isomorphic to the degrees of constructibility of reals in some generic extension of L.

And there is a lot of further work on this topic by Groszek, Slaman and others.

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  • $\begingroup$ The inner models arising in a forcing extension over $L$ by a forcing notion $\mathbb{B}$ are precisely isomorphic to the collection of complete subalgebras of $\mathbb{B}$. $\endgroup$
    – Joel David Hamkins
    Sep 21, 2013 at 13:57
  • $\begingroup$ The comment seems a bit fishy. What if $\Bbb B$ is the completion of a trivial forcing which just chooses an ordinal below $\omega_1$? Certainly this has a lot of nontrivial complete subalgebras (defined by subsets of atoms) but in fact there are no inner models, because we didn't add anything to the universe. $\endgroup$
    – Asaf Karagila ♦
    Sep 21, 2013 at 14:00
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    $\begingroup$ Yes, you are right. I should assume that $\mathbb{B}$ is atomless (and I had meant that $\mathbb{B}$ is a complete Boolean algebra). $\endgroup$
    – Joel David Hamkins
    Sep 21, 2013 at 14:14
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    $\begingroup$ Joel, you may want to add a small remark to the theorem, indicating that adding a Cohen real results in a partial rather than linear order. $\endgroup$
    – Andrés E. Caicedo
    Sep 21, 2013 at 14:43
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The answer to the second question is negative. Take, for example, Sacks forcing over $L$. The result is $L[r]$ where $r$ is a real number and the following property is true: $$\forall x(x\in L\lor r\in L[x])$$

This shows that there is no intermediate model between $L$ and an extension by one Sacks real. This property is called minimality, and it is not unique to the Sacks real.

The first answer is also negative, as others have said. One can force over $L$ both a Cohen real $c$ and a random real $r$ (as a product of the forcing, that is), and then $L[r],L[c]$ are two different inner models of $L[r\times c]$ which are incomparable.

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