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The question is whether, when you add a Cohen subset to a cardinal $\kappa$, that cardinal becomes a characteristic of the resulting forcing extension $V[G]$. Or can there be strange instances in which the very same model is realized as a Cohen subset forcing extension over different ground models with different cardinals?

To be precise, can it happen that $M[G]=N[H]$, where $M$ and $N$ are transitive models of ZFC and $G$ is $M$-generic for the forcing to add a Cohen subset to some cardinal $\kappa$, that is, using $\text{Add}(\kappa,1)^M$, and $H$ is similarly $N$-generic to add a Cohen subset to some other cardinal $\delta$, using $\text{Add}(\delta,1)^N$?

For a more concrete version of the question, imagine that we have added a Cohen real $c$ and form the extension $M[c]$; could it be that this model might also be realized as $N[A]$ for some other ground model $N$, where $A$ is an $N$-generic Cohen subset of $\omega_1^N$? Note that $M\neq N$ since it must be that $c\in N$ as the higher forcing does not add reals. For my application, I need to understand the case where the two cardinals are both inaccessible cardinals (if not much more). Also, it is not difficult to identify general situations where this kind of thing is impossible. What I really want to know is if it can ever happen at all.

I conjecture that this situation is impossible, and that indeed, when you add a Cohen subset to a cardinal, you have in particular made that cardinal definable, as "the cardinal for which the universe was just obtained by adding a Cohen subset to it".

The question is really a part of the subject known as set-theoretic geology, but it has recently arisen in another project of mine.

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Are you only interested in the case where you specifically force to get just a Cohen subset, i.e. only $Add(\kappa,1)$, or could you allow forcings which generically adjoin a Cohen subset as a by-product of adjoining some other generic object? – Everett Piper Feb 1 '13 at 21:35
Everett, in that generality, the phenomenon could happen, for example, if you allowed the product $\text{Add}(\kappa,1)\times\text{Add}(\delta,1)$. So mainly, I am interested in just $\text{Add}(\kappa,1)$. But actually, I have in mind more general kinds of forcing, such as $\text{Add}(\kappa,\kappa^+)$ or $\text{Coll}(\kappa,\kappa^+)$, as well. – Joel David Hamkins Feb 1 '13 at 22:36
Joel, an observation: if $M=W[A]$ is already an $Add(\delta,1)$ extension for some $\delta>\kappa$, then forcing to add a subset to $\kappa$ is equivalent to forcing over $W$ with the product, and so the answer to your question is 'yes'. If we specifically disallow $M$ of this form, then we have $M[G]=N[A]$ for which there is no $W$ with $W[A][G]=M[G]=N[A]$, which is looking an awfully lot like a counterexample to directedness of grounds (I may be ignoring some subtleties here, however.) – jonasreitz Feb 2 '13 at 3:55
Jonas, I don't think your example works, since if $B$ is the subset of $\kappa$ you are proposing to add, then although $W[A][B]$ is an extension by adding a subset to $\kappa$, nevertheless $W[B][A]$ is not an extension of $W[B]$ by $\text{Add}(\delta,1}^{W[B]}$, since $B$ added new conditions to the partial order. Or have I misunderstood you? – Joel David Hamkins Feb 2 '13 at 12:39
...$Add(\delta,1)^{W[B]}$... – Joel David Hamkins Feb 2 '13 at 12:40
up vote 4 down vote accepted

The main part of this question is answered by theorem 10 of my joint paper with Bagaria, Tasprounis and Usuba, below.

From J. Bagaria, J. D. Hamkins, K. Tsaprounis, T. Usuba, Superstrong and other large cardinals are never Laver indestructible.

Let $\mathcal{C}(\kappa)$ assert that $\kappa$ is a regular cardinal and the universe was obtained by forcing over some ground $W$ to add a Cohen subset to $\kappa$, that is, "$V=W[G]$ for some ground $W$ and some $W$-generic $G\subset\text{Add}(\kappa,1)^W$."

Theorem. If $\mathcal{C}(\gamma)$ and $\mathcal{C}(\kappa)$ hold, where $\gamma<\kappa$, then $2^\gamma=\kappa$. Consequently,

  • There are at most two regular cardinals satisfying property $\mathcal{C}$.

  • There is at most one inaccessible cardinal satisfying property $\mathcal{C}$.

  • If $\mathcal{C}(\kappa)$ holds, then $\kappa$ is $\Delta_3$-definable as $$\text{``the smallest regular cardinal with property $\mathcal{C}$,''}$$ or as $$\text{``the second regular cardinal with property $\mathcal{C}$.''}$$

  • If $\mathcal{C}(\kappa)$ holds and $\kappa$ is inaccessible, then $\kappa$ is $\Pi_2$-definable as $$\text{``the inaccessible cardinal with property $\mathcal{C}$.''}$$

Furthermore, these definitions work also in $V_\theta$, whenever $\theta$ is a $\beth$-fixed point of cofinality larger than $2^{2^\kappa}$ or for which $V_\theta$ satisfies $\Sigma_2$-collection.

In other words, if the universe can be realized both as a forcing extension by adding a Cohen subset to $\gamma$ and also as a forcing extension (over a different ground) by adding a subset to $\kappa$, and $\gamma\lt\kappa$, then $\kappa=2^\gamma$. It follows that there can be at most two cardinals that arise in this way, and furthermore, that in the case of an inaccessible cardinal, there is at most one over which the universe was obtained by adding a Cohen subset.

We do not know, however, whether the situation of $2^\gamma=\kappa$ can actually arise and perhaps it is true in general that there can be at most one cardinal to which one has added a Cohen subset.

The main theorem of the paper is that superstrong and other large cardinal notions (including almost huge cardinals, huge cardinals, superhuge cardinals, rank-into-rank cardinals, extendible cardinals, $1$-extendible cardinals, $0$-extendible cardinals, weakly superstrong cardinals, uplifting cardinals, pseudo-uplifting cardinals, superstrongly unfoldable cardinals, $\Sigma_n$-reflecting cardinals, $\Sigma_n$-correct cardinals and $\Sigma_n$-extendible cardinals, for $n\geq 3$) are never Laver indestructible. The method of proof of the main theorem applies, however, to answer this question, and I view this answer really as an explanation of the main nonindestructibility result, for the reasons that I had explained the conjecture in the question here.

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Nice!${}{}{}{}$ – Andrés Caicedo Jul 15 '13 at 2:36
Thanks, Andres, I hadn't expected that we could settle this, but it turns out that the methods do work out. But meanwhile, there is still something open, so I hope that others may press ahead... – Joel David Hamkins Jul 15 '13 at 2:39
I've accepted this answer, but I shall be delighted to accept another answer, if someone can settle the general case. – Joel David Hamkins Jul 15 '13 at 17:21
Joel, your result seems to imply that a modified Silver iteration of suitable length $\lambda$ will ``kill" all the large cardinals you mention above (and more) that are below $\lambda$ (in the ground model $V$). Maybe you'll have some of these large cardinals $<\lambda$ in $V[G]$, but they won't have had that property in $V$. Is this right? Can you say more about how long iterations will interact with this phenomenon? – Everett Piper Jul 18 '13 at 0:35
Or are products and iterations irrelevant (you seemed to indicate this in a response above) in the sense that the definability of the cardinal in question is only relevant at that stage of the iteration? – Everett Piper Jul 18 '13 at 0:36

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