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Here's a very silly mistake I made recently: I claimed that if $\mathbb{P}\in L(\mathbb{R})$ is a forcing which adds a real, then $$(*)\quad L(\mathbb{R})^{V^\mathbb{P}}=L(\mathbb{R})^\mathbb{P}.$$ While it is true (EDIT: no it isn't, see the answer below) that $L(\mathbb{R})^\mathbb{P}=L(\mathbb{R})^{(L(\mathbb{R})^\mathbb{P})}$, what I wrote above is nonsense: $V$ may have names for reals which are not in $\mathbb{R}$.

Indeed, from large cardinals we can prove that $(*)$ fails - if it held, then examining $Col(\omega,\kappa)$ for $\kappa$ regular in $V$, we'd have that $L(\mathbb{R})$ contains a proper class of measurables; and this is known to not be the case, assuming large cardinals (specifically, from large cardinals there are no measurable cardinals in $L(\mathbb{R})$ above $\Theta$).

My question is around principles of the form $(*)$. Specifically, for a definable class of forcings $\mathcal{C}$ which add a real, let (FC)$_\mathcal{C}$ ("forcing conflation") be the statement $(*)$ restricted to forcings in $\mathcal{C}\cap L(\mathbb{R})$. I'm interested in when (FC)$_\mathcal{C}$ is compatible with large cardinals - specifically, a proper class of Woodins - and when it adds large cardinal strength to ZFC+proper class of Woodins.

To make this concrete, I'll ask:

Is (FC)$_{proper}$ consistent with ZFC + "There is a proper class of Woodins," relative perhaps to even stronger large cardinal hypotheses? If so, does ZFC + "There is a proper class of Woodins" + (FC)$_{proper}$ have consistency strength beyond a proper class of Woodins?

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$(FC)_{\text{proper}}$ is false. If there are infinitely many Woodin cardinals with a measurable above, then after adding a single Cohen real, we do not have $L(\mathbb R)[c] = L(\mathbb R)^{V[c]}$. The reason is that $L(\mathbb R)[c]$ does not satisfy $\text{AD}$: in $L(\mathbb R)[c]$, the set of ground model reals is uncountable, but does not contain a perfect subset. (See Hamkins's answer to the question Is there a perfect set of ground model reals in the Cohen extension? which uses at worst countable choice.) On the other hand, $L(\mathbb R)^{V[c]}$ satisfies $\text{AD}$ by our large cardinal hypothesis.

Note in this situation that $L(\mathbb R)^{V[c]}=L(\mathbb R)^{L(\mathbb R)[c]}\subsetneq L(\mathbb R)[c]$, since nice names for reals are all coded by reals. Therefore it is not true in general that $L(\mathbb R)^{L(\mathbb R)[r]} = L(\mathbb R)[r]$ (or that $L(\mathbb R)[r]\subseteq L(\mathbb R)^{V[r]}$) when $r$ is a generic real. The problem this time is that $\mathbb R$ (as computed in the ground model) need not be in $L(\mathbb R)^{L(\mathbb R)[r]}$.

The large cardinal assumption can actually be reduced to the assumption that $\text{AD}$ holds in $L(\mathbb R)$, because it can be proved under this hypothesis that $ L(\mathbb R)^{V[c]} = L(\mathbb R)^{L(\mathbb R)[c]}$ satisfies $\text{AD}$. You might be interested in Kechris and Woodin's paper Generic Codes, in which they prove this fact and analyze forcing over $L(\mathbb R)$ with Levy collapses, showing that for $\kappa<\delta^2_1$ a reliable ordinal, if $G\subseteq \text{Col}(\omega, \kappa)$ is $L(\mathbb R)$-generic then in $L(\mathbb R)[G]$, there is a definable elementary embedding from $L(\mathbb R)$ into $L(\mathbb R)^{L(\mathbb R)[G]}$

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  • $\begingroup$ Welcome to MathOverflow! And thank you for this clarifying post. I wonder whether one can prove $L(\mathbb{R})^{V[c]}\subsetneq L(\mathbb{R})[c]$ under considerably weaker assumptions? $\endgroup$
    – Joel David Hamkins
    Commented Jan 16, 2017 at 12:04
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    $\begingroup$ Suppose there is no inner model with a Woodin cardinal. Let $K$ denote the Jensen-Steel core model, and suppose $c$ is Cohen over $K$. One theorem of Steel implies that $K^{K[c]} = K$. Another states that $K\cap \textit{HC}$ is definable in $L_{\omega_1}(\mathbb R)$. Thus $K\cap \textit{HC} = K^{K[c]}\cap \textit{HC}^{K[c]}\in L(\mathbb R)^{K[c]}$. It follows that $L(\mathbb R)^K\subseteq L(\mathbb R)^{K[c]}$, and so the two models are equal. So in some sense no large cardinal hypothesis below a Woodin suffices to show $L(\mathbb R)^{L(\mathbb R)[c]}\subsetneq L(\mathbb R)[c]$. $\endgroup$
    – Gabe Goldberg
    Commented Jan 16, 2017 at 22:30
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    $\begingroup$ The statement "$L(\mathbb R)^{V[c]}\subsetneq L(\mathbb R)[c]$" has no consistency strength beyond ZFC plus an inaccessible. Suppose $G$ is $V$-generic for $\text{Col}(\omega,{<}\kappa)$. We have $\text{Col}(\omega,{<}\kappa) \simeq \text{Col}(\omega,{<}\kappa)\times \text{Add}(\omega,1)$. So by Solovay's theorem, $V[G]$ and $V[G][c]$ both satisfy the statement "Every uncountable set of reals in $L(\mathbb R)$ has a perfect subset," and this suffices to run my argument. Thanks for welcoming me, although actually I just accidentally created a second account. $\endgroup$
    – Gabe Goldberg
    Commented Jan 16, 2017 at 22:33
  • $\begingroup$ Thanks very much. The moderators can merge your accounts, if you just let them know the usernames. $\endgroup$
    – Joel David Hamkins
    Commented Jan 16, 2017 at 22:40
  • $\begingroup$ When I said in my first comment that "the two models are equal" I meant the two models $L(\mathbb R)^{K[c]}$ and $L(\mathbb R)^K[c]$. (I merged the accounts, by the way, but they were both of the form "user $n$".) $\endgroup$
    – Gabe Goldberg
    Commented Jan 17, 2017 at 0:26

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