Suppose $m, n\in\omega$ and $\kappa$ is a cardinal. Then $\kappa$ is $\Pi^m_n$-indescribable if every $\Pi^m_n$-sentence true about $\kappa$ is true about some $\lambda<\kappa$; formally, if for every $\Pi_n$-sentence $\varphi$ in the language of set theory with a unary predicate and every $A\subseteq V_\kappa$, there is some $\lambda<\kappa$ such that $$(V_{\kappa+m}, \in, A)\models\varphi\implies (V_{\lambda+m}, \in, A\cap V_\lambda)\models\varphi.$$ For example, the $\Pi^1_1$-indescribable cardinals are exactly the weakly compact cardinals. (Note that we can go beyond $\Pi^m_n$-describability - quite a ways, even - but let's ignore that for now.)

I'm interested in a version of indescribability where we pay attention to what can happen in generic extensions of $V$ (I don't think this really qualifies as a "generic" version indescribability, but there's a vague connection). Specifically, for an ordinal $\alpha$ and a set $C\subseteq V_\alpha$, write "$(V_\alpha, \in, C)\models_f\varphi$" if $((V_\alpha)^{V[G]}, \in, C)\models\varphi$ for every $G$ which is set-generic over $V$ (note that the forcing for which $G$ is generic need not be an element of $V_\alpha$). Now say that a cardinal $\kappa$ is $\Pi^m_n$-forcing indescribable if for every $\Pi_n$-sentence $\varphi$ in the language of set theory with a unary predicate and every $A\subseteq V_\kappa$, there is some $\lambda<\kappa$ such that $$(V_{\kappa+m}, \in, A)\models_f\varphi\implies (V_{\lambda+m}, \in, A\cap V_\lambda)\models_f\varphi.$$ Note that $A$ is a fixed set in the ground model.

My question is, roughly how "big" are the forcing-indescribable cardinals? For example, it is not clear to me what the relationship is between $\Pi^1_1$-indescribable (= weakly compact) cardinals and $\Pi^1_1$-forcing indescribable cardinals. I believe that in $L$, $\Gamma$-forcing indescribability implies $\Gamma$-indescribability (since we can replace $\varphi$ with "$\varphi$ holds in $L$"), but I don't see the converse, or how this holds in general $V$.

  • $\begingroup$ Wouldn't it be possible to somehow mess things up just by collapsing $V_{\kappa+m}$ to be countable? $\endgroup$
    – Asaf Karagila
    Oct 14, 2016 at 19:49
  • $\begingroup$ @AsafKaragila In what way? Even if $V_{\kappa+m}$ is countable, $(V_{\kappa+m})^{V[G]}$ will have lots of interesting structure . . . $\endgroup$ Oct 14, 2016 at 19:51
  • $\begingroup$ Consider the name for the generic for the collapse of $V_\kappa$ to be countable. This is a subset of $V_\kappa$, so it's fine. Now state that this is a generic name for a surjection from the natural numbers onto the universe. Try to reflect that after forcing. $\endgroup$
    – Asaf Karagila
    Oct 14, 2016 at 19:53
  • $\begingroup$ @AsafKaragila But the statement "$A$ is a name for a surjection from the naturals onto the universe" isn't true in all forcing extensions of $V$ ($A$ doesn't change but $V_\kappa$ does), so $V_\kappa\not\models_f$ it. $\endgroup$ Oct 14, 2016 at 19:55
  • $\begingroup$ Right. Do you even have some non-trivial examples of $\varphi$ such that $(V_\kappa,\in,A)\models_f\varphi$? (And by non-trivial I mean avoiding things like $V\neq L$ or something like that.) $\endgroup$
    – Asaf Karagila
    Oct 14, 2016 at 19:58

1 Answer 1


Let me start things off by providing an upper bound. The bound is very large, however, and I expect that it can be improved, perhaps dramatically. But at least it shows the consistency of your large cardinal relative to some other well-studied large cardinals.

Theorem. If $\kappa$ is $1$-$C^{(2)}$-extendible, then it is forcing $\Pi^m_n$-indescribable for every $m,n$.

Definition. A cardinal $\kappa$ is $1$-$C^{(2)}$-extendible, if there is an elementary embedding $j:V_{\kappa+1}\to V_{\theta+1}$, with critical point $\kappa$, such that the target $j(\kappa)=\theta$ is $\Sigma_2$-correct in $V$, meaning $V_\theta\prec_{\Sigma_2} V$.

This is a fairly strong large cardinal notion, far stronger than the totally indescribable cardinals you mention in your question. For example, every 1-extendible cardinal is superstrong and much more. But these cardinals are weaker than Vopěnka's principle.

Proof. Assume that $\kappa$ is $1$-$C^{(2)}$-extendible. So there is an elementary embedding $j:V_{\kappa+1}\to V_{\theta+1}$ with critical point $\kappa$ and $j(\kappa)=\theta$ is $\Sigma_2$-correct in $V$.

Suppose now that $A\subset V_\kappa$ and $\langle V_{\kappa+m},\in,A\rangle\models_f\varphi$, which means that $\langle V[G]_{\kappa+m},\in,A\rangle\models\varphi$ for every forcing extension $V[G]$. This is a $\Sigma_2$ property about $A$ and $\kappa$, since any violation of it would be revealed inside some large enough $V_\eta$, using forcing inside that $V_\eta$.

Thus, by $\Sigma_2$-correctness, we see that $V_\theta$ agrees that $\langle V_{\kappa+m},\in,A\rangle\models_f\varphi$. Since $A=j(A)\cap\kappa$, we may pull this back by elementarity to conclude that there is some $\lambda<\kappa$ with $\langle V_{\lambda+m},\in,A\cap V_{\lambda+m}\rangle\models_f\varphi$ inside $V_\kappa$. But $\kappa$ itself must also be $\Sigma_2$-correct, and so actually $\langle V_{\lambda+m},\in,A\cap V_{\lambda+m}\rangle\models_f\varphi$ in $V$, as desired. QED

I'll think some more about lower bounds and about pulling down the strength of the hypothesis.

Update. I've realized that we can improve the upper bound as follows. We don't really need the "$+1$", since that actually provided a uniform version of the phenomenon, with the same embedding working for every $A$.

Thomas Johnstone and I defined that a cardinal $\kappa$ is uplifting, if it is inaccessible and $V_\kappa\prec V_\theta$ for cofinally many inaccessible cardinals $\theta$. (J. D. Hamkins, T. Johnstone, Resurrection axioms and uplifting cardinals) A boldface version is that $\kappa$ is strongly uplifting, if for every $A\subset V_\kappa$ there are cofinally many inaccessible cardinals $\theta$ for which $\langle V_\kappa,\in,A\rangle\prec\langle V_\theta,\in,A^*\rangle$ for some $A^*\subset V_\theta$. (J. D. Hamkins, T. Johnstone, Strongly uplifting cardinals and boldface resurrection) These have diverse equivalent formulations, as I mention on the linked blog post, connected with strengthenings of the strongly unfoldable cardinals.

Let me now strengthen this a little more, for the present application, with the following new large cardinal concept.

Definition. A cardinal $\kappa$ is strongly $C^{(n)}$-uplifting, if for every $A\subset V_\kappa$ there is a $\Sigma_n$-correct cardinal $\theta$ and $A^*\subset V_\theta$ with $\langle V_\kappa,\in,A\rangle\prec\langle V_\theta,\in,A^*\rangle$.

This is what we really needed in the theorem above.

Theorem. If $\kappa$ is strongly $C^{(2)}$-uplifting, then it is forcing $\Pi^m_n$-indescribable for every $m$ and $n$.

Proof. Argue as in the first theorem above, but now we have only $\langle V_\kappa,\in,A\rangle\prec\langle V_\theta,\in,A^*\rangle$, instead of $j$. If $\langle V_\kappa,\in,A\rangle+m\models_f\varphi$, then this will be true inside $V_\theta$ since it is $\Sigma_2$-correct, and so $V_\theta$ thinks that this holds on an initial segment of $A^*$, and so we get $\lambda<\kappa$ with $\langle V_\lambda,\in,A\cap V_\lambda\rangle\models_f\varphi$ inside $V_\kappa$, which is right about this since $\kappa$ is itself $\Sigma_2$-correct. QED

I think the strongly $C^{(2)}$-uplifting cardinals are comparatively weak, and absolute to $L$, but I'll think more about it.

  • $\begingroup$ That Wikipedia article is really annoying in switching between "extendible" and "extensible" arbitrarily. Sometimes even in the same sentence. $\endgroup$
    – Asaf Karagila
    Oct 15, 2016 at 15:46
  • $\begingroup$ I think it is not arbitrary, but is using that difference to indicate $\forall\eta$. I hadn't seen this usage much before, but it cites Kanamori for it. Is that usage in Kanamori? I always just call them "extendible" rather than "extensible". $\endgroup$ Oct 15, 2016 at 15:55
  • $\begingroup$ I don't think "extensible" appears in Kanamori, and it really the last two anonymous edits that introduced that. In any case, I think that it's terrible grammar to switch from US to UK spelling of the same word in the middle of the sentence... $\endgroup$
    – Asaf Karagila
    Oct 15, 2016 at 16:43
  • $\begingroup$ If that is true, I think it should be edited back to "extendible." $\endgroup$ Oct 15, 2016 at 17:25
  • 1
    $\begingroup$ @JoelDavidHamkins Re: lower bounds, I've been thinking and I can't even show that every $\Pi^1_1$-forcing indescribable cardinal is weakly compact. Am I missing something, or is this indeed nontrivial? If it is nontrivial, do you know the answer? (I would love it if the least $\Pi^1_1$-forcing indescribable can consistently be much lower than the least weak compact.) $\endgroup$ Oct 15, 2016 at 18:56

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