Consider the following result (which is Lemma 2.8 in Mitchell and Steel's paper on Fine Structure and Iteration Trees):

Lemma 2.8 Let $\pi \colon \mathcal{H} \to \mathcal{M}$ be generalized $r \Sigma_{k}$ elementary, where $\mathcal M$ is a ppm (not of type III) and $1 \leq k < \omega$. Suppose that $\rho_{k}^{\mathcal{M}} \subseteq \mathcal{H}$ and $\pi \restriction \rho_{k}^{\mathcal{M}} = \operatorname{id}$. Suppose also that $\pi(r)$ is the $k$th standard parameter of $(\mathcal{M}, \pi(q))$ and that $\pi(r)$ is $k$-solid and $k$-universal over $(\mathcal{M}, \pi(q))$. Then

  1. $\rho_{k}^{\mathcal{H}} = \rho_{k}^{\mathcal{M}}$,
  2. $r$ is the $k$th standard parameter of $(\mathcal{H},q)$ and
  3. $r$ is $k$-universal over $(\mathcal{H}, q)$.

The proofs of items $1.$ and $2.$ are included in this paper and in both cases it seems that the $k$-solidity of $\pi(r)$ over $(\mathcal{M}, \pi(q))$ isn't actually needed. Hence I decided to prove item $3.$ to see how $k$-solidity comes into play. However, if my argument is correct, it doesn't rely on $k$-solidity either.

Here is my proof of item $3.$:

Proof (of $3.$). Let $A \in \mathcal{H}$ be such that $A \subseteq \rho_{k}^{\mathcal{H}}$. Then $\pi(A) \cap \rho_{k}^{\mathcal{M}} \in \mathcal M$. By the $k$-universality of $\pi(r)$ over $(\mathcal{M}, \pi(q))$ there is hence some generalized Skolem term $\tau \in S_{\kappa}$ and some $\vec{\alpha} \in ^{< \omega} \rho_{k}^{\mathcal{M}}$ s.t. $$ \pi(A) \cap \rho_{k}^{\mathcal{M}} = \tau^{\mathcal{M}}[\vec{\alpha}, \pi(r), \pi(q)] \cap \rho_{k}^{\mathcal{M}}. $$ Let $B := \tau^{\mathcal{H}}[\vec{\alpha},r,q] \cap \rho_{k}^{\mathcal{H}}$. Combining the fact that $\pi$ is generalized $r \Sigma_{k}$-elementary, $\rho_{k}^{\mathcal{H}}= \rho_{k}^{\mathcal{M}} \subseteq \mathcal{H}$ and $\pi \restriction \rho_{k}^{\mathcal M} = \operatorname{id}$ we have that \begin{align*} \mathcal{H} \models A \cap \rho_{k}^{\mathcal{H}} = B &\iff \mathcal{H} \models A \cap \rho_{k}^{\mathcal{H}} = \tau^{\mathcal H}[\vec{\alpha}, r,q] \cap \rho_{k}^{\mathcal{H}} \\ &\iff \mathcal{M} \models \pi(A) \cap \rho_{k}^{\mathcal{M}} = \tau^{\mathcal{M}}[\vec{\alpha},\pi(r),\pi(q)] \cap \rho_{k}^{\mathcal{M}}. \end{align*} Since the last line is true, it follows that $A \cap \rho_{k}^{\mathcal{H}} =\tau^{\mathcal{H}}[\vec{\alpha},r,q] \cap \rho_{k}^{\mathcal{H}}$. Thus $r$ is indeed $k$-universal over $(\mathcal{H}, q)$. Q.E.D.

Question: Did I miss something and this result actually relies on the $k$-solidity of $\pi(r)$ over $(\mathcal{M},\pi(q))$ or can this assumption be dropped?

If $k$-solidity is needed, I'd like to understand where exactly in the proof it is used and ideally I'd like to see an example in which Lemma 2.8 fails without $k$-solidity.

PS: I am aware that this question isn't exactly 'ongoing research' and I strongly considered posting it over at MSE. However, since the group of people able to answer this question is more likely to be encountered here and since a somewhat similar question has been asked and well-received here, I decided to go with mathoverflow.

  • 3
    $\begingroup$ Seriously? You've never posted anything here before? (And also, "grad-level" questions of this nature are fine on MathOverflow. Especially when it's clear that it's not trying to have someone do your homework for you. You can find discussions on this topic on the MathOverflow Meta site.) $\endgroup$
    – Asaf Karagila
    Dec 2, 2016 at 15:27
  • 1
    $\begingroup$ It doesn't need solidity. I also did a write-up of that lemma in my master's thesis, if you're interested. It's Lemma 2.20 in dansnielsen.files.wordpress.com/2016/09/thesis_v2_240816.pdf. $\endgroup$ Dec 3, 2016 at 14:22
  • $\begingroup$ Thanks @Dan. I suspected as much, spelled out the entire proof earlier today and figured out the same. Unless you want to post your proof as an answer, I'll answer my own question later today. $\endgroup$ Dec 3, 2016 at 14:33
  • 1
    $\begingroup$ @Stefan As I'm working with slightly different definitions, it's probably easier for you to just copy your proof onto here - so be my guest :) $\endgroup$ Dec 4, 2016 at 0:07

1 Answer 1


In my question I already verified that (assuming 1. and 2.) item 3. is provable without assuming that $\pi(r)$ is $k$-solid over $(\mathcal{M},q)$. To see that we can actually drop $k$-solidity in this lemma, it hence suffices to see that 1. and 2. also don't require that $k$-solidity of $\pi(r)$.

If $\rho_{k}^{\mathcal{M}} = \operatorname{Ord}^{\mathcal{M}}$, then $\pi = \operatorname{id}$ and the lemma trivially holds. Thus assume that $\rho_{k}^{\mathcal{M}} < \operatorname{Ord}^{\mathcal{M}}$.

  1. Let $\alpha \leq \rho_{k}^{\mathcal M}$. Since $\pi \restriction \rho_{\kappa}^{\mathcal{M}} = \operatorname{id}$ and since $\pi$ is generalized $r \Sigma_{k}$-elementary, we have - up to a slight abuse of notation - \begin{align*} \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) &= \{(\phi, \vec{a}, s) \mid \vec{a} \in ^{< \omega}{\alpha} \wedge \mathcal{H} \models \phi[\vec{a}, s] \} \\ &= \{(\phi, \vec{a}, s) \mid \vec{a} \in ^{< \omega}{\alpha} \wedge \mathcal{M} \models \phi[\vec{a}, \pi(s)] \} \end{align*} In particular, for any $\phi \in r \Sigma_{k}$ and any $\vec{a} \in ^{< \omega}{\alpha}$, we have $$ (\phi, \vec{a}, s) \in \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{s \}) \iff (\phi, \vec{a}, \pi(s)) \in \operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}). $$ By enlarging $\alpha$, if necessary, we may assume that $\alpha$ is primitive recursively closed and hence uniformly code \begin{align*} \{(\phi, \vec{a}) \mid (\phi, \vec{a}, s) \in \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) \} \\ = \{(\phi, \vec{a}) \mid (\phi, \vec{a}, \pi(s)) \in \operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}) \} \end{align*} as a subset $A \subseteq \alpha$. Since $\alpha < \rho_{k}^{\mathcal{M}}$, we have $\operatorname{Th}_{k}^{\mathcal{M}}(\alpha \cup \{ \pi(s) \}) \in \mathcal{M}$ and hence $A \in \mathcal{M}$. By the strong acceptability of $\mathcal{M}$ - observing that $\rho_{k}^{\mathcal{M}}$ is an $\mathcal{M}$-cardinal - this yields $$ A \in \mathcal{J}_{\rho_{k}^{\mathcal{M}}}^{\mathcal M} = \left( H_{\rho_{k}^{\mathcal{M}}} \right)^{\mathcal{M}} \overset{\pi \restriction \rho_{k}^{\mathcal{M}} = \operatorname{id}}{=} \left( H_{\rho_{k}^{\mathcal{M}}}\right)^{\mathcal{H}} \subseteq \mathcal{H}. $$ Therefore $$ \operatorname{Th}_{k}^{\mathcal{H}}(\alpha \cup \{ s \}) = \{ (\phi, \vec{a}, s) \mid \langle \phi, \vec{a} \rangle \in A \} \in \mathcal{H} $$ and $\rho_{k}^{\mathcal{M}} \leq \rho_{k}^{\mathcal{H}}$.

    On the other hand, suppose that $\rho_{k}^{\mathcal{M}} < \rho_{k}^{\mathcal{H}}$. Then $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{M}} \cup \{(r,q)\}) \in \mathcal{H}$. Let $A \subseteq \rho_{k}^{\mathcal{M}}$, $A \in \mathcal{H}$ be a uniform code for $$ \{ (\phi, \vec{a}) \mid (\phi, \vec{a}, (r,q)) \in \operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{M}} \cup \{(r,q)\})\}. $$ Then $A = \pi(A) \cap \rho_{k}^{\mathcal{M}} \in \mathcal{M}$ witnesses (as above) that $\operatorname{Th}_{k}^{\mathcal{M}}(\rho_{k}^{\mathcal{M}} \cup \{ \pi(r), \pi(q)\}) \in \mathcal{M}$. This contradicts the fact that $\pi(r)$ is the $k$th standard parameter of $(\mathcal{M},\pi(q))$!

  2. The proof above also shows that $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup \{ (r,q)\}) \not \in \mathcal{H}$. Hence it suffices to show that for all $s <_{\operatorname{lex}} r$ $$ \operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup \{ (s,q)\}) \in \mathcal{H}. $$ So, fix $s <_{\operatorname{lex}} r$. Then $\pi(s)<_{\operatorname{lex}} \pi(r)$ and hence $$ \operatorname{Th}_{k}^{\mathcal{M}}(\rho_{k}^{\mathcal{H}} \cup \{ (\pi(s),\pi(q))\}) \in \mathcal{M}. $$ Let $A \subseteq \rho_{k}^{\mathcal{M}}$, $A \in \mathcal{M}$ be the code of this fact as above. Since $\pi(r)$ is $k$-universal over $(M, \pi(q))$ there is some $\tau \in \operatorname{Sk}_{k}$ and $\vec{a} \in ^{< \omega}{\rho_{k}^{\mathcal{M}}}$ such that $A = \tau^{\mathcal{M}}[\vec{a}, \pi(r), \pi(q)] \cap \rho_{k}^{\mathcal{M}}$. Let $$ B = \tau^{\mathcal{H}}[\vec{a}, \pi(r), \pi(q)] \cap \rho_{k}^{\mathcal{M}}. $$ Since $\pi$ is generalized $r \Sigma_{k}$-elementary we have, for all $\xi < \rho_{k}^{\mathcal{M}} = \rho_{k}^{\mathcal{H}}$ $$ \mathcal{M} \models \xi \in \tau^{\mathcal{M}}[\vec{a},\pi(r),\pi(q)] \iff \mathcal{H} \models \xi \in \tau^{\mathcal{H}}[\vec{a},r,q]. $$ Thus $B = A \in \mathcal{H}$ witnesses that $\operatorname{Th}_{k}^{\mathcal{H}}(\rho_{k}^{\mathcal{H}} \cup \{(s,q)\}) \in \mathcal{H}$ and hence the $<_{\operatorname{lex}}$-minimality of $r$.

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