This question assumes some familiarity with Jensen's fine structure analysis of the constructible universe L (https://en.wikipedia.org/wiki/Jensen_hierarchy, http://www.math.cmu.edu/~laiken/papers/FineStructure.pdf).

Everything to follow is in L.

Given some $J_\alpha$, the n-th projectum of $J_\alpha$, $\rho_n(J_\alpha)$ is defined as follows: the least $\rho\leq \alpha$ such that there exists a subset of $\omega\cdot \rho$ which is $\Sigma_n(J_\alpha)$ but not in $J_\alpha$. Another equivalent characterization is that $\rho_n$ is the least $\delta\leq \alpha$ such that there exists a $\Sigma_n(J_\alpha)$ that maps $\omega\cdot \delta$ onto $J_\alpha$. Of course if $1<\rho_n<\alpha$, then $J_\alpha\models \rho_n \text{ is a cardinal}$ so $\omega\cdot \rho_n = \rho_n$.

The exercise is asking to produce any arbitrary pattern of the projectums. More concretely like the following, exhibit a $J_\alpha$ such that $\rho_k(J_\alpha)=\alpha$ for $k=0,1,2,3$, $\alpha>\rho_4(J_\alpha)>\rho_5(J_\alpha)$, and for all $j\geq 6$ $\rho_j=\rho_5$.

What I can do now is to produce one drop (I feel if somehow I know how to produce two drops then I am done). More precisely, consider $J_{\omega_2}$. Let $\xi$ be the least ordinal in $J_{\omega_2}$ which is not $\Sigma_4$-definable from $\omega_1$. Take the $\Sigma_4$ Skolem Hull in $J_{\omega_2}$ with parameters from $\omega_1 \cup \{\xi\}$, denoted by $Hull_{\Sigma_4}^{J_{\omega_2}}(\omega_1\cup \{\xi\}) \simeq_\pi J_{\beta}=Hull_{\Sigma_4}^{J_{\beta}}(\omega_1\cup \{\pi(\xi)\})$ by condensation via $\pi$. Then it's not hard to verify that $\rho_4(J_\beta)=\omega_1$ (with standard parameter $\{\pi(\xi)\}$) and $\rho_k(J_\beta)=\beta, k<4$ by elementarily.

But it is obvious that the above construction also yields that $\rho_k(J_\alpha)=\omega_1$ for all $k\geq 4$ by cardinality considerations. My feeling is that I should probably produce those projectums starting from $\rho_5$ (i.e. backtrack). But I don't see how, so far, to get another projectum drop. Thanks in advance!

  • $\begingroup$ What is $\alpha$ in the fourth paragraph? Also, is this an actual exercise? If so, perhaps you could mention where it's from? $\endgroup$
    – user7835
    Feb 7, 2016 at 2:27
  • $\begingroup$ $\alpha$ is just any ordinal. Well there was some example in Jensen's original paper that produced (roughly) the pattern I mentioned above. I heard about this from a conversation but so far I couldn't either refute or confirm this. $\endgroup$
    – Jing Zhang
    Feb 7, 2016 at 2:31
  • $\begingroup$ Okay. Judging from the rest of the paragraph I thought that $\alpha$ might have been $\omega_2$. $\endgroup$
    – user7835
    Feb 7, 2016 at 2:34
  • $\begingroup$ @tci sorry I found that (I looked at the wrong paragraph) they are all $\beta$. See my edits. $\endgroup$
    – Jing Zhang
    Feb 7, 2016 at 2:38
  • 1
    $\begingroup$ Did you look at "Patterns of Projecta" By Adam Krawczyk? $\endgroup$
    – William
    Apr 30, 2016 at 23:09

1 Answer 1


Let me include an argument here. Goal: given $s\in 2^{<\omega}$ we will produce a $J$-structure such that the projecta behave exactly as indicated by the string $s$ (1 means drop and 0 means stay).

Assume $V=L$. For any $s\in 2^{<\omega}$, for all infinite cardinals $\kappa$, there exists $\kappa \leq \alpha <\kappa^+$ such that in $J_\alpha$ the projecta pattern is $s$.

Fix $\kappa$. We need to demonstrate how to produce one more drop. Proceed by induction on the length of $s$, say $n$. Let $\beta\geq \kappa^+$ be such that $J_\beta$ has projecta pattern $s$. Our goal is to produce $\kappa\leq \alpha <\kappa^+$ such that the projecta pattern is $s^\frown \langle 1\rangle$, aka one more drop of projecta. Consider $A=Hull_{n+1}^{J_\beta}(\kappa \cup S)$ where $$S=\{\kappa\}\cup \{\rho_i(J_\beta): i\leq n\} \cup \{p_i: i\leq n\}\cup \{\gamma_i: i\leq n+1\}$$ where $\gamma_i$ is the parameter in $J_\beta$ for the $\Sigma_{i}$-Skolem function for $J_\beta$. We know $A\prec_{n+1} J_\beta$. What is left to check $\rho_{n+1}(A)=\kappa$ and $\rho_i(A)=\rho_i(J_\beta)$ for all $i\leq n$. Then the transitive collapse of $A$ will work.

$\rho_{n+1}(A)\leq\kappa$ because there exists a $\Sigma_{n+1}$ definable function that maps a subset of $\kappa$ onto $A$. But $\kappa$ is a cardinal. It follows that $\rho_{n+1}(A)=\kappa$.

Fix any $i\leq n$, we need to show $\rho_i(A)=\rho_i(J_\beta)$ (this also applies to the degenerate case where $\rho_i(J_\beta)=\beta$). $\rho_i(A)\leq \rho_i(J_\beta)$, since $h_i$ maps a subset of $\rho_i(J_\beta)$ onto $A$ with parameter in $A$, we know by soundness that $\rho_i(A)\leq \rho_i(J_\beta)$. Conversely, some $\Sigma_i(A)$ $h'_i$ maps $\rho_i(A)$ onto $A$. We have $A\models \forall x \exists y<\rho_i(A) h'_i(x)=y$. Note the sentence is $\Pi_{i+1}$ with parameters in $A$. By $\Sigma_{n+1}$ elementarity we know that $J_\beta \models \forall x \exists y<\rho_i(A) h'_i(x)=y$. Hence $\rho_i(J_\beta)\leq \rho_i(A)$.

  • $\begingroup$ In the definition of $S$, what is $p_i$? (Appearing in "$\{p_i:i\leq n\}$") $\endgroup$
    – C7X
    Jul 23, 2022 at 19:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .