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Given $J_\alpha$ denotes the $1+\alpha$th term in the Jensen Hierarchy and $J=\bigcup_{\alpha\in On}J_\alpha$, are there any known large cardinals $\mathfrak{K}$ such that $\text{ZFC}+\mathfrak{K}$ is consistent and $\text{ZFC}+\mathfrak{K}\vdash J\neq L$? (assuming ZFC is consistent)

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    $\begingroup$ (1) The term large cardinal usually implies a property whose consistency strength exceeds that of ZFC itself, so just assuming ZFC is consistent will not be enough to give you the large cardinal's consistency. Secondly, the whole point of the Jensen hierarchy is to rebuild L, not to build a whole other model. You can add predicates and whatnot, and then the story changes, but you can also add predicates and whatnot to L itself. $\endgroup$
    – Asaf Karagila
    Mar 17, 2018 at 11:20
  • $\begingroup$ If ZFC+K is consistent, then ZFC is consistent, but that statement is not necessarily the other way around. Even if J is meant to rebuild L, is that enough to conclude J=L? I agree that J=L is reasonable, but can it be proven by some theory? If V=L then sentences about L are such for the entire set-theoretic universe, so they cannot be proven in ZFC. But ZFC+K for some large cardinal K might disprove L=J, regardless of whether K exists. This is precisely what I'm trying to figure out. $\endgroup$ Mar 17, 2018 at 11:52
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    $\begingroup$ ZFC and much weaker theories prove J=L outright. The point of the J hierarchy is that it builds the same L, but with a different hierarchy, which has a better interaction with certain rudimentary set-theoretic constructions (en.wikipedia.org/wiki/Jensen_hierarchy#Rudimentary_functions). So $J_\alpha$ is often not the same as $L_\alpha$, but there is a class club of ordinals $\alpha$ where $J_\alpha=L_\alpha$. $\endgroup$ Mar 17, 2018 at 12:49
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    $\begingroup$ @JoelDavidHamkins Unless this question gets closed before you see this, consider moving your comment to the answer box. As far as I can see, you've completely answered the question. The only thing that might still be added is something more specific about "much weaker theories" (I imagine KP is more than enough), but even without such specifics, your answer suffices. $\endgroup$ Mar 17, 2018 at 14:15
  • $\begingroup$ @AndreasBlass OK, I posted an answer, including a link to Jensen's paper, where I think the issue is clearly stated. $\endgroup$ Mar 17, 2018 at 14:40

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Unfortunately, your question seems to be based on a misunderstanding of the $J$ hierarchy. There is no such large cardinal hypothesis.

The reason is that already ZFC and indeed much weaker theories such as KP prove $J=L$ outright. The point of the $J$ hierarchy was not to construct a different universe, but rather to reorganize the same constructible universe $L$ by means of a different stratification into levels, the $J_\alpha$ hiearchy, so as to realize a better interaction with certain rudimentary set-theoretic constructions (see http://en.wikipedia.org/wiki/Jensen_hierarchy#Rudimentary_functions). This change in the hierarchy allowed certain observations to be made about the fine structure of the constructible universe, and for these reason, set theorists studying the fine structure of $L$ typically prefer to use the $J_\alpha$ hierarchy.

From Ronald Jensen, Fine structure of the constructible hierarhcy, ANNALS OF MATHEMATICAL LOGIC 4 (1972) pp. 229-308.

We have found it convenient to replace the usual $L_\alpha$ hierarchy by a new hierarchy $J_\alpha$. We define $J_{\alpha+1}$ not as the collection of definable subsets of $J_\alpha$, but as the closure of $J_\alpha\cup\{J_\alpha\}$ under a class of functions which we call "rudimentary". These are just the functions obtained by omitting the recursion schema from the usual list of schemata for primitive recursive set functions. In a sense they form the smallest class of functions $\mathcal{R}$ such that there is a smooth definability theory for transitive domains closed under $\mathcal{R}$. The main difference between the two hierarchies is that $J_\alpha$ has rank $\omega\alpha$ rather than $\alpha$. However, the subsets of $J_\alpha$ which are elements of $J_{\alpha+1}$ are just the definable ones, $J_{\alpha+1}$ is, so to speak, the result of "stretching" the collection of the definable subsets of $J_\alpha$ upwards $\omega$ levels in rank without adding new ones. The exact correspondence between the two hierarchies is given by: $$J_0 = L_0 = \emptyset; \quad L_{\omega+\alpha}=V_{\omega+\alpha}\cap J_{1+\alpha}.$$ Thus $J_\alpha=L_\alpha$ whenever $\omega\alpha=\alpha$.

So although $J_\alpha$ is often not the same as $L_\alpha$, nevertheless there is a class club of ordinals $\alpha$ where $J_\alpha=L_\alpha$, and so $J=L$ in the end.

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    $\begingroup$ Thank you for the answer! I now understand that $J=L$ is in fact much simpler than I thought it was. I assumed that because $J$ and $L$ are very large structures (possibly, entire set-theoretic universes) proving $J=L$ would be as hard as proving something like $L=V$, but this makes more sense. $\endgroup$ Mar 22, 2018 at 21:34

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