24 votes
Accepted

Does $V = \textit{Ultimate }L$ imply GCH?

In his slide Absolutely ordinal definable sets John Steel writes: At the same time, one hopes that V = ultimate L will yield a detailed fine structure theory for V, removing the incompleteness that ...
Mohammad Golshani's user avatar
16 votes
Accepted

For which ordinals do we have $V_\alpha = L_\alpha$?

Yes; in fact the first $\alpha>\omega$ with $V_\alpha=L_\alpha$ has cofinality $\omega$. To obtain this $\alpha$, define $f:Ord\to Ord$ by $f(\xi)=$ the smallest $\eta$ such that $V_\xi\subseteq L_\...
Andreas Blass's user avatar
14 votes
Accepted

If $L_\alpha \vDash ZFC$, then do we have $L_{\alpha+1} \vDash \alpha\text{ is inaccessible}$?

Yes. The elements of $L_{\alpha+1}$ are exactly those subsets of $L_\alpha$ which are definable from parameters over $L_\alpha$. But $L_\alpha\models\mathrm{ZFC}$, so from here we can just use the ...
Farmer S's user avatar
  • 8,627
14 votes
Accepted

Can $L$ be defined without parameters?

Yes, the parameter-free version of $L$ gives rise to the same constructible universe $L$. You will still get all of $L$ this way, but it will come more slowly. The reason is that at stage $\alpha+1$, ...
Joel David Hamkins's user avatar
13 votes
Accepted

Is $\mathsf{ZFC+V=L}$ consistently $\omega$-complete?

Claim: $T+$"$T$ is $\omega$-complete" is inconsistent. For suppose it's consistent and now work in a model $V$ of this theory. Let $T^+$ be the resulting completion of $T$ (i.e. the unique ...
Farmer S's user avatar
  • 8,627
13 votes
Accepted

A doubt about the Gödel condensation lemma

Your mistake is that taking the Mostowski collapse does not preserve elementarity. We do have a countable transitive $A$ and a countable $M$ with $$A\cong M\preccurlyeq V_{\omega_1},$$ where $M$ comes ...
Noah Schweber's user avatar
12 votes
Accepted

Is the power set axiom essential for constructing L?

KP alone - which is vastly weaker than the theory in question - proves the sentence "For every ordinal $\alpha$, $L_\alpha$ exists," since it is strong enough to enable effective transfinite recursion....
Noah Schweber's user avatar
12 votes
Accepted

Is the smallest $L_\alpha$ with undefinable ordinals always countable?

${\mathfrak t}$ is the least $\beta$ such that there is a $\gamma<\beta$ with $L_\gamma \prec L_\beta$. That ${\mathfrak t} \leq$ the least such $\beta$ is obvious. On the other hand, if $X \subset ...
Ralf Schindler's user avatar
11 votes
Accepted

Can only the constructible sets be proven to exist in $ZF$ without benefit of extra assumptions?

$\sf ZF$ can prove that $\mathcal P(\omega)$ exists, but it cannot prove whether or not every subset of $\omega$ is constructible. In other words, there are sets which provably exist, but it is not ...
Asaf Karagila's user avatar
  • 37.8k
11 votes
Accepted

Stability for the Gödel and Jensen hierarchies

As remarked the non-trivial direction is to show $L_\sigma\prec_{\Sigma_1}L_\gamma$ implies $J_\sigma\prec_{\sigma_1}J_\gamma$. Let's take the extreme case that $\gamma=\sigma+1$. Suppose $J_{\sigma+1}...
Philip Welch's user avatar
  • 4,909
11 votes
Accepted

Is this relation about elementary embedding transitive?

EDIT: If the codomain $N$ is allowed to be illfounded, then the answer is yes. (Here if $\kappa=\mathrm{crit}(j)$ then this will mean that $\kappa\subseteq N$, but it might be that $N$ is illfounded ...
Farmer S's user avatar
  • 8,627
10 votes

Can only the constructible sets be proven to exist in $ZF$ without benefit of extra assumptions?

There are several issues with this question; Andres has pointed out one glaring one, which is the conflation of theories and models. There is another issue, however, around what you are asking in the ...
Noah Schweber's user avatar
9 votes

Does $V = \textit{Ultimate }L$ imply GCH?

During this year's conferene on inner model theory in Münster, Gabriel Goldberg proved that the so-called Ultrapower Axiom implies that $\mathrm{GCH}$ holds above a supercompact cardinal (and since ...
Stefan Mesken's user avatar
9 votes

What is the largest large-cardinal hypothesis consistent with $ZFC + V=L$?

For your first question: If $\alpha$ is countable, then $\alpha$-Erdos cardinals are consistent with $V=L$ (if they are consistent at all). On the other hand, the existence of an $\omega_1$-Erdos ...
Will Brian's user avatar
  • 17.3k
8 votes
Accepted

What is the largest large-cardinal hypothesis consistent with $ZFC + V=L$?

Re: ordinary mathematics, I think your assumptions are incorrect (although the question is interesting on its own, since ZFC+V=L is a natural set theory). ZFC+V=L is (most, including Simpson, would ...
Noah Schweber's user avatar
8 votes
Accepted

Is there a 'Constructible Universe' that is a submodel of a non-transitive model of $ZF$?

What $\sf ZF$ actually proves is that there is a class called $L$, and for every axiom of $\sf ZFC$, the relativization of that axiom holds in $L$. Therefore, the fact we are talking about non-...
Asaf Karagila's user avatar
  • 37.8k
8 votes

On the utility of Silver machines

The existence of morasses using Silver machine in L is proved in the PhD thesis of Thomas Lloyd Richardson, ``Silver Machine Approach to the Constructible Universe''. See also Silver machines and ...
Mohammad Golshani's user avatar
8 votes
Accepted

Height of diamond

It depends on what diamond sequence you have. $V=L$ proves there is a $\diamondsuit$-sequence of $L$-rank $\omega_1+1$: recall the famous way of proving the existence of $\diamondsuit$-sequence. The ...
Hanul Jeon's user avatar
  • 2,754
8 votes
Accepted

Can local $0^\#$ exists in L?

Take a countable elementary submodel of $L_\kappa[0^\#]$, code that into a real, note that "There is a real coding a well-founded model of $V=L[0^\#]$" is a $\Sigma^1_2$ statement, remember ...
Asaf Karagila's user avatar
  • 37.8k
8 votes

Can local $0^\#$ exists in L?

(EDIT: Now edited to compute the precise value of $\alpha$.) @AsafKaragila already answered the question, but this is answering the follow-up question od @Reflecting_Ordinal in the comments to Asaf's ...
Farmer S's user avatar
  • 8,627
8 votes

Elementary countable submodels in Gödel's universe

No, there are many instances of $L_\alpha\prec L_\beta$ without $L_\alpha\prec L_{\omega_1}$. Here is one easy way to construct one. Consider the smallest $\alpha$ that has an elementary extension $L_\...
Joel David Hamkins's user avatar
8 votes

Elementary countable submodels in Gödel's universe

As Asaf and Joel have observed, the answer to your question is negative. However, there is a sense in which being an elementary submodel of $L_{\omega_1}$ is the only way to "persistently" ...
Noah Schweber's user avatar
8 votes
Accepted

Forcing a unique $\Delta_3^1$ generic real

Since the first question has already been answered in the EDIT at the end of the question, I will focus on the second question. The short answer to second question is: Yes, a good source is Sy ...
Ali Enayat's user avatar
  • 16.7k
8 votes
Accepted

Inner model with a $\mathit{\Delta}^1_3$-good well-ordering of the reals

The situation is a bit more complicated than you might hope because of the periodicity phenomena in the projective hierarchy. For odd $n$, assuming $\mathbf{\Delta}^1_{n-1}$-determinacy, the set $Q_n$ ...
Gabe Goldberg's user avatar
7 votes
Accepted

On a particular proof of "if the sharp of every real exists and every club contains a club constructible from a real, then $\delta^1_2 = \omega_2$"

Let $\kappa$ be a sufficiently large regular cardinal. Take a countable elementary substructure $H$ of $H(\kappa)$ containing $z$ and $<_\alpha$ with $\omega_1\cap H\in D$. Let $\pi : M\to H(\kappa)...
Gabe Goldberg's user avatar
7 votes
Accepted

If we have a class like $L$ but allowing a set number of unbounded quantifiers, is it strict superset of $L$?

Every $L^{\Sigma_n}$, for $n\geq 2$ will be the same as HOD, the class of hereditarily ordinal-definable sets. This is a consequence of the Myhill-Scott theorem, which asserts that if you form the ...
Joel David Hamkins's user avatar
7 votes
Accepted

Can countable ordinals start gaps of every order in the constructible universe?

I presume that by "starts a gap (of order $n$)" you mean "starts a gap of positive length (of order $n$)", since length $0$ is trivial. Note that the answers given by Monroe Eskew ...
Farmer S's user avatar
  • 8,627
7 votes
Accepted

Why do $\pi$ and $\bigcup$ commute for Gödel-closed extensional classes?

Here is what I wrote when I solved this problem a few years ago: As $M$ is extensional, the transitive collapse is an isomorphism. The statement $C=G_i(A,B)$ can be expressed by a $\Delta_0$ formula $...
Pace Nielsen's user avatar
  • 17.9k
7 votes
Accepted

Elementary countable submodels in Gödel's universe

Very clearly not. Take some countable elementary submodel $M_0$ of $L_{\omega_2}$, and take $M_1$ to be another one, but with $M_1$ a end extension of $M_0$. We can find such models by first finding ...
Asaf Karagila's user avatar
  • 37.8k
7 votes
Accepted

Are all constructible from below sets parameter free definable?

There are two issues with your question. First, your statement "in other words" is not correct, since there are theories whose models have the property that whenever a statement holds of ...
Joel David Hamkins's user avatar

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