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1 vote
2 answers
228 views

Can this semi-constructible structure satisfy existence of a measurable cardinal?

If we add a primitive unary function symbol $\mathfrak L$ to the first order language of set theory. Axiom of semi-constructibility: if $\phi^\alpha (y,x_1,\ldots,x_n)$ is a formula in which all and ...
Zuhair Al-Johar's user avatar
7 votes
1 answer
556 views

Does this ZFC+V=L like theory, have a limit on large cardinal properties?

Let $\sf T$ be a theory that has as axioms every axiom of $\sf ZFC$, and every theorem of $\sf ZFC + [V=L]$ that is neither provable nor disprovable by $\sf ZFC$, whose addition or addition of its ...
Zuhair Al-Johar's user avatar
1 vote
1 answer
112 views

Constructible cardinality downslides and their consistency strengths?

Posting "Large cardinals and constructible universe" mentions that $\omega_1^L < \omega_1$ if we assume Ramsey cardinal. My question can we have more downslides like for example $\omega_2^...
Zuhair Al-Johar's user avatar
4 votes
1 answer
158 views

Impact of coining $L$ in $\mathcal L_{\omega_1, \omega}$ on which large cardinals it can satisfy?

What happens to the constructible universe $L$ if we build it in the first of infinitary languages $\mathcal L_{\omega_1, \omega}$? Would the usual limitation of $L$ not satisfying existence of a ...
Zuhair Al-Johar's user avatar
3 votes
0 answers
123 views

At which large cardinal property this second order ordinal arithmetic stops?

Language: Second order logic, with as usual predicates written in upper case, and objects in lower case. Let $<$ be a primitive constant binary relation symbol. Equality between objects is ...
Zuhair Al-Johar's user avatar
9 votes
2 answers
426 views

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

Assume $0^\#$ exists and there is an inaccessible cardinal. Are there two transitive sets $M,N$ s.t. $M\in N,M\vDash ZF+V=L[0^\#],N\vDash ZF+V=L$?
Reflecting_Ordinal's user avatar
7 votes
1 answer
395 views

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

Here we choose the definition of "is a cardinal" as there is no surjective map from a smaller ordinal to it. It's easy to prove that, if $L_{\alpha+1} \vDash\ \alpha\text{ is inaccessible}$, ...
Reflecting_Ordinal's user avatar
6 votes
0 answers
202 views

Consistency strength of Sy Friedman's result about admissibility spectrum

A result by Sy Friedman in his book "fine structure and class forcing", is that, assume $0^\sharp$ exists, there exists a real number R such that the ordinals admissible in R (called $\...
Reflecting_Ordinal's user avatar
3 votes
1 answer
236 views

smallest ordinal $\alpha$ such that $L \cap P(L_\alpha)$ is uncountable

Let $V$ denote the von Neumann universe and $L$ Gödel's constructible universe. For any set $X$, let $P(X)$ denote the power set of $X$. Assume that $0^\sharp$ exists (and ZFC). What is the smallest ...
Jesse Elliott's user avatar
14 votes
2 answers
426 views

Consequences of existence of a certain function from $\omega_1$ to $\omega_1$

In his book [1], Paul Larson remarks (Remark 1.1.22) that in L there is a function $h:\omega_1\rightarrow\omega_1$ such that for any countable elementary submodel $X$ of $V_\gamma$ (where $\gamma$ is ...
Todd Eisworth's user avatar
6 votes
0 answers
451 views

Inaccessible cardinals and the perfect set property for coanalytic sets

I am wondering who proved the following fact: ($\ast$) If $\omega_1$ is not inaccessible in $L$, then there is an uncountable coanalytic set of reals without a perfect subset. I have been unable to ...
Trevor Wilson's user avatar
7 votes
1 answer
398 views

How similar are large cardinals, over $L$?

EDIT: Joel's answer shows that no $\Sigma_2$ large cardinal property will do the job - however, $\Pi_2$ properties (such as unfoldability and its relatives) may still be useful. Throughout this ...
Noah Schweber's user avatar
2 votes
0 answers
147 views

When do wide initial segments ruin admissibility?

Suppose $\alpha$ is admissible and $\beta<\alpha$. We know that $L_\alpha$ is an admissible set (by definition), but adding subsets of $\beta$ to $L_\alpha$ might break admissibility: while set ...
Noah Schweber's user avatar
8 votes
1 answer
623 views

Is every ordinal potentially definable?

It is easy to see that, if $V\models\alpha>\omega_1^{CK}$, then $\alpha$ is not recursive in any forcing extension of $V$. The argument goes as follows: The relation "$\Phi_e=r$" is $\Pi^0_2$. The ...
Noah Schweber's user avatar
5 votes
1 answer
489 views

Simplest non-constructible set of integers compatible with the nonexistence of $0^\sharp$?

What is the simplest non-constructible set of integers (say, in the analytical hierarchy) that is compatible with the nonexistence of $0^\sharp$? In particular, can there still be a non-constructible $...
Jesse Elliott's user avatar
14 votes
3 answers
777 views

When is $A$ "$L$-ish" whenever $B$ is "$L$-ish"?

My question is about a variant of the usual notion of relative constructibility, $\le_c$ (which an earlier version of this question confusingly denoted "$\le_L$"), in set theory. Fix a ...
Noah Schweber's user avatar
23 votes
1 answer
3k views

Gödel's Constructible Universe in Infinitary Logics (A Possible Approach to HOD Problem)

Gödel's constructible universe ($L$) is defined using definable power set operator in first order logic ($\mathcal{L}_{\omega ,\omega}$). One can produce such a universe in infinitary logics in ...
user avatar
13 votes
1 answer
971 views

V=HOD & The Height of the Large Cardinal Tree

As we know the assumption $V=L$ adds a restriction on the height of the large cardinal tree. Also there is a strict border like $0^{\sharp}$ exists such that all large cardinal axioms which are ...
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