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The answer is the ordinal $\omega_1^{ck}$, named after Church and Kleene, which is defined to be the supremum of the ordinals coded by a computable relation on $\mathbb{N}$. It happens also to be the …

Let me try to help by explicating the argument Noah mentioned. I think this is part of logic folklore—it amounts at bottom to the facts that every permutation of a pure set is an isomorphism, and isom …

This question continues the theme of some recent questions on implicit definability. … The main original question was whether implicit definability is transitive. Main Question 1. Is the implicitly-definable-over relation transitive? …

I like this question very much. For 1, you can add a relation symbol for every set in the family, and this will of course suffice to define every set in the family, while creating no additional def …

I view it as a kind of lucky miracle that ordinal-definability doesn't stumble on this problem.) So this version aligns with the usual version of HOD. Conclusion. …

Yes, because there is a definable ordinal pairing function. Specifically, if you want to get the set $\{y\in V_\theta\mid V_\theta\models\varphi(y,\alpha)\}$, then let $\beta=\langle\theta,\alpha\rang …

The answer is no. Indeed, one can rarely move from consistency to truth in this way. For a counterexample, let $Q(v)$ be the property "CH holds and $v$ is an ordinal." If CH holds, then $Q$ expresses …

$\newcommand{\N}{\mathbb{N}}$My question, more precisely, is: Question. Is there a set $B\subset \N\times\N$, such that the set of indices where it is arithmetically definable, that is, $\{ n\in\N \mi …

(Model theorists please note that this is implicit definability in a model, which is not the same as the notion used in Beth's implicit definability theorem.) … Implicit definability is a very weak form of second-order definability, one which involves no second-order quantifiers. …

Contrary to my initial expectation, the answer is Yes. This answer is based on the idea of Clemens Grabmayer, which makes the observation that addition $+$ is definable from multiplication $\cdot$ and …

Note that we may refer to $\Sigma_2$-truth since there is a universal truth predicate for truth of bounded complexity (so there will be no issues with Tarski's theorem on the non-definability of truth) … And finally, asserting that the elements of $A$ are really not ordinal-definable has complexity $\Pi_2$, since "$x\in\text{OD}$'' has complexity $\Sigma_2$, as any instance of ordinal-definability reflects …

I am also interested in the case of $\Pi_2$-definability. Question 2. Is there a model of $\text{ZFC}+V\neq\text{HOD}$ in which every $\Pi_2$-definable set has an ordinal-definable element? …

But there can be no definable class $T$ that is a truth predicate, because of Tarski's theorem on the non-definability of truth. …

The answer to question 1 is no. Let $C$ be the class of all countable sets. If the answer were affirmative, we would get the relation $E$ and theroy $T$ so that $a\in C$ if and only if $\langle a,E\up …

It's a nice question. This Boolean algebra, known as the Lindenbaum algebra, is a countable atomless Boolean algebra — it is atomless because we can always take the conjunction of any formula with a n …