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(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 …

Here is one way to do it. 2-colorability case. First let's warm up with the 2-colorability case. Notice that odd-length cycles are not 2-colorable, since the colors have to alternate as you go around …

$\newcommand{\N}{\mathbb{N}}$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 …

A partial order relation $\leq$ on a set $A$ is a binary relation that is reflexive, transitive, and antisymmetric. A preorder relation $\unlhd$ (also sometimes known as a quasi order or pseudo order) …

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 $\alpha+1$, …

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? …

In the case of ordinal definability, the structures are $\alpha\mapsto V_\alpha$. … than ordinal-definability. …

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 …

There is no such phenomenom for $n\geq 2$. The reason is that if a model of ZFC has a definable choice function, of any complexity, then it actually has one of complexity $\Delta_2$. This is because t …

It is the same in Peano arithmetic, where the standard language is $\{+,\cdot,0,1,<\}$ for the standard model $\langle\mathbb{N},+,\cdot,0,1,<\rangle$, even though $0$, $1$, and $<$ are definable from …

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 …

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 …

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 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? …