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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) …

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 …

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 …

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 …

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 …

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 …

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 …

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 $\alpha+1$, …

And yes, clearly every model of your theory is pointwise definable (in the first-order language), because that is precisely what your axiom of definability asserts. …

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 every paramet …

Let me say first that your concept is similar in spirit to the notion of sententially categorical cardinal appearing in my joint paper J. D. Hamkins and R. Solberg, Categorical large cardinals and th …

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. …

The answer to question 1 is no, because the theory of the minimal transitive model itself is parameter-free definable as that theory, and in the other answer I explained why this theory is not an elem …

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 …

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? …