Does every nonempty definable finite set have a definable member?

2012-09-18T06:31:12Z

I asked this on MSE yesterday ( http://math.stackexchange.com/q/197873/39378 ) but no one has answered it yet. I hope it's not too soon to post it here.

Here are a few ways to formalize the question, so you can pick your favorite and answer it. Assume whatever large cardinals you like.

(1) Is it consistent with ZFC that there is an inaccessible cardinal $\delta$ and a nonempty finite set that is first-order definable without parameters over $(V_\delta,\in)$ but has no elements that are first-order definable without parameters over $(V_\delta,\in)$?

(2) Is there any model of ZFC that has a finite nonempty set, first-order definable without parameters over the model, with no element that is first-order definable without parameters over the model?

(3) Is it consistent with ZFC that there is an ordinal-definable finite nonempty set with no ordinal-definable member? (I am aware of the question http://mathoverflow.net/questions/17608/a-question-about-ordinal-definable-real-numbers, but that question asks about sets of real numbers and I already know the answer to my question for sets of real numbers, or indeed for sets of subsets of any ordinal, because they are definably linearly ordered.)

(4) Any of the above formulations with ZFC replaced by ZF.

Answer by François G. Dorais for Does every nonempty definable finite set have a definable member?

2012-09-18T14:29:05Z

I believe the answers to these questions are all positive. This kind of problem was discussed by Groszek and Laver in Finite groups of OD-conjugates [Period. Math. Hungar. 18 (1987), 87-97, MR0895774]. Answering a question of Mycielski, they show that there can be two sets of reals $x,y$ such that $\lbrace x,y\rbrace$ is ordinal definable but neither $x$ nor $y$ is ordinal definable. They also prove a lot of other interesting things about OD conjugates.

Here is the brief argument from the intro to that paper. Suppose $u, v$ are two mutually Sacks generic reals over $L$. Both $u$ and $v$ have minimal degree over $L$. Let $x$ and $y$ be the $L$-degrees of $u$ and $v$ respectively. Then $x$ and $y$ satisfy the same formulas with ordinal parameters because Sacks forcing is homogeneous. However, $\lbrace x, y \rbrace$ is definable (without parameters) since these are the only two minimal $L$-degrees in $L[u,v]$.

Does every nonempty definable finite set have a definable member? - MathOverflow

Does every nonempty definable finite set have a definable member?

Answer by François G. Dorais for Does every nonempty definable finite set have a definable member?