# Does every nonempty definable finite set have a definable member?

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 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.

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Here is a related question on pointwise algebraic models of set theory, in which every set is an element of a finite definable set: mathoverflow.net/questions/71537/…;. – Joel David Hamkins Sep 18 '12 at 11:02
Also this question: mathoverflow.net/questions/10413/… – François G. Dorais Sep 18 '12 at 14:18
@Joel That is interesting. It looks like what that says in general is that given a nonempty definable finite set $S$ one may assume by shrinking $S$ that every element $A \in S$ is definable from some element $x \in A$. In the present situation we probably don't know that $x$ is definable, so there is no obvious induction on rank and indeed the problem seems to be nontrivial already at very low rank. For example, what if $S$ is a finite set of countable sets of reals, e.g., a finite set of Turing degrees or constructibility degrees or $E_0$-classes? – Trevor Wilson Sep 18 '12 at 15:04
Suppose our finite set is the set of the square roots of -1. This two-element set is certainly definable in ZF. But does there exist a definition in the (primitive) language of ZF that is satisfied by i alone or by -i alone? (Note that the symbols i and -i do not belong to the (primitive) language of ZF but must be defined in that language before they can appear in definitions). – Garabed Gulbenkian Sep 18 '12 at 19:59
@Garabed I assume you are talking about the set of square roots of $-1$ in $\mathbb{C}$. Whether this pair has a definable member depends on how you define $\mathbb{C}$. In the usual construction of $\mathbb{C}$ by pairs of real numbers, both imaginary units are definable. However, using François's answer below, one can show that it is consistent that there is a definable field isomorphic to $\mathbb{C}$ with no definable imaginary unit. See my answer to math.stackexchange.com/q/181464/39378 for a proof. – Trevor Wilson Sep 18 '12 at 21:54

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 $L[u,v]$.
I don't think $u$ and $v$ (or any two distinct reals) can satisfy the same formulas with ordinal parameters. What about formulas like "my 17th component is 0"? Nevertheless, the argument is probably OK since the $L$-degrees $x$ and $y$ do seem to satisfy the same formulas with ordinal parameters. – Andreas Blass Sep 18 '12 at 14:56
Absolutely, Andreas, I just corrected. (Of course, the proof in the paper did not claim that $u$ and $v$ satisfy the same formulas with ordinal parameters.) – François G. Dorais Sep 18 '12 at 15:20
@Trevor: $x$ and $y$ are sets of reals. – François G. Dorais Sep 18 '12 at 15:28