Does every nonempty definable finite set have a definable member? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:13:33Z http://mathoverflow.net/feeds/question/107441 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107441/does-every-nonempty-definable-finite-set-have-a-definable-member Does every nonempty definable finite set have a definable member? Trevor Wilson 2012-09-18T06:31:12Z 2012-09-18T15:18:26Z <p>I asked this on MSE yesterday ( <a href="http://math.stackexchange.com/q/197873/39378" rel="nofollow">http://math.stackexchange.com/q/197873/39378</a> ) but no one has answered it yet. I hope it's not too soon to post it here.</p> <p>Here are a few ways to formalize the question, so you can pick your favorite and answer it. Assume whatever large cardinals you like.</p> <p>(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 $(V_\delta,\in)$?</p> <p>(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?</p> <p>(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 href="http://mathoverflow.net/questions/17608/a-question-about-ordinal-definable-real-numbers" rel="nofollow">http://mathoverflow.net/questions/17608/a-question-about-ordinal-definable-real-numbers</a>, 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.)</p> <p>(4) Any of the above formulations with ZFC replaced by ZF.</p> http://mathoverflow.net/questions/107441/does-every-nonempty-definable-finite-set-have-a-definable-member/107474#107474 Answer by François G. Dorais for Does every nonempty definable finite set have a definable member? François G. Dorais 2012-09-18T14:29:05Z 2012-09-18T15:18:26Z <p>I believe the answers to these questions are all positive. This kind of problem was discussed by Groszek and Laver in <em>Finite groups of OD-conjugates</em> [Period. Math. Hungar. 18 (1987), 87-97, <a href="http://www.ams.org/mathscinet-getitem?mr=895774" rel="nofollow">MR0895774</a>]. 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.</p> <p>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 $L[u,v]$.</p>