Denote Zermelo Fraenkel set theory without choice by ZF. Is the following true: In ZF, every definable non empty class A has a definable member; i.e. for every class $A = \lbrace x : \phi(x)\rbrace$ for which ZF proves "A is non empty", there is a class $a = \lbrace x : \psi(x)\rbrace$ such that ZF proves "a belongs to A"?



Definable collections without definable members (in ZF)
**Update.** (June, 2017) François Dorais and I have completed a paper that ultimately grew out of this questions and several follow-up questions. 



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> F. G. Dorais and J. D. Hamkins, [When does every definable nonempty set have a definable element?](http://jdh.hamkins.org/when-does-every-definable-nonempty-set-have-a-definable-element/) ([arχiv:1706.07285](https://arxiv.org/abs/1706.07285))
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> > ***Abstract.** The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi\_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma\_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$.*
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> Read more [at the blog post](http://jdh.hamkins.org/when-does-every-definable-nonempty-set-have-a-definable-element/).
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Click on the history to see the original answer. Related questions and answers appear at:


* [Can $V\neq\HOD$, if every $\Sigma\_2$-definable set has an ordinal-definable element?](https://mathoverflow.net/q/180850/1946)
* [Is it consistent with ZFC (or ZF) that every definable family of sets has at least one definable member?](https://mathoverflow.net/q/180734/1946)
* We also make use of the $\Sigma\_2$ conception of [Local properties in set theory](https://jdh.hamkins.org/local-properties-in-set-theory).


