Suppose we have extended $ZF$ by adding to $ZF$ an unary function symbol $arb$ (an arbitrary element of a set) and a corresponding axiom "For every non-empty set $S$, $arb(S)$ is in $S$".
Will be the resulting theory a conservative extension of $ZF$?



Notation arb(x)
It depends how the remaining axioms of ZF are altered (or not). For example, do we broaden the scheme of replacement to apply to formulas of set theory which contain the symbol "$arb$"? If so, then as Zhen Lin says the resulting theory proves the axiom of choice: given a collection of sets $\mathcal{A}=\lbrace A\_i: i\in I\rbrace$, applying replacement to the formula "$x=arb(S)$" as $S$ ranges over the elements of $\mathcal{A}$ yields a choice function. If we don't do this - that is, if we have replacement and separation only for sentences of the language of set theory without "$arb$" - then we do in fact get a conservative extension: given any model $\mathcal{M}$ of ZF, let $arb$ pick any element from each element of $\mathcal{M}$ (I guess I'm using Choice in the ambient universe to do this, but if all we care about is not proving any new theorems, then this can be gotten rid of by Shoenfield absoluteness); the resulting extension is a model of ZF with $arb$.


