most recent 30 from http://mathoverflow.net 2013-06-19T21:04:08Z http://mathoverflow.net/feeds/question/118815 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118815/small-implications-of-the-axiom-of-replacement arsmath 2013-01-13T15:10:37Z 2013-01-13T17:25:39Z

<p>The axiom of replacement implies the existence of sets larger than usual in mathematical practice, but can be used to prove theorems about sets of real numbers, such as Borel determinacy. This is interesting because it suggests there's some sort of recursive procedure that makes sense for sets of reals, but is not provable in ZC alone. This procedure seems like it would be of independent interest from the question of the existence of sets beyond $V_{\omega + \omega}$ in the cumulative heirarchy.</p> <p>Is there a weaker axiom or recursive set of axioms that can be added to ZC that imply exactly the implications of replacement that hold for sets in $V_{\omega + \omega}$, one that explains the kind of additional constructions that replacement permits you to make? </p>

http://mathoverflow.net/questions/118815/small-implications-of-the-axiom-of-replacement/118822#118822 Goldstern 2013-01-13T16:05:50Z 2013-01-13T16:05:50Z

<p>The set $M$ of all formulas $\varphi$ that are of the form $V_{\omega+\omega} \vDash \psi$ is certainly recursive. Now the set <code>$N:=\{ \varphi\in M: ZFC \vdash \varphi\}$</code> is c.e.</p> <p>A standard trick gives an equivalent set $N'$ which is recursive (decidable): replace the $n$-th formula in $N$ (in any computable enumeration) by an equivalent formula that is much longer that all previous ones. </p> <p>Is this set $N'$ what you are looking for? I realize that it does not have the nice form you probably wanted. </p>