In the constructible universe , $L$, there is a definable well-ordering of the entire universe. The ordering, as I mention in this MO answer, This universe is that built up in transfinite stages $x$ preceeds L_\alpha$, and the ordering has$x\lt_L y$when$x$is constructed at an earlier stageof the$L$hierarchy, or else they are constructed at the same stage, but$x$is constructed at that stage by an earlier definition, or with the same definition, but with earlier parameters. If one only wants to well-order the real numbers, then one can I also explain this in this MO answer. One may extract from this definition a$\Delta^1_2$ordering rather low-complexity definable well-ordering of the reals by capturing the countable pieces of the$L$hieararchy by reals. The reason That isthat , if$x$is a real number of$L$, then it appears at some countable stage$L_\alpha$for a countable ordinal$\alpha$, and the entire structure$L_\alpha$is countable, and hence itself coded by a real. Here, we code a set by a real in any of the standard ways, for example, by coding a well-founded extensional relation on$\omega$whose Mostowski collapse is the given set. Furthermore, the$L$-order is absolute to any$L_\alpha$, since$L_\alpha$knows about the$L_\beta$-heirarchy for$\beta<\alpha$. Also, if a countable structure is well-founded and thinks$V=L$, then it is$L_\alpha$for some$\alpha$. Note that if a real$z$codes a first order structure$M$, then the question of whether$M$satisfies a first order assertion is an arithmetic statement in$z$, since we need only quantify over the coded elements, which are coded by natural numbers. • There is some countable ordinal$\alpha$such that$L_\alpha$satisfies$x\lt_L y$. • For every countable ordinal$\alpha$, if$x$and$y$are reals in$L_\alpha$, then$L_\alpha$satisfies$x\lt_L y$. • All reals$z$coded coding well-founded structures$L_\alpha$in which$x$and$y$are reals satisfy$x\lt_L y$. • The third fourth statement has complexity$\Sigma^1_2$, since being-well-founded is$\Pi^1_1$. Similarly the fourth fifth statement has complexity$\Pi^1_2$, so overall the ordering is$\Delta^1_2$. The end result is that in the universe$L$, there is a low-complexity definable well-ordering of the reals. In this universe, therefore, all of the supposedly non-constructive applications of AC turn out to be completely definable. 2 added 84 characters in body In the constructible universe, there is a definable well-ordering of the entire universe. The ordering, as I mention in this MO answer, is that$x$preceeds$y$when$x$is constructed at an earlier stage of the$L$hierarchy, or else they are constructed at the same stage, but$x$is constructed at that stage by an earlier definition, or with the same definition, but with earlier parameters. If one only wants to well-order the real numbers, then one can extract from this definition a$\Delta^1_2$ordering of the reals. The reason is that if$x$is a real number of$L$, then it appears at some countable stage$L_\alpha$for a countable ordinal$\alpha$, and the entire structure$L_\alpha$is countable, and hence itself coded by a real. Furthermore, the$L$-order is absolute to any$L_\alpha$, since$L_\alpha$knows about the$L_\beta$-heirarchy for$\beta<\alpha$. Also, if a countable structure is well-founded and thinks$V=L$, then it is$L_\alpha$for some$\alpha$. Putting all this together, we get that the following are equivalent for any two reals$x$and$y$: •$x\lt_L y$in the$L$order. • There is some countable ordinal$\alpha$such that$L_\alpha$satisfies$x\lt_L y$. • There is a real$z$coding a well-founded structure that thinks$V=L$(and so this structure must be some$L_\alpha$) in which$x$and$y$are reals and the structure satisfies$x\lt_L y$. • All reals$z$coded well-founded structures$L_\alpha$in which$x$and$y$are reals satisfy$x\lt_L y$. The third statement has complexity$\Sigma^1_2$, since being-well-founded is$\Pi^1_1$. Similarly the fourth statement has complexity$\Pi^1_2$, so overall the ordering is$\Delta^1_2. \Delta^1_2$. 1 In the constructible universe, there is a definable well-ordering of the entire universe. The ordering, as I mention in this MO answer, is that$x$preceeds$y$when$x$is constructed at an earlier stage of the$L$hierarchy, or else they are constructed at the same stage, but$x$is constructed at that stage by an earlier definition, or with the same definition, but with earlier parameters. If one only wants to well-order the real numbers, then one can extract from this definition a$\Delta^1_2$ordering of the reals. The reason is that if$x$is a real number of$L$, then it appears at some countable stage$L_\alpha$for a countable ordinal$\alpha$, and the entire structure$L_\alpha$is countable, and hence itself coded by a real. Furthermore, the$L$-order is absolute to any$L_\alpha$, since$L_\alpha$knows about the$L_\beta$-heirarchy for$\beta<\alpha$. Also, if a countable structure is well-founded and thinks$V=L$, then it is$L_\alpha$for some$\alpha$. Putting all this together, we get that the following are equivalent for any two reals$x$and$y$: •$x\lt_L y$in the$L$order. • There is some countable ordinal$\alpha$such that$L_\alpha$satisfies$x\lt_L y$. • There is a real$z$coding a well-founded structure that thinks$V=L$(and so this structure must be some$L_\alpha$) in which$x$and$y$are reals and the structure satisfies$x\lt_L y$. • All reals$z$coded well-founded structures$L_\alpha$in which$x$and$y$are reals satisfy$x\lt_L y$. The third statement has complexity$\Sigma^1_2$, since being-well-founded is$\Pi^1_1$. Similarly the fourth statement has complexity$\Pi^1_2$, so overall the ordering is$\Delta^1_2.