TITLE: In $L$, does there exist a definable non-principal ultrafilter on $\mathbb{N}$
QUESTION [7 upvotes]: The axiom of constructibility $V=L$ leads to some very interesting consequences, one of which is that it becomes possible to give explicit constructions of some of the "weird" results of AC. For instance, in $L$, there is a definable well-ordering of the real numbers (since there is a definable well-ordering of the universe).
Since AC holds true in $L$, the ultrafilter lemma must be true. Does this mean that a definable non-principal ultrafilter on $\mathbb{N}$ exists in $L$, given by an explicit formula?
If so, what is the formula?

REPLY [8 votes]: Let me emphasize what Noah de-emphasized (by putting it in parentheses) in his answer: Instead of the "really cheap" idea of the first free ultrafilter in $L$-order, one can use the $L$-order of subsets of $\mathbb N$ to greedily construct an ultrafilter. The $L$-ordering of the subsets of $\mathbb N$ is a $\Delta^1_2$ relation with some additional properties (I believe Moschovakis uses the terminology $\Delta^1_2$-good well-ordering). One can then deduce that the greedily constructed ultrafilter is a $\Delta^1_2$ set of subsets of $\mathbb N$.
Since free ultrafilters on $\mathbb N$ never have the Baire property and are never Lebesgue measurable, this shows that the classical results, that analytic and coanalytic sets have the Baire property and are measurable, cannot be extended higher in the projective hierarchy in ZFC (without additional hypotheses like large cardinals or determinacy).
Similar greedy constructions based on the $L$-ordering of the reals produce $\Delta^1_2$ examples of lots of other sets of reals that, in the absence of $V=L$, would be obtained by applying the axiom of choice and thus might not be definable at all.  Examples include Hamel bases over $\mathbb Q$, Vitali sets, and Bernstein sets.