DISCLAIMER: All pointclasses considered here are boldface.


Most of the time, when doing descriptive set theory, we want the projective sets to "behave well;" for example, maybe we don't want there to be nonmeasurable projective sets, or projective well orderings of $\mathbb{R}$, etc. Generally, this means making some (fairly conservative) large cardinal assumption, or equivalent.


At the far opposite end of things is the axiom that all sets are constructible, $V=L$. This axiom implies that there is a projective - in fact, $\Delta^1\_2$ - well-ordering of the reals, and so projective sets become bad very early in the hierarchy. 


My question is about the state of affairs when $V=L$ holds. My motivation is simply that I don't feel I have a good grasp on basic concepts in descriptive set theory, and the following seemed like a good test problem to assign myself; but I have thought about it for a while without making progress, so I'm asking here:


Let $\oplus$ be one of the usual pairing operators on $\omega^\omega$. For the purposes of this question, we say that a pointclass $\Gamma\subseteq \mathcal{P}(\omega^\omega)$ has the uniformization property if whenever $A\in \Gamma$, there is some $B\in \Gamma$ such that:


* $B\subseteq A$, and
* Whenever $x\oplus y\in A$, there is a unique $z$ such that $x\oplus z\in B$.


That is, we view $A$ as coding a relation on $\omega^\omega\times \omega^\omega$, and $B$ is the graph of a function contained in $A$. (This is not usually how uniformization is presented, but it's equivalent for all intents and purposes.) My question is then:



> 
> Assume $V=L$. Let $D$ be the set of (boldface) $\Delta^1\_2$ elements of $\omega^\omega$; does $D$ have the uniformization property?
> 
> 
> 


Now, it seems clear to me that $D$ should **not** have the uniformization property. [EDIT: As Joel's answer below shows, this is completely wrong.] The counterexample should be just the $\Delta^1\_2$ well-ordering $\prec$ given by the assumption that $V=L$: uniformizing $\prec$ requires us to choose, for each real $r$, a real $s$ such that $r\prec s$; and although $\prec$ is $\Delta^1\_2$, the usual way of doing this - choosing the immediate $\prec$-successor of $r$ - is no longer $\Delta^1\_2$.


However, I don't know how to show that $\prec$ - or any other $\Delta^1\_2$ set - cannot be uniformized in $\Delta^1\_2$. I suspect I'm just missing something fairly simple.




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Note: it is known that the boldface pointclasses $\Pi^1\_1$ and $\Sigma^1\_2$ have the uniformization property, and assuming large cardinals, the uniformization property can be further propagated to every pointclass $\Pi^1\_{2n+1}$, $\Sigma^1\_{2n}$. On the other hand, the class $\Delta^1\_1$ of Borel sets lacks the uniformization property, provably in $ZFC$.



Does the boldface class $\Delta^1_2$ have the uniformization property? (assuming $V=L$)
Unless I am mistaken, it seems to me that $\Delta^1\_2$ does have the uniformization property in $L$. 


For any set $A$ in $\Delta^1\_2$, let $B$ select the $L$-least witness on each slice. So $B$ unifomizes $A$, and the graph of $B$ appears to be $\Delta^1\_2$, by the following reasoning:


* $x\oplus z\in B$ if and only if it is in $A$, and for every well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.
* $x\oplus z\in B$ if and only if it is in $A$, and there is a well-founded countable model $M$ of $V=L$ containing $x$ and $z$, if $y$ is a real in $M$ preceding $z$, then $x\oplus y\notin A$.


The point here is that the countable well-founded models are correct about the $L$-predecesors of the reals that they can see. So we can use any or all of them when verifying that $z$ is least such that some $\Delta^1\_2$ property holds. Note that the "for every real $y$ in $M$" is merely a natural number quantifier, since $M$ is coded as a countable structure. So the first of these characterizations is $\Pi^1\_2$ and the second is $\Sigma^1\_2$, and so it is $\Delta^1\_2$ overall. 


