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. 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.

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 $ZFC$.

I suspect that the assumption $V=L$ is unnecessary to show that $\Delta^1_2$ lacks the uniformization property, but I am less certain about this.

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.

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 $ZFC$.