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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 $ZFC$.

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  • $\begingroup$ Regarding your proposed counterexample, why isn't the immediate successor function in your context $\Delta^1_2$? After all, $s$ is the immediate successor of $r$ if and only if every (so $\Pi^1_2$) countable well-founded model of V=L containing both of them thinks it is, if and only if there is (so $\Sigma^1_2$) a countable well-founded model of V=L containing both of them that thinks it is. $\endgroup$ – Joel David Hamkins Aug 17 '13 at 21:38
  • $\begingroup$ Basically, the situation is that any property that can be verified inside any or all countable $L_\alpha$ with the parameters will be $\Delta^1_2$, since one can say either that is is true in one of them (giving the $\Sigma^1_2$ form) or in all of them (giving the $\Pi^1_2$ form). $\endgroup$ – Joel David Hamkins Aug 17 '13 at 22:43
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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.

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  • $\begingroup$ Well, that certainly explains why I couldn't prove my claim. :) (I've changed the title of the question, just to avoid future confusion.) Thanks for the clear answer! $\endgroup$ – Noah Schweber Aug 18 '13 at 13:21
  • $\begingroup$ I guess the same idea works for $\Delta^1_n$ for any $n$. $\endgroup$ – Joel David Hamkins Aug 18 '13 at 13:34
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Assuming $V=L$ then we have $AC$ and $CH$, so every set of reals is at most $\aleph_1$ Suslin. So we can find scales for them and uniformize them.In particular every $\Delta^1_2$ set of reals can be uniformized. As Joel said in the comment above this works for all $\Delta^1_n$ under $V=L$.

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  • $\begingroup$ Under $V=L$, Uniformization fails for any $\bf\Pi^1_n$ and $\varPi^1_n$ with $n\ge 2$, but rather surprisingly every lightface nonempty $\varPi^1_n$ set contains a $\varPi^1_n$ singleton by an old result of Harvey Friedman $\endgroup$ – Vladimir Kanovei Jul 9 '16 at 21:16

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