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I'm pretty sure this is an easy descriptive set theory question that I'm just blanking on.

Is it consistent with large cardinals - say, with a measurable - that every (nonempty) $\Sigma^1_4$ class of reals has a $\Delta^1_4$ member?

My recollection is that there is an easy negative answer, but I can't remember the details.

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This is true in $L$ by Exercise $5\,A.4$ in Moschovakis $\text{Descriptive Set Theory}$.

In fact for all $n \geq 2$, $\Delta_n^1$ is a basis for $\Sigma_n^1$. The following is a sketch:

Using the $\Delta_2^1$ well-ordering of $({}^\omega 2)^L$, you can show that $\Sigma_n^1$ has the uniformization property.

Given $A$ which is $\Sigma_n^1$, consider $\{0\} \times A$ which is still $\Sigma_n^1$. Uniformize this, to get a $\Sigma_n^1$ singleton, $A^* = \{(0, y)\}$. Then

$$n \in y \Leftrightarrow (\exists z)(n \in z \wedge (0,z) \in A^*) \Leftrightarrow (\forall z)((0,z) \in A^* \Rightarrow n \in z)$$

Therefore, $y$ is $\Delta_n^1$ and $y \in A$.


If your question is whether it is consistent relative large cardinal that "there is a measurable cardinal and $\Delta_4^1$ is a basis for $\Sigma_4^1$", this is also true:

Moschovakis 6C.6 shows that $\Delta_{2n}^1$ determinacy implies $\Delta_{2n + 2}^1$ is a basis for $\Sigma_{2n + 2}^1$.

As above, this follows from $\Delta_{2n}^1$ determinacy implying $\Sigma_{2n + 2}^1$ uniformization.


Also Martin and Solovay in $\text{A Basis Theorem for$\dots$}$ showed that if $\mu$ is a measure on a measurable cardinal, then $L[\mu][c]$ where $c$ is generic any small forcing (say Cohen forcing), then there is a $\Pi_3^1$ (in particular $\Sigma_4^1$) set without any ordinal definable reals.

So the presence of a measurable cardinal is also not enough to imply your property.

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  • $\begingroup$ I think I may have misunderstood the question. Maybe you meant: is it consistent that "there exists a measurable cardinal and $\Delta_1^1$ is a basis for $\Sigma_4^1$"? $\endgroup$ – William Oct 3 '15 at 12:08
  • $\begingroup$ @William The second part of your answer is what I was looking for. My memory, it seems, is far off (especially at midnight :P). Thanks! $\endgroup$ – Noah Schweber Oct 3 '15 at 20:19

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