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I note that the ordinals of L are the same as V, so I guess that it has no $\Pi_1^1$ consequences. On the other hand Wikipedia tells me that it asserts the existance of a $\Delta_2^1$ non-measurable set of reals. "Measurable" involves third-order concepts but I know that there's often a "coding trick" that gets around that sort of thing, so I guess it has some analytic consequences.

Of course I am guessing -- I am not very good at this stuff. But I am curious. What's the least point in the analytic hierarchy where V=L matters (if any)?

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    I had no idea what this question was about. I still don't really, but in case anybody else with similar ignorance and curiosity would like to avoid trying to Google L and V: L is the [constructible universe](http://en.wikipedia.org/wiki/Constructible_universe) and V is the [Von Neumann universe](http://en.wikipedia.org/wiki/Von_Neumann_universe)2010-10-15

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Shoenfield's absoluteness theorem ensures that any $\Sigma^1_2$ statement (and therefore, any $\Pi^1_2$ one) has the same truth value in $L$ as in $V$.

On the other hand, $\Sigma^1_3$ statements differ in general whether they are in $V$ or in $L$. For a silly example, $\exists x\in{\mathbb R}\,(x\notin L)$ can be rewritten as a $\Sigma^1_3$ statement, which is false in $L$ but true in general. Given a $\Sigma^1_2$ formula $\phi(x,y)$, the statement ``$\phi$ defines a well-ordering of the reals'' is $\Pi^1_3$. Again, there is a specific such $\phi$ for which this is true in $L$ and false in general.

("In general" means here that it can be made false by forcing. There are deeper discrepancies if one allows large cardinals into the picture.)

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    Thanks for the link and the examples. It follows that a $\Pi_3^1$ sentence true in V is true in L, therefore even $\Delta_3^1$ sentences must be invaraint between L and V. And this is tight because e.g. $\mathbb{R} \subset L$ is $\Pi_3^1$ by some coding trick (I am trying to find it now as an exercise...)2010-10-15