It is easy to see that, if $V\models\alpha>\omega_1^{CK}$, then $\alpha$ is not recursive in any forcing extension of $V$. The argument goes as follows:

  • The relation "$\Phi_e=r$" is $\Pi^0_2$.

  • The predicate "$r$ codes an ill-founded order" is $\Sigma^1_1$.

  • So if $V[G]\models $"$\Phi_e$ is a well-ordering," then so did $V$; and moreover, $V[G]$ and $V$ interpret $\Phi_e$ in the same way.

In the presence of generic absoluteness, we can extend this. For instance, it follows from Projective Determinacy that the set of ordinals which have projective copies in some set-generic extension of $V$ is precisely the set of ordinals which have projective copies already in $V$.

My question is about what happens when we don't have generic absoluteness.

Is it consistent with $ZFC$ that: "For every ordinal $\alpha$, there is a set-generic extension of the universe in which $\alpha$ has a projectively definable copy?"

If so, what is the least $n$ such that it is consistent with ZFC that every ordinal is "potentially $\Pi^1_n$"?

I am especially interested in what we know if $V=L$. In particular, I don't know that "there is a non-potentially-$\Pi^1_3$ ordinal" is even consistent with $ZFC+V=L$!

Conversely, I would be delighted (and amazed) if "Every ordinal is potentially $\Pi^1_3$" had nonzero consistency strength over $ZFC$, despite contradicting large cardinals.


1 Answer 1


Great question!

The answer is yes. Start with $V=L$, and fix any ordinal $\alpha$. Now go to the forcing extension $V[G]$ obtained by collapsing $\aleph_\alpha^L$ to $\omega$, so that $(2^\omega)^{V[G]}=\aleph_{\alpha+1}^L=\omega_1^{V[G]}$. In $V[G]$, the ordinal $\alpha$ is definable in the structure $H_{\omega_1}^{V[G]}$, since this structure sees all countable objects, including all $L_\beta[G]$ for $\beta<\omega_1^{V[G]}=\aleph_{\alpha+1}^L$, and therefore $H_{\omega_1}^{V[G]}$ can identify the unique $\alpha$ for which $\omega_1=\aleph_{\alpha+1}^L$. This definition is projective, since we need to quantify only over countable objects.

I'll have to think a bit more to identify the exact complexity of the definition.

  • $\begingroup$ My answer is about defining the order-type, which is somewhat different than what you asked, since to have a projective copy we'd need to pick out a particular copy. I'll think a bit more about the connection. $\endgroup$ Oct 11, 2015 at 23:17
  • $\begingroup$ Awesome! But yeah, the problem is normally I would use large cardinals to go from a projective definition to a projective copy . . . :P $\endgroup$ Oct 11, 2015 at 23:19
  • $\begingroup$ My answer shows that if you start in $L$, then for any ordinal $\alpha$ there is a forcing extension $V[G]$ in which the collection of reals coding relations on $\omega$ with order type $\alpha$ is projectively definable. But you want there to be a particular such real that is (lightface) projectively definable? For large $\alpha$ that won't be true in my model for any real, by homogeneity considerations. If I can use $G$ as a parameter, though, then I can use the $L[G]$ order. $\endgroup$ Oct 11, 2015 at 23:23
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    $\begingroup$ Concerning homogeneity: suppose $\alpha$ is uncountable and there is a real $x$ coding $\alpha$ that is definable in my model $V[G]$. Since the forcing is homogeneous, it follows that $\text{HOD}^{V[G]}\subset V$, and so $x\in V$, contradicting $\alpha$ uncountable. And I agree with your point about parameters (which I had also realized shortly after posting my comment). $\endgroup$ Oct 12, 2015 at 11:39
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    $\begingroup$ I think it is worth investigating Joel's idea in a different generic extension of L. Harrington has a paper on Long Projective Wellordings. He can force over L to increase the size of the continuum (C) and ensure that there is a lightface projective well ordering of the reals and also ensure that every set of reals of size less than C is boldface projective, $\Pi^1_2$ I think. If we use Harrington's method to make C $\aleph_{\alpha+1}$ then we should be able projectively define the ordering on the real numbers that occupy the positions of cardinals. $\endgroup$ Oct 30, 2015 at 13:47

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