Cross posted from MSE at commenter's suggestion: https://math.stackexchange.com/questions/2269319/definability-of-truth-in-l-without-0

I'm interested in the relationship between the existence of $0^{\#}$ and the definability of truth in $L.$ My advisor showed me an argument that you don't need $0^{\#}$ for truth in $L$ to be definable (from a Mahlo cardinal you can make a model where regular cardinals $\kappa$ such that $L_{\kappa} \prec L$ are precisely those where GCH fails). Of course, this definition requires quantification over the set-theoretic universe, which is much less impressive than the $\Delta^1_3$ definability of $0^{\#},$ and I don't see how to "mark" countable ordinals in the way forcing failures of GCH can mark regular cardinals.

So my question is, how low complexity can a definition of truth (without parameters) in $L$ be if $0^{\#}$ doesn't exist? Can there still be a definition with only real quantification? Could it even be $\Delta^1_3?$

  • 2
    $\begingroup$ I think we can decrease the complexity a bit as follows. Given a regular cardinal $\kappa$ such that $L_\kappa \prec L$, we may force with the Levy collapse to make $\kappa = \omega_1$. Then $L_{\omega_1} \prec L$, and the first-order theory of $L_{\omega_1}$ is $\Sigma_\omega(H_{\omega_1},\in)$ and therefore $\Sigma^1_\omega$. $\endgroup$ May 8, 2017 at 18:01
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    $\begingroup$ You're right, the reason he made the definition this way was to create a predicate for truth with parameters in $L.$ So I guess the question is, for truth without parameters, can we make it $\Delta_n^1$ for some $n?$ $\endgroup$ May 8, 2017 at 18:40
  • $\begingroup$ For future reference, you can flag the question on MSE and ask for it to be migrated, instead of cross posting. It's cleaner that way... $\endgroup$
    – Asaf Karagila
    May 9, 2017 at 4:32
  • $\begingroup$ Alright duly noted. $\endgroup$ May 9, 2017 at 4:33

1 Answer 1


The answer is yes, the theory of $L$ can be definable by a low-complexity definition quantifying over reals, even when $0^\sharp$ does not exist.

Here is one way to achieve this. Let me assume that the theory of $L$ is an element of $L$. This happens, for example, if $L_\kappa\prec L$ for some ordinal $\kappa$, because in this case the theory of $L$ is the same as the theory of $L_\kappa$, which is an element of $L$.

So let $t$ be the theory of $L$, which I have assumed is (coded by) a real in $L$. This real is therefore the $\alpha^{th}$ real in the $L$-order, and in order to define the theory $t$, it will suffice to define the ordinal $\alpha$.

Let $L[G]$ be a forcing extension of $L$ forcing to collapse $\aleph_{\alpha}^L$ to $\omega$. So in $L[G]$, the true $\omega_1$ is the same as $\omega_{\alpha+1}^L$, and we can determine this inside $H_{\omega_1}$. In that structure, we can define the class of ordinals that are cardinals in $L$, and there will be exactly $\alpha$ of them.

This makes the theory of $L$ definable inside $V=L[G]$ by a formula quantifying only over reals, and it will be $\Delta^1_n$ for some smallish $n$.

Let's try to find out how complex the definition is. My proposed definition is that in $V=L[G]$, the theory of $L$ is the theory coded by the real $t$ which is the $\alpha^{th}$ real in the $L$ order, where $\alpha$ is the number of infinite $L$-cardinals that are countable in $V$.

So, $t$ is as desired if there is a countable transitive model $L_\beta$ that thinks $t$ is the $\alpha^{th}$ real and which has exactly $\alpha$-many infinite cardinals, which do not get collapsed in any larger countable transitive $L_\gamma$, whereas all larger countable ordinals above those ordinals do get collapsed in some larger $L_\gamma$.

What is the complexity? It seems to be $\Sigma^1_4$. My initial thought that it might be $\Sigma^1_3$ are not right, as explained in the comments, since in that case it would be upward absolute by further forcing, but it clearly is not, since we could collapse more cardinals and thereby change the meaning of $\alpha$.

  • $\begingroup$ Are you computing the complexity of $t \subset \omega$ or of $\{t\} \subset \mathcal{R}?$ In the case of the former, $\Pi_3^1$ would follow from $\Sigma_3^1$ simply by checking if $L \models \neg \varphi.$ $\endgroup$ May 9, 2017 at 1:20
  • $\begingroup$ Ah, let's do it that way then. $\endgroup$ May 9, 2017 at 1:23
  • $\begingroup$ Now it seems this definition would not be invariant under forcing. If we ask for a definition of truth in $L$ which is invariant under forcing, would that require $0^{\#}$ to exist? $\endgroup$ May 9, 2017 at 1:56
  • $\begingroup$ ^ Actually I'm unsure about this. If your definition really is $\Sigma_3^1,$ then it is invariant under forcing right? That seems hard to believe. $\endgroup$ May 9, 2017 at 1:59
  • $\begingroup$ It's definitely not invariant under forcing, since we could collapse more cardinals, and this would change $\alpha$. So you are right, it can't be $\Sigma^1_3$. I guess it is $\Delta^1_4$. $\endgroup$ May 9, 2017 at 2:26

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