# Can $L$ be defined without parameters?

If we omit parameters in the definition of $$L$$ would the result still be $$L$$?

That is, we define a successor stage $$L_{\alpha+1}$$ in the constructible universe $$L$$, without including parameters; as:

$$L_{\alpha+1} =\{\{ y \mid y \in L_\alpha \land (L_\alpha, \in) \models \phi(y) \}\mid \phi \text { is a first order formula}\}$$

• This gives only countably many sets at each stage. Jun 16 at 12:16
• Joel Hamkins has provided a nice answer. For posterity, I will just add that the very last exercise (#23, p.183) of Kunen's set theory text (the 1980 ed.) asks the reader to show that the parameter-free version of L coincides with the usual one. I don't know who first noticed this fact. Jun 16 at 14:19
• @MonroeEskew At limit stages you collect what you have so far. Jun 17 at 0:35

Yes, the parameter-free version of $$L$$ gives rise to the same constructible universe $$L$$. You will still get all of $$L$$ this way, but it will come more slowly.
The reason is that at stage $$\alpha+1$$, you in effect have $$\alpha$$ as a parameter, since this is definable as the largest ordinal. So you can use $$\alpha$$ as a parameter. Every finite sequence of ordinals is coded by a single ordinal, and so in this way you can have any finitely many ordinal parameters, and this is enough to pick out any object in the resulting hierarchy. So eventually you will get every set in $$L$$.
It seems that the two hierarchies will catch up to each other at every cardinal stage, $$L_\kappa$$ for any cardinal $$\kappa$$. At the finite stages this is clear, since the finite sets in $$L_n$$ are all definable. For any ordinal $$\xi$$, if the set $$L_\xi$$ appears at some stage $$\lambda$$, then it will be definable in the later stages from ordinal parameter $$\xi$$, and any definable subset of it will be definable in $$L_\xi$$ from some ordinal parameter $$\beta$$, and hence will be definable at stage $$\alpha+1$$ in your hierarchy, if $$\alpha$$ codes the pair $$\langle\xi,\beta\rangle$$. Inductively, this will all happen before the next cardinal stage, and therefore $$L_{\xi+1}$$ will arrive also before the next cardinal stage.
• It seems one needs to find $\alpha$ such that the definition of $L_\xi$ is absolute to $L_\alpha$. Or is this not really an issue? Jun 18 at 6:14
• I agree, but $L_\xi$ is definable in all the later limit stages of the OPs construction in which $\xi$ is definable, since the set $L_\xi$ itself validates all the earlier stages of construction, and if $\xi$ is a successor then the top level will be definable, since the truth predicate of $L_\xi$ is definable. Jun 18 at 7:36