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



Can $L$ be defined without parameters?
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.


