In $L(\mathbb{R})$, presumably $L_0(\mathbb{R})$ is the transitive closure of "$\mathbb{R}$", but what exactly is meant by "$\mathbb{R}$"? (i.e. which set theoretic representation of “the reals” is $\mathbb{R}$ intended to denote in this context?)
My default guess would be that "$\mathbb{R}$" means $\mathcal{P}(\omega)$ (the universe's power set of $\omega$), in which case $L_0(\mathbb{R}) = \mathcal{P}(\omega)$. But I could imagine other representations of "$\mathbb{R}$":
- With rational numbers represented as pairs of finite ordinals, the elements of "$\mathbb{R}$" could be sets of such rationals (i.e. literal Dedekind cuts)
- This article talks about "$\mathbb{R}$" meaning $\omega^{\omega}$ (which doesn't really make sense to me, but there it is...)
- In set theory with "atoms" (a.k.a. "urelements"), I suppose "$\mathbb{R}$" could be a set of atoms.
I suppose in principle there might be multiple valid choices, but is there an established convention for what "$\mathbb{R}$" is usually understood to mean for $L(\mathbb{R})$?
The point is that it doesn't really matter. Of course, if you chose that $\Bbb R$ is represented by a set of functions which are injections from $\mathcal P(\omega)$ into the ordinals, then formally speaking $L(\Bbb R)$ will include well-orderings of $\mathcal P(\omega)$. Clearly, this is not what we mean to investigate.
Taking either $\mathcal P(\omega)$, or $2^\omega$ or $\omega^\omega$, or even a subset of $\mathcal P(\omega)$ which is order-isomorphic to the reals, or any "reasonable variation thereof" will result in the same model. Why? Because we can translate one structure to the other in a very definable way. So whichever coding you take for the real numbers, as long as it is reasonable, will include all of the above.
For sake of concreteness, we will often think about $\mathcal P(\omega)$ as representing the real numbers. It is usually easy enough, since it's already a transitive set, to simply build the rest of the $L$-hierarchy on top of that.