I asked this on math.stackexchange but did not receive an answer, so I'm asking here.

Assume large cardinals. Can we have $\omega_2^{L(\mathbb{R})}=\omega_2$?

Note that $\omega_1=\omega_1^{L(\mathbb{R})}$ always: clearly $\omega_1\ge\omega_1^{L(\mathbb{R})}$, and since $L(\mathbb{R})$ contains every real, any countable ordinal in $V$ is countable in $L(\mathbb{R})$. Also, note that we can trivially have $\omega_2^{L(\mathbb{R})}<\omega_2$: just force with $Col(\omega_1,\omega_2)$. $L(\mathbb{R})$ doesn't change, but the old $\omega_2$ is collapsed.

As Asaf Karagila pointed out at my MSE question, in general $\omega_n^{L(\mathbb{R})}$ is singular in $L(\mathbb{R})$, and so can't be $\omega_n$. However, this doesn't work for $n=2$: $\omega_2^{L(\mathbb{R})}$ is measurable in $L(\mathbb{R})$.

More generally, I'm curious for what ordinals $\alpha$ we may have $\omega_\alpha=(\omega_\alpha)^{L(\mathbb{R})}$.


Should be possible, at least for $\omega_2$. Under large cardinals, $\omega_2^{L(\mathbb{R})}$ is the supremum of the lengths of the boldface $\Delta^1_2$-prewellorderings of the reals (i.e., $\delta^1_2$), which is the same whether it's computed in V or in $L(\mathbb{R})$. Woodin showed that, for example, if the nonstationary ideal on $\omega_1$ is $\omega_2$-saturated and $P(\omega_1)^\sharp$ exists, then $\delta^1_2$ is $\omega_2$.

  • $\begingroup$ Nice! I didn't know that - do you have a citation where I can find the proof? $\endgroup$ – Noah Schweber Dec 19 '16 at 22:11
  • $\begingroup$ Yes, it's in Woodin's book (The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal), Theorem 3.17. $\endgroup$ – Juan Dec 19 '16 at 22:15
  • $\begingroup$ Awesome, thanks! And it looks like the original computation of $\delta^1_2$ under large cardinals was due to Martin, unpublished. Do you have an MSE account? If so, if you post this answer below my MSE question I'll also accept it there (otherwise I'll post my own answer, citing this one, and make it cw; I just don't want that question to remain on the "open" list). $\endgroup$ – Noah Schweber Dec 19 '16 at 22:15
  • $\begingroup$ Oh, yes, forgot about that! I don't have an MSE account (that I know of), but you can post the answer there if you want to. $\endgroup$ – Juan Dec 19 '16 at 22:23
  • $\begingroup$ I didn't know you have to have an account to post questions and answers on Stack Exchange sites. I thought you only need an account to post comments and vote. $\endgroup$ – bof Dec 19 '16 at 22:41

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