Suppose $V=L$ + reasonable hypotheses (e.g. "ZFC has a countable transitive model"). Call a countable ordinal $\alpha$ *memorable* if for some countable $\beta$, $\alpha$ is definable *without parameters* in every $L_\gamma$ with $\beta<\gamma<\omega_1$.

My question is:

Are there uncountably many memorable ordinals?

Some comments, right off the bat:

The answer is trivially "no" for

*uniformly*memorable ordinals, that is, ordinals which are coboundedly definable without parameters*by the same formula*. However, since the defining formula is allowed to vary with $\gamma$, this doesn't work.Towards a positive answer, note that there are uncountably many countable $\theta$ such that $L_\theta$ is pointwise definable (call such an ordinal "insightful"); this was proved by Hamkins, Linetsky, and Reitz. However, this doesn't actually resolve the issue: for a fixed $\alpha$, there may be many countable $\eta>\alpha$ with no insightful ordinals $>\alpha$ which are definable without parameters in $L_\eta$ (e.g. if there is no greatest insightful ordinal $<\eta$, this seems a distinct possibility).

Interestingly, this question has possibly interesting variants even if $L$ is a very tiny subclass of $V$! Given any hierarchy $(M_\alpha)_{\alpha\in\omega_1}$ with $M_\alpha\cap ON=\alpha$, we can ask whether uncountably many $\alpha$ satisfy "For all sufficiently large $\beta<\omega_1$, $\alpha$ is parameter-free definable in $M_\beta$." Now, we can trivially construct examples where the answer is "yes": namely, let $M_\alpha$ be $L_{\eta_\alpha}$ where $\eta_\alpha$ is the $\alpha$th insightful ordinal; so the general existence question isn't interesting. However, I would be curious if there are any "natural" hierarchies of $M_\alpha$s which do satisfy this property; especially if their union is $H_{\omega_1}$ in some $V$ which is very far from $L$.

memorable after all... :P – Asaf Karagila Jan 8 '17 at 20:09that