Memorable ordinals

It is a very nice question, but unfortunately, the answer is no.

Theorem. There are only countably many memorable ordinals.

Proof. Let $\delta$ be a countable ordinal with $L_\delta\prec L_{\omega_1}$. I claim that every memorable ordinal is less than $\delta$.

To see this, suppose that $\alpha$ is memorable, as witnessed by $\beta$. Fix an ordinal $\gamma$ above both $\delta$ and $\beta$ with $L_\gamma\prec L_{\omega_1}$. It follows that $L_\delta\prec L_\gamma\prec L_{\omega_1}$. But since $\beta\leq\gamma$, we know $\alpha$ is definable in $L_\gamma$ and hence (since there are no parameters) definable in $L_\delta$. So $\alpha<\delta$, as desired. QED

REPAIRED ARGUMENT

The memorables are strictly bounded between the first $\Sigma_2$-stable in $\omega_1$ and the first $\Sigma_3$-stable in $\omega_1$. The least non-memorable is thus a $(\Sigma_2\wedge \Pi_2)$-singleton.

Let $\delta_0$ be this first non-memorable. The previous erroneous argument yields a characterisation or perhaps a paraphrase of $\delta_0$.

For $\beta<\omega_1$ let

(A) $H(\beta)$ be the Skolem Hull inside $L_\beta$ of the empty set.

(B) Let $\omega_1(\beta):= (\omega_1)^{L_\beta}$ if the latter is defined, $= \beta$ otherwise.

Then (i) $H(\beta)$ is the set of pointwise definable objects in $L_\beta$. (ii) $H(\omega_1(\beta))=$ $L_\tau$ for some $\tau\leq \omega_1(\beta)$. (iii) For unboundedly many $\beta$, $\beta = \omega_1(\beta)$.

Claim Let $\delta_1 =$ the least $\delta< \omega_1$ so that for unboundedly many $\beta$ $H(\beta)=L_{\delta_1}$. Then $\delta_1=\delta_0$.

Proof: It is easy to argue that $\delta_1$ is defined. Then, by definition $\delta_1$ is not memorable (as $\delta_1 \notin H(\beta)$ for arbitrarily large $\beta$). So it suffices to show that $\tau<\delta_1 \rightarrow \tau$ is memorable.

By definition, and countability, of $\delta_1$:

$\exists \beta_0\forall \beta> \beta_0\forall\tau<\delta_1\,\, H(\beta)\neq L_\tau \quad (*).$

As $\beta \longrightarrow \omega_1$ so does $\omega_1(\beta) \longrightarrow \omega_1$ non-decreasingly. Thus there is $\beta_1>\beta_0$ so that:

$\forall\beta>\beta_1\,\, \omega_1 (\beta)>\beta_0.$

Then, using (ii), for any $\beta>\beta_1\,\, H(\omega_1(\beta))=L_\gamma$ for some $\gamma \leq\omega_1(\beta)$ but by $(*)$ $\delta_1\leq \gamma.$ Thus any $\tau< \delta_1$ is pointwise definable in $H(\omega_1(\beta))$ and so (using a definition of $\omega_1$), it is pointwise definable in $L_\beta$. So all $\tau$ less than $\delta_1$ are memorable as required. $\quad $ QED.

So something similar would work if CH holds, or in CH models. Eg let $A\subseteq\omega_1$ be such that $H_{\omega_1}=L_{\omega_1}[A]$. Consider the least $\delta$ with $L_\delta[A\cap \delta]\prec L_{\omega_1}[A]$.