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 $L$.

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    I used to remember all those memorable ordinals. But I guess they weren't that memorable after all... :P – Asaf Karagila Jan 8 '17 at 20:09
up vote 13 down vote accepted

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

  • Nice, thank you! – Noah Schweber Jan 8 '17 at 19:16
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    I think we should aspire to give a characterization of the memorable ordinals. – Joel David Hamkins Jan 8 '17 at 19:32
  • Notice that the model $L_\delta$ consists exactly of the definable elements of $L_{\omega_1}$ and hence is itself pointwise definable. – Joel David Hamkins Jan 8 '17 at 19:32
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    Clearly, we are thinking in tune here with the overlapping comments. – Joel David Hamkins Jan 8 '17 at 20:00
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    For downward closure: if $\alpha$ is memorable, then eventually in $L_\gamma$ it is definable and also $\alpha$ is countable there. So the L-least real coding $\alpha$ can now be used to define any $\alpha'<\alpha$. So they form an initial segment of the ordinals. – Joel David Hamkins Jan 8 '17 at 20:22

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 $L_\delta[A\cap \delta]\prec L_{\omega_1}[A]$.

  • Very nice, Philip, thank you for posting! Your argument seems right, but this causes me to be confused. Isn't "$\alpha$ is memorable" a $\Sigma_2$ expressible assertion in $L_{\omega_1}$? In that case, then if $L_\delta\prec L_{\omega_1}$, there would have to be non-memorable ordinals below $\delta$, which contradicts your theorem. And $\Sigma_3$-correct $\delta$ should be enough for that. What have I got wrong? – Joel David Hamkins Jan 13 '17 at 13:51
  • You are of course right, Joel, there have to be non-memorables below $\delta$. My mistake was the final ''Hence'' (I really should check my work! Elementarity in every element of an increasing tower does not imply elementarity in the union.) – Philip Welch Jan 13 '17 at 14:23
  • I guess your argument does show: the supremum of the memorable ordinals is exactly the first ordinal $\tau$ with uncountably many elementary extensions below $L_{\omega_1}$. Right? – Joel David Hamkins Jan 13 '17 at 17:11
  • Incidentally, this kind of characterization of an ordinal, in terms of its number and kind of elementary extensions, is here in Bonn referred to as a Welch-style characterization. Peter Koepke had said a few days ago, Perhaps we should aim at a Welch-style characterization, and now you've come through! – Joel David Hamkins Jan 13 '17 at 19:02
  • I guess $\tau$ is the least ordinal with uncountably many elementary extensions $L_\tau\prec L_\gamma$, period. – Joel David Hamkins Jan 13 '17 at 22:21

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