Timeline for Memorable ordinals
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2017 at 7:52 | comment | added | Joel David Hamkins | I think my remarks about about $2$-stable and $L_\delta\prec_{\Sigma_2} L_{\omega_1}$ should have been instead about $L_\delta\prec_{\Sigma_3} L_{\omega_1}$. We only know that the memorable ordinals are bounded below the least $\Sigma_3$-correct ordinal in $L_{\omega_1}$. | |
| Jan 10, 2017 at 2:59 | comment | added | Noah Schweber | I wonder if we can get every $\Delta_2$-definable ordinal as well. The outline I see is the following. Let $\varphi\in\Sigma_2$, $\psi\in\Pi_2$ each define a unique ordinal $\alpha$. Then sufficiently large $\gamma$ see (i) a witness to $\varphi(\alpha)$ and (ii) for each $\beta<\alpha$, a witness to $\neg\psi(\beta)$; so in $L_\gamma$, $\alpha$ is the least ordinal satisfying $\varphi$. But taking this apart, this relies on all sufficiently large $\gamma$ seeing $\varphi\iff\psi$; so really this only works for "provably" $\Delta_2$ ordinals. Is this fixable? | |
| Jan 8, 2017 at 20:27 | comment | added | Joel David Hamkins | The $\Sigma_1$-definable objects are eventually defined by the same formula every time. | |
| Jan 8, 2017 at 20:22 | comment | added | Joel David Hamkins | 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. | |
| Jan 8, 2017 at 20:19 | comment | added | Joel David Hamkins | Oh, I see why you did that now. | |
| Jan 8, 2017 at 20:15 | comment | added | Noah Schweber | (Re: your penultimate comment:) I don't quite see that - what if $L_\gamma$ sees a $\Sigma_1$ formula which holds of $\theta$, and thinks it defines $\theta$ since it doesn't see the other solution(s)? I feel like we need to go above just finding witnesses to the smaller ordinals' $\Sigma_1$-definability. | |
| Jan 8, 2017 at 20:15 | comment | added | Joel David Hamkins | So we've trapped the memorable ordinals strictly above 1-stable and below 2-stable. | |
| Jan 8, 2017 at 20:14 | comment | added | Joel David Hamkins | Yes, I agree. The way I think about it is: the least $\alpha$ that isn't $\Sigma_1$-definable in $L$ is revealed as such once you get above the witnesses required to $\Sigma_1$-define the smaller ordinals. So this defines $\alpha$ in all further $L_\gamma$. | |
| Jan 8, 2017 at 20:12 | comment | added | Noah Schweber | it can tell that everything $<\theta$ isn't $\Sigma_1$-definable, and it sees every other potential $\Sigma_1$ definition of $\theta$ either not hold of $\theta$, or hold of at least two ordinals and hence not define $\theta$. Is this right? | |
| Jan 8, 2017 at 20:11 | comment | added | Noah Schweber | Hm, I think the least non-$\Sigma_1$-definable ordinal $\theta$ will be memorable. Here's what I'm thinking: for each $\Sigma_1$ formula $\varphi$, let $\beta_\varphi$ be (i) the least $\alpha$ such that $L_\alpha$ sees two ordinals satisfying $\varphi$, or (ii) $0$, otherwise. If we take $\gamma=\sup_{\varphi\in\Sigma_1}\beta_\varphi$ plus enough so that $L_\gamma$ sees that every ordinal $<\theta$ is $\Sigma_1$-definable, I think every $L_\eta$ with $\eta>\gamma$ sees $\theta$: (cont'd) | |
| Jan 8, 2017 at 20:09 | comment | added | Joel David Hamkins | Are they closed downwards? | |
| Jan 8, 2017 at 20:05 | comment | added | Noah Schweber | Well, I think $L_\delta$ (for $L_\delta\prec L_{\omega_1}$) is right about memorability unless my quantifier counting is off. Given that, the least unmemorable $\gamma$ - being $<\delta$ - will be definable in $L_\delta$. However, since $\delta$ itself isn't memorable, we can't "move this definition up," so this doesn't give a contradiction immediately. Can we get around this? | |
| Jan 8, 2017 at 20:00 | comment | added | Joel David Hamkins | Clearly, we are thinking in tune here with the overlapping comments. | |
| Jan 8, 2017 at 19:59 | comment | added | Noah Schweber | Hah, comment overlap! Yes, I suspect that's a characterization although I don't immediately see how to prove it. | |
| Jan 8, 2017 at 19:59 | comment | added | Noah Schweber | Hehe, I was just about to ask whether the memorable ordinals could be cofinal in $\delta$! Thinking about the satisfaction relation, is $\prec_{\Sigma_3}$ enough to bound them? (I'm not sure about my quantifier counting.) | |
| Jan 8, 2017 at 19:58 | comment | added | Joel David Hamkins | Indeed, since "$\alpha$ is memorable" is $\Sigma_2$ expressible in $L_{\omega_1}$, they will all be below the first $\delta$ where $L_\delta\prec_{\Sigma_2} L_{\omega_1}$. Perhaps this will be a characterization? | |
| Jan 8, 2017 at 19:56 | comment | added | Joel David Hamkins | Since "$\alpha$ is memorable" is first-order expressible in $L_{\omega_1}$, it follows from $L_\delta\prec L_{\omega_1}$ that they must be bounded below $\delta$. | |
| Jan 8, 2017 at 19:32 | comment | added | Joel David Hamkins | Notice that the model $L_\delta$ consists exactly of the definable elements of $L_{\omega_1}$ and hence is itself pointwise definable. | |
| Jan 8, 2017 at 19:32 | comment | added | Joel David Hamkins | I think we should aspire to give a characterization of the memorable ordinals. | |
| Jan 8, 2017 at 19:16 | comment | added | Noah Schweber | Nice, thank you! | |
| Jan 8, 2017 at 19:16 | vote | accept | Noah Schweber | ||
| Jan 8, 2017 at 18:34 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |