Timeline for Elementary submodels of V
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 26, 2014 at 20:51 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Corrected the correct/reflecting terminology to current usage.
|
| Aug 3, 2012 at 13:41 | vote | accept | curious | ||
| Aug 3, 2012 at 13:41 | vote | accept | curious | ||
| Aug 3, 2012 at 13:41 | |||||
| Aug 3, 2012 at 11:39 | comment | added | Joel David Hamkins | ....And on the other hand, the least totally indescribable cardinal $\kappa$ is definable in $V_{\kappa+\omega}$ and is therefore not $\Sigma_2$-reflecting, since if $\theta\geq\kappa+\omega$, then $V_\theta$ thinks there is a totally indescribable cardinal but no $\theta'\lt\kappa$ thinks that is true. | |
| Aug 3, 2012 at 11:38 | comment | added | Joel David Hamkins | Everett, my $\theta$ was arbitrary, and $\theta=\delta$ was allowed. Meanwhile, the two notions of reflection are orthogonal. For example, the existence of $\Sigma_2$ reflecting cardinals is provable in ZFC and thus much weaker than the totally indescribable cardinals in consistency strength (even inaccessible $\Sigma_2$-reflecting cardinals are weaker). | |
| Aug 3, 2012 at 7:01 | comment | added | Everett Piper | The thought I have is to let $\delta=\theta$ and suppose $V_\delta\models \varphi[x]$ for $\varphi$ arbitrary and $x\in V_\delta$. Then $\{x\}\subset V_\delta$ and $V_\delta\models \varphi^*[\{x}\]$ where $\varphi^*$ is $$\forall y (y\in\{x\})\wedge\varphi[y]$$. If $\delta$ is totally indescribable, doesn't it then follow that $V_\alpha\models \varphi^*[\{x\}]$ for some $\alpha<\delta$? | |
| Aug 3, 2012 at 6:49 | comment | added | Everett Piper | Joel, relating to your ending remark, would you mind elaborating a little on the difference between $\Sigma_2$-reflecting cardinals and the notion of totally indescribable cardinal? It almost seems like the equivalence you mention allows the first $\theta$ to be $\delta$ itself. But your final sentence indicates that $\theta$ is strictly above $\delta$. Do you mean to have $\theta$ strictly larger than $\delta4? | |
| Aug 2, 2012 at 22:16 | comment | added | Joel David Hamkins | One way of observing that the theory $V_\delta\prec V$ does not prove Con(ZFC) is that if it did, then this proof would use only finitely many of the assertions in the scheme $V_\delta\prec V$, but any finitely many of those assertions hold for some $V_\delta$ by the reflection theorem, and so this would mean that ZFC itself would imply Con(ZFC), which contradicts the incompleteness theorem unless ZFC is inconsistent. | |
| Aug 2, 2012 at 15:01 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 71 characters in body
|
| Aug 2, 2012 at 13:30 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 1578 characters in body
|
| Aug 2, 2012 at 12:49 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |