Timeline for answer to Indescribability of cardinals and categoricity of $V_\kappa$ by Joel David Hamkins
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Jul 8, 2013 at 0:02 | comment | added | Joel David Hamkins | ...I had meant $\Pi^1_0$-indescribable, rather than $\Pi^1_0$-reflecting. | |
| Jul 7, 2013 at 16:37 | vote | accept | Asaf Karagila♦ | ||
| Jul 7, 2013 at 15:41 | comment | added | Joel David Hamkins | As for your first question, yes, the first inaccessible is $\Pi^0_2$-categorical. And it seems to be $\Pi^1_0$-reflecting, since the subscript $0$ means that there actually are no second-order quantifiers in the statement $\psi$ to be reflected, and so we may just apply Lowenheim-Skolem to find $\alpha\lt\kappa$ with $\langle V_\alpha,R\cap V_\alpha\rangle\prec\langle V_\kappa,R\rangle$. | |
| Jul 7, 2013 at 15:36 | comment | added | Joel David Hamkins | Second, yes, the first inaccessible above the least measurable are also below the first $\Sigma_2$-correct ordinal, since they are $\Sigma_2$-definable. These are rather small in terms of the much larger large cardinals. Consistency-wise, $\Sigma_2$-reflecting ordinals are very weak, but their size in a given universe is governed in effect by what cardinals exist there. | |
| Jul 7, 2013 at 15:34 | comment | added | Joel David Hamkins | First, I think the least inaccessible is $\Pi^0_2$-categorical, that is, $n=2$ is enough, since the least inaccessible $\kappa$ is unique such that $V_\kappa\models\text{ZFC}_2+$"there is no inaccessible cardinal", and this latter assertion is $\Pi_2$ expressible in set theory, and hence $\Pi^0_2$ expressible as an assertion about $V_\kappa$. | |
| Jul 7, 2013 at 15:16 | comment | added | Asaf Karagila♦ | Joel, I don't understand something. If the least inaccessible is $\Pi^0_n$ categorical (where $n=5$ is probably enough), how can it be $\Pi^1_0$ indescribable? Also, we can describe the first inaccessible larger than the first measurable cardinal, "there is a maximal inaccessible cardinal and it is the only measurable", and that too is a $\Pi^0_n$ statement (because measurable are first-order definable). So either $\Sigma_2$-correct ordinals are very very very large (regardless to the shape of the universe), or I'm missing a point. | |
| Jul 7, 2013 at 14:52 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jul 7, 2013 at 14:06 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jul 7, 2013 at 14:00 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |