Timeline for Computing $L$-rank (constructible universe)
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2013 at 17:55 | comment | added | Monroe Eskew | Oh I see, sorry. | |
| Nov 16, 2013 at 9:13 | comment | added | Philip Welch | @Monroe: I was discussing in the parenthetical remark ``such $\alpha$'' with a new real appearing definably over $L_\alpha$. Then one will also appear over $L_{\alpha +1}$, ... , $L_{\alpha+\omega}$... :) | |
| Nov 16, 2013 at 8:58 | vote | accept | phil | ||
| Nov 16, 2013 at 1:10 | comment | added | Monroe Eskew | Or rather $L_{\alpha+2}$. | |
| Nov 16, 2013 at 0:58 | comment | added | Monroe Eskew | It seems like you said if $\alpha$ is a countable limit, then a new real appears in $L_{\alpha+1}$. | |
| Nov 15, 2013 at 20:00 | history | edited | Philip Welch | CC BY-SA 3.0 |
added 236 characters in body
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| Nov 15, 2013 at 19:10 | comment | added | Philip Welch | @Monroe: which part of my answer is incorrect? | |
| Nov 15, 2013 at 18:33 | comment | added | phil | Thank you for your answer. Could you give me an explanation for your last statement? I can't see why equality can't hold if $\rho(x)=\gamma+1$. | |
| Nov 15, 2013 at 18:29 | comment | added | Monroe Eskew | Actually this is not totally correct. It's true that new reals are constructed arbitrarily high, but surprisingly, there are also arbitrarily long gaps where no new reals appear. Let $\alpha < \omega_1$ be arbitrary, and let $\omega_1 > \beta > \alpha$ be such that there is an elementary $j : L_\beta \to L_{\omega_2}$ with $cr(j) = \gamma \geq \alpha$. Then for all $\delta < \gamma$, $L_{\omega_2} \models$ "There are no new reals constructed between $\omega_1$ and $\omega_1 + \delta$." By elementarity and absoluteness, there are no new reals added between $\gamma$ and $\gamma + \delta$. | |
| Nov 15, 2013 at 18:08 | history | answered | Philip Welch | CC BY-SA 3.0 |