Timeline for A question about the Ordinal Definable elements of Power Sets
Current License: CC BY-SA 3.0
3 events
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| Mar 22, 2015 at 23:04 | comment | added | Joel David Hamkins | You can have $X$ itself definable, if you let $X$ consist of the minimal rank sets not in $\text{OD}$. If there are any non-OD sets at all, then this is an uncountable definable set with only countably many (actually zero) ordinal definable elements. | |
| Mar 22, 2015 at 22:00 | comment | added | Joel David Hamkins | Nice observation. For your question, if $X\subset V-\text{OD}$, then $\text{OrDef}(X)=\emptyset$, and if $V\neq\text{HOD}$, there are uncountable such $X$. Perhaps you want $X$ itself to be OD, or to allow $X$ as a parameter in the definitions? | |
| Mar 22, 2015 at 21:29 | history | answered | Vladimir Kanovei | CC BY-SA 3.0 |