sigma-algebra generated by OD sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:25:40Z http://mathoverflow.net/feeds/question/103223 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103223/sigma-algebra-generated-by-od-sets sigma-algebra generated by OD sets Trevor Wilson 2012-07-26T17:44:59Z 2012-07-26T19:49:42Z <p>Assume $V=L(\mathbb{R})$ and the Axiom of Determinacy. Is every set of reals generated by ordinal-definable sets of reals under the operations of countable union and intersection?</p> <p>The class of sets generated in this way is Wadge-cofinal and not wellorderable (it contains $\lbrace x\rbrace$ for every $x \in \mathbb{R}$) so there don't seem to be obvious limitations on its extent.</p> <p>This question came up when I was trying to answer Asaf Karagila's "bonus question" here: <a href="http://mathoverflow.net/questions/102000/generating-family-for-the-lebesgue-sigma-algebra" rel="nofollow">http://mathoverflow.net/questions/102000/generating-family-for-the-lebesgue-sigma-algebra</a></p> http://mathoverflow.net/questions/103223/sigma-algebra-generated-by-od-sets/103237#103237 Answer by Joel David Hamkins for sigma-algebra generated by OD sets Joel David Hamkins 2012-07-26T19:49:42Z 2012-07-26T19:49:42Z <p>This does not answer the question, but for every set $A\subset\mathbb{R}$ in $L(\mathbb{R})$, you can get $A\times\{r\}$ in your algebra, for some real $r$, and this does not use AD.</p> <p>The reason is that if $V=L(\mathbb{R})$, then every set is definable from an ordinal and a real. So if $A\subset\mathbb{R}$, then $A=\{ x\in\mathbb{R}\mid \varphi(x,\alpha,r)\}$ for some definition $\varphi$ and parameters $\alpha\in\text{Ord}$, $r\in\mathbb{R}$. For any finite binary sequence $t$, let $B_t=\{(x,s)\mid t\subset s,\varphi(x,\alpha,s)\}$, which is an ordinal definable subset of the plane. Finally, observe that $A\times\{r\}=\{(x,r)\mid \varphi(x,\alpha,r)\}$ is precisely $\bigcap_{t\subset r} B_t$, a countable intersection of OD sets.</p>