Measures that are not OD - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:44:00Z http://mathoverflow.net/feeds/question/102376 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102376/measures-that-are-not-od Measures that are not OD Trevor Wilson 2012-07-16T17:55:56Z 2012-07-16T18:43:47Z <p>Is anything known about the consistency strength of the statement:</p> <p>"There is a normal measure (on a cardinal) that is not ordinal-definable"?</p> <p>In particular, is it consistent relative to the existence of a measurable cardinal? It looks like it's consistent relative to the existence of a supercompact cardinal. If $\kappa$ is supercompact then we can force to make it Laver indestructible.</p> <p>So assume that $\kappa$ is still $(\kappa+2)$-strong after we add $(2^{2^\kappa})^+$ many Cohen subsets of $\kappa^+$, more than the number of measures on $\kappa$ in $V$. Solovay proved that if $\kappa$ is $(\kappa+2)$-strong then for every set $X \in V_{\kappa+2}$ there is a normal measure on $\kappa$ whose ultrapower contains $X$. So letting $X$ range over the Cohen subsets of $\kappa^+$ that we added, a counting argument shows that we must get some normal measures on $\kappa$ that are not in $V$. Cohen forcing is homogeneous, so these measures cannot be ordinal-definable. I don't know how strong this kind of indestructibility is, or whether it's necessary.</p> <p>I am also interested to know anything about countably complete measures on any set that are not ordinal-definable from that set.</p> http://mathoverflow.net/questions/102376/measures-that-are-not-od/102377#102377 Answer by Andreas Blass for Measures that are not OD Andreas Blass 2012-07-16T18:12:03Z 2012-07-16T18:12:03Z <p>It was proved, long ago, that one can force over a model with a measurable cardinal $\kappa$ and get a model with lots (<code>$2^{2^\kappa}$</code>, I think) of normal measures. I believe that (1) most of those measures won't be OD in that model and (2) the relevant paper is </p> <p>Kenneth Kunen and Jeff Paris, Boolean extensions and measurable cardinals, Annals Math. Log., Vol. 2 (1971), pp. 359-377.</p> <p>Unfortunately, I'm traveling and would find it difficult to check these things right now.</p> http://mathoverflow.net/questions/102376/measures-that-are-not-od/102378#102378 Answer by Joel David Hamkins for Measures that are not OD Joel David Hamkins 2012-07-16T18:20:24Z 2012-07-16T18:43:47Z <p>This is equiconsistent with a measurable cardinal. </p> <p>Start with a measurable cardinal $\kappa$ in $V$, and assume without loss of generality that $2^\kappa=\kappa^+$. Indeed, we might as well assume $V=L[\mu]$ is the canonical inner model of one measurable cardinal. Next, perform the Easton support iteration of forcing that adds a Cohen subset to every inaccessible $\gamma\leq\kappa$, and let $V[G*g]$ be the corresponding forcing extension, where $G$ is the forcing up to $\kappa$ and $g$ is the stage $\kappa$ forcing. The standard lifting arguments show that $\kappa$ remains measurable in $V[G*g]$. </p> <p>Specifically, fix any ultrapower $j:V\to M$ by a normal measure on $\kappa$ in $V$. The forcing $j(P)$ is isomorphic to $P*P_{\rm tail}$, and one may find in $V[G*g]$ an $M$-generic filter $j(G*g)\subset j(P)$ satisfying the lifting criterion, and thereby lift the embedding to $j:V[G][g]\to M[j(G)][j(g)]$. The filter $g$ is used at stage $\kappa$ in $j(G)$. There are in fact numerous lifts of $j$ to the forcing extension, and since these are all still ultrapowers by normal measures in $V[G*g]$, this is a model where $\kappa$ carries $2^{2^\kappa}$ many normal measures.</p> <p>Each of these measures is determined by and determines the filter $j(G*g)\subset j(P)$ that was used in the construction. Since the forcing is almost homogeneous, it follows that the $\text{HOD}^{V[G*g]}\subset \text{HOD}^V$, and moreover even if we add $G$ as a parameter, we have $\text{HOD(G)}^{V[G][g]}\subset\text{HOD(G)}^{V[G]}$, since the stage $\kappa$ forcing is almost homogeneous. Thus, in particular, if one of the measures in $V[G][g]$ is ordinal definable, then so would be the corresponding $j(G)$, and so we would have $j(G)\in V[G]$. But $g$ appears explicitly at stage $\kappa$ of $j(G)$, and so this is impossible. </p> <p>This argument therefore shows that $V[G][g]$ is a model where $\kappa$ is measurable, carries $2^{2^\kappa}$ many normal measures, and none of these measures is ordinal definable there.</p>