Measures that are not OD - MathOverflow

Measures that are not OD

2012-07-16T17:55:56Z

Is anything known about the consistency strength of the statement:

"There is a normal measure (on a cardinal) that is not ordinal-definable"?

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.

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.

I am also interested to know anything about countably complete measures on any set that are not ordinal-definable from that set.

Answer by Andreas Blass for Measures that are not OD

2012-07-16T18:12:03Z

It was proved, long ago, that one can force over a model with a measurable cardinal $\kappa$ and get a model with lots ( $2^{2^\kappa}$ , 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

Kenneth Kunen and Jeff Paris, Boolean extensions and measurable cardinals, Annals Math. Log., Vol. 2 (1971), pp. 359-377.

Unfortunately, I'm traveling and would find it difficult to check these things right now.

Answer by Joel David Hamkins for Measures that are not OD

2012-07-16T18:20:24Z

This is equiconsistent with a measurable cardinal.

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 $V[G*g]$.

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.

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.

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.