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.

  • 2
    $\begingroup$ Trevor, wouldn't the Kunen-Paris construction (Boolean extensions, and measurable cardinals, AML 2(4) (1971), 359-377) already accomplish this, for similar reasons? $\endgroup$ – Andrés E. Caicedo Jul 16 '12 at 18:12
  • $\begingroup$ Andres, with merely a measurable cardinal, one cannot necessarily add so many subsets to $\kappa$ and preserve measurability. So that particular counting argument won't work here, but a modified argument does seem to work, and the argument I give in my answer shows how to make all the measures non-OD. $\endgroup$ – Joel David Hamkins Jul 16 '12 at 19:20

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.

  • $\begingroup$ This is theorem 21.3 in Jech's Set Theory book. How can a new measure be OD? if it were wouldn't it already be in the ground model (by sigma_1_1 absoluteness considerations?) $\endgroup$ – Eran Jul 17 '12 at 8:20
  • $\begingroup$ Eran, a new measure can be made OD by further forcing---for example, code it into the continuum function. Every model of set theory has a forcing extension satisfying V=HOD, so in principle, any set can be made ordinal-definable. The Shoenfield $\Sigma^1_2$ absoluteness theorem refers to absoluteness for assertions quantifying over the reals, whereas the OD concept allows definitions quantifying over the entire universe. $\endgroup$ – Joel David Hamkins Jul 18 '12 at 1:52
  • $\begingroup$ Thanks Joel. The reference for this forcing is McAloon? also, on a related note, Kunen's Set Theory exercise VII E5 explains why a simple cohen real is not ordinal definable. $\endgroup$ – Eran Jul 18 '12 at 18:22

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.

  • 2
    $\begingroup$ This forcing is a Silver-style iteration, but one can also use as Andreas describes the Kunen-Paris forcing, which is an Easton product forcing, using $\gamma$ in a fixed unbounded measure zero set. The OD part of the argument would be basically the same for the two choices. $\endgroup$ – Joel David Hamkins Jul 16 '12 at 18:26
  • $\begingroup$ Thanks for you answer, Joel. I accepted Andreas's because it looks like it came first, but what you wrote is very helpful. I did not expect such quick responses! $\endgroup$ – Trevor Wilson Jul 16 '12 at 18:29
  • $\begingroup$ Since the Kunen-Paris forcing has no stage $\kappa$ forcing, I've realized that the arguments are actually different in the two cases. Indeed, I'm not seeing just now how to get the stronger result that no measure is OD using only the Kunen-Paris forcing, but I expect that this should be possible. $\endgroup$ – Joel David Hamkins Jul 16 '12 at 19:22

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