My questions concern the following quote from “The HOD Dichotomy”, page 8.

"… notice that $\ cof(\omega)\cap\lambda$ belongs to $HOD$ even though it might mean something else there. Also, $\{S\subseteq\lambda\mid S\in HOD \text{ and S is stationary}\}$ belongs to $HOD$ even though there might be sets which are stationary in $HOD$ but not actually stationary. In any case, $HOD$ can recognise when a given $S\in HOD$ is stationary in $V$...”

My three (probably naive) questions are:

  1. Given an ordinal $\lambda$, what could $cof(\omega)\cap\lambda$ mean in $HOD$ that it doesn’t mean in $V$?

  2. Perhaps a set $C$ could be club in a given ordinal $\kappa$ in $HOD$, and hence a set $S\in HOD$ could be stationary there, though $S$ is not stationary in $V$. Could there be a set $C$ club in some $\kappa$ which is not actually club in $\kappa$?

  3. How can $HOD$ recognize when a particular set $S$ is stationary in $V$ if it isn't stationary in $HOD$?

Is there some kind of absoluteness between $HOD$ and $V$ involving club and stationary sets that simply "falls out of" the notion ordinal or hereditarily ordinal definable that I have missed?


The broader point here is that $\text{HOD}$ has all the sets of ordinals that are definable in $V$, and in this way it is able to know some things about what is going on in $V$, even if it cannot see the full reasons for those facts. For example, $\text{HOD}$ has the function giving the cofinality in $V$ of any ordinal, simply because this function (up to any given $\lambda$) is ordinal definable in $V$, even when cofinality is not absolute between $\text{HOD}$ and $V$; and similarly, $\text{HOD}$ has the set of its subsets of $\kappa$ that are stationary in $V$, even when stationarity is not absolute between $\text{HOD}$ and $V$.

In (1), for example, the point is that $\text{cof}(\omega)\cap\lambda$ is the set of ordinals below $\lambda$ that have cofinality $\omega$ in $V$, and this is a definable set which therefore must be in $\text{HOD}$. But there is no reason to think that these ordinals all have cofinality $\omega$ in $\text{HOD}$, since having cofinality $\omega$ is not necessarily absolute between $V$ and $\text{HOD}$. Indeed, one can easily collapse cardinals to $\omega$, which will make many new ordinals of cofinality $\omega$, but since this forcing is weakly homogeneous, the $\text{HOD}$ of the forcing extension will not think they have cofinality $\omega$. So the set $\text{cof}(\omega)\cap\lambda$ is in $\text{HOD}$, but it isn't necessarily the same as $\text{cof}(\omega)^{\text{HOD}}\cap\lambda$, that is, as the set of ordinals below $\lambda$ that have cofinality $\omega$ in $\text{HOD}$. But the former set will include the latter set, because if an ordinal has cofinality $\omega$ in $\text{HOD}$, then it really has cofinality $\omega$.

For (2), being club is absolute between $\text{HOD}$ and $V$, since for $C\subset\kappa$ to be club means that it is closed and unbounded in $\kappa$, and if you think about it, you will realize that $V$ and $\text{HOD}$ cannot disagree about whether a set is closed or whether it is unbounded in $\kappa$. It follows from this that being stationary is downward absolute, since if a set $S$ really meets every club, then it will also meet every club in $\text{HOD}$, since those clubs really are clubs. But being stationary is not necessarily upward absolute, $\text{HOD}$ may have fewer club sets, and in this way it can be easier for a set to be stationary in $\text{HOD}$ than in $V$. This can definitely happen, if you force over $L$, say to destroy the stationarity of the $L$-least stationary co-stationary subset $S\subset\omega_1$. This forcing is homogeneous, and so the $\text{HOD}$ of the extension $L[G]$ is just $L$ again, where $S$ is stationary, even though it isn't stationary in $L[G]$.

For (3), what Woodin means is just that the set $\{S\subset\lambda\mid S\in\text{HOD}\text{ and }S\text{ is stationary in }V\}$ is a set in $\text{HOD}$, and so $\text{HOD}$ has access to this set. So $\text{HOD}$ can know which of its stationary sets are actually stationary in $V$, since the set of those sets is definable in $V$ and hence an element of $\text{HOD}$, and $\text{HOD}$ can tell which are the elements of that particular set.

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  • $\begingroup$ @JDH. Thank you for your answer. One interesting fact that I glean from it is that stationary sets are definable, whereas there are club sets that are not not (at least from ordinal parameters). Is this correct? Since you mention the forcing which kills a stationary/co-stationary set, is this the only way to conclude that there are club sets which are not definable, or is there an easier argument to see this? $\endgroup$ – Everett Piper Jun 10 '14 at 2:51
  • $\begingroup$ No, not every stationary set is definable. But the collection of sets in HOD that are stationary in V is definable, and this is what he is using in connection with (3). $\endgroup$ – Joel David Hamkins Jun 10 '14 at 2:52
  • $\begingroup$ Meanwhile, yes, the only way for stationarity to be non-absolute is for the smaller model to have a set that it thinks is stationary, when the larger model has a club disjoint from it. So the larger model has a club that is not in the smaller model, which is disjoint from the stationary set. The club-shooting forcing is a way of making this kind of situation happen. $\endgroup$ – Joel David Hamkins Jun 10 '14 at 2:54
  • $\begingroup$ Right. That was sloppy of me. I'll need to think about your great answer some more. In the meantime, I guess I'm trying to articulate a question about the definability (not necessarily using only ordinal parameters) of clubs and stationary sets. $\endgroup$ – Everett Piper Jun 10 '14 at 3:02
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    $\begingroup$ A typo: the thing you say is a set in your last paragraph is a proper class, because you didn't say "$S \subseteq \lambda$". By the way, this makes me wonder: can $\text{HOD}$ can define the class of $V$-stationary sets? At a glance this seems to me like it might be stronger than the fact that this class is amenable to $\text{HOD}$, and I'm not sure if you (and Woodin) meant to make this stronger claim. $\endgroup$ – Trevor Wilson Jun 10 '14 at 3:58

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