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Joel David Hamkins
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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\in\text{HOD}\mid S\text{ is stationary in }V\}$$\{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.

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 $\{S\in\text{HOD}\mid S\text{ is stationary in }V\}$ 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.

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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Joel David Hamkins
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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 $\{S\in\text{HOD}\mid S\text{ is stationary in }V\}$ 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.

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.

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 $\{S\in\text{HOD}\mid S\text{ is stationary in }V\}$ 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.

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 $\{S\in\text{HOD}\mid S\text{ is stationary in }V\}$ 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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Joel David Hamkins
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ForThe 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.

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 $V$, butand this is not necessarily thea definable set of ordinals belowwhich therefore must be in $\lambda$$\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 $\text{HOD}$. So this is a definable set ofIndeed, one can easily collapse cardinals to $\omega$, which will make many new ordinals of cofinality $\omega$, andbut since this forcing is what puts it intoweakly homogeneous, the $\text{HOD}$, but there of the forcing extension will not think they have cofinality $\omega$. So the set $\text{cof}(\omega)\cap\lambda$ is in general no reason to think that$\text{HOD}$, but it isisn'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 $\text{HOD}$. It is just aBut the former set of ordinalswill include the latter set, and it includes everybecause if an ordinal below $\lambda$ in $\text{HOD}$ that has cofinality $\omega$ in $\text{HOD}$, butthen it may also include other ordinals, which havereally has cofinality $\omega$ in $V$ but not in $\text{HOD}$.

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 $\kappa$. Meanwhile,It follows from this that being stationary is not necessarilydownward absolute, since if a set is stationary when$S$ really meets every club, then it has nonempty intersection withwill also meet every club in $\text{HOD}$, butsince those clubs really are clubs. But being stationary is not necessarily upward absolute, $\text{HOD}$ may simply have fewer club sets, and soin this way it iscan 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 $L[G]$. Meanwhile, being stationary is downward absolute, and so any set that is stationary in $V$ will also be stationary in $\text{HOD}$.

For (3), what Woodin means is just that the set $\{S\in\text{HOD}\mid S\text{ is stationary in }V\}$ 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.

For (1), the point is that $\text{cof}(\omega)\cap\lambda$ is the set of ordinals below $\lambda$ that have cofinality $\omega$ in $V$, but this is not necessarily the set of ordinals below $\lambda$ that have cofinality $\omega$ in $\text{HOD}$, since having cofinality $\omega$ is not necessarily absolute between $V$ and $\text{HOD}$. So this is a definable set of ordinals, and this is what puts it into $\text{HOD}$, but there is in general no reason to think that it is the same as $\text{cof}(\omega)^{\text{HOD}}\cap\lambda$, that is, the set of ordinals below $\lambda$ that have cofinality $\omega$ in $\text{HOD}$. It is just a set of ordinals, and it includes every ordinal below $\lambda$ in $\text{HOD}$ that has cofinality $\omega$ in $\text{HOD}$, but it may also include other ordinals, which have cofinality $\omega$ in $V$ but not in $\text{HOD}$.

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 $\kappa$. Meanwhile, being stationary is not necessarily absolute, since a set is stationary when it has nonempty intersection with every club, but $\text{HOD}$ may simply have fewer club sets, and so it is 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 $L[G]$. Meanwhile, being stationary is downward absolute, and so any set that is stationary in $V$ will also be stationary in $\text{HOD}$.

For (3), what Woodin means is just that the set $\{S\in\text{HOD}\mid S\text{ is stationary in }V\}$ 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.

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.

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 $\{S\in\text{HOD}\mid S\text{ is stationary in }V\}$ 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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Joel David Hamkins
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Joel David Hamkins
  • 236.5k
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  • 777
  • 1.4k
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