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Working in $\sf ZF$

Define: $W_0 = \emptyset \\ W_{\alpha+1} = H_{\leq |W_\alpha|} =\{x \mid \forall y: y \in \operatorname {trcl} (\{x\}) \ |y| \leq |W_\alpha| \} \\ W_\lambda= \bigcup W_{\alpha < \lambda}, \text { for limit ordinal } \lambda$

Where cardinality "| |" is define after Scott.

This cumulative size hierarchy is also indexed by ordinals.

Now if we define ordinal definable sets in terms of those stages istead of the usual $V_\alpha$ stages of the cumulative hierarchy. That is, we use exactly the same definition of ordinal definability but only replace the symbol $V$ by $W$. Designate that as $\operatorname {OD}^*$, then:

Is $\sf HOD=HOD^*$?

Working in $\sf ZF$

Define: $W_0 = \emptyset \\ W_{\alpha+1} = H_{\leq |W_\alpha|} =\{x \mid \forall y: y \in \operatorname {trcl} (\{x\}) \ |y| \leq |W_\alpha| \} \\ W_\lambda= \bigcup W_{\alpha < \lambda}, \text { for limit ordinal } \lambda$

Where cardinality "| |" is defined after Scott.

This cumulative size hierarchy is also indexed by ordinals.

Now if we define ordinal definable sets in terms of those stages istead of the usual $V_\alpha$ stages of the cumulative hierarchy. That is, we use exactly the same definition of ordinal definability but only replace the symbol $V$ by $W$. Designate that as $\operatorname {OD}^*$, then:

Is $\sf HOD=HOD^*$?

Working in $\sf ZF$

Define: $H_0 = \emptyset \\ H_{\alpha+1} = H_{\leq \alpha} =\{x \mid \forall y: y \in \operatorname {trcl} (\{x\}) \ |y| \leq |H_\alpha| \} \\ H_\lambda= \bigcup H_{\alpha < \lambda}, \text { for limit ordinal } \lambda$

Where cardinality "| |" is define after Scott.

This cumulative size hierarchy is also indexed by ordinals.

Now if we define ordinal definable sets in terms of those stages istead of the usual $V_\alpha$ stages of the cumulative hierarchy. That is, we use exactly the same definition of ordinal definability but only replace the symbol $V$ by $H$. Designate that as $\operatorname {OD}^*$, then:

Is $\sf HOD=HOD^*$?

Working in $\sf ZF$

Define: $W_0 = \emptyset \\ W_{\alpha+1} = H_{\leq |W_\alpha|} =\{x \mid \forall y: y \in \operatorname {trcl} (\{x\}) \ |y| \leq |W_\alpha| \} \\ W_\lambda= \bigcup W_{\alpha < \lambda}, \text { for limit ordinal } \lambda$

Where cardinality "| |" is define after Scott.

This cumulative size hierarchy is also indexed by ordinals.

Now if we define ordinal definable sets in terms of those stages istead of the usual $V_\alpha$ stages of the cumulative hierarchy. That is, we use exactly the same definition of ordinal definability but only replace the symbol $V$ by $W$. Designate that as $\operatorname {OD}^*$, then:

Is $\sf HOD=HOD^*$?

Define: $H_0 = \emptyset \\ H_{\alpha+1} = H_{\leq \alpha} =\{x \mid \forall y: y \in \operatorname {trcl} (\{x\}) \ |y| \leq |H_\alpha| \} \\ H_\lambda= \bigcup H_{\alpha < \lambda}, \text { for limit ordinal } \lambda$

This cumulative size hierarchy is also indexed by ordinals.

Now if we define ordinal definable sets in terms of those stages istead of the usual $V_\alpha$ stages of the cumulative hierarchy. That is, we use exactly the same definition of ordinal definability but only replace the symbol $V$ by $H$. Designate that as $\operatorname {OD}^*$, then:

Is $\sf HOD=HOD^*$?

Working in $\sf ZF$

Define: $H_0 = \emptyset \\ H_{\alpha+1} = H_{\leq \alpha} =\{x \mid \forall y: y \in \operatorname {trcl} (\{x\}) \ |y| \leq |H_\alpha| \} \\ H_\lambda= \bigcup H_{\alpha < \lambda}, \text { for limit ordinal } \lambda$

Where cardinality "| |" is define after Scott.

This cumulative size hierarchy is also indexed by ordinals.

Now if we define ordinal definable sets in terms of those stages istead of the usual $V_\alpha$ stages of the cumulative hierarchy. That is, we use exactly the same definition of ordinal definability but only replace the symbol $V$ by $H$. Designate that as $\operatorname {OD}^*$, then:

Is $\sf HOD=HOD^*$?