$M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$.
$M$ has a definable well-ordering of the universe.
Every definable nonempty set in $M$ has a definable element.
Every definable nonempty set in $M$ has an ordinal-definable
element.
Every $(\Sigma_2\wedge\Pi_2)$$\Pi_2$-definable nonempty set in $M$ has an
an ordinalordinal-definable element.
Every ordinal-definable nonempty set in $M$ has an
ordinal-definable element.
($5\to 1$) For theThe stronger implication — thanks to Françoishas now undergone a few
in the comments! — we analyze the complexity of theimprovements, so let me discuss it. I had proposed considering as
definition ofabove the set $A$ in the previous case. The main observationof all minimal-rank non-OD sets, which is
that $x\in\text{OD}$ isdefinable and nonempty in any model of $\Sigma_2$$V\neq\text{HOD}$, as every instance ofbut has
definability is witnessed by reflection inside some $V_\theta$no ordinal-definable elements. I had guessed that $\Sigma_5$ would be
Thesufficient to define $\Sigma_2$ assertions are precisely$A$. In the semi-local
propertiescomments, thoseFrançois refined this, arguing that can bethis set was actually
verified in some rank initial segment of the universe$\Sigma_3$-definable and indeed $V_\theta$$\Delta_3$-definable.
If $V\neq\text{HOD}$ Using his
idea, then the setI was able to push this down to show that $A$ of minimal-rank non-OD setsis
is characterized$\Sigma_2\wedge\Pi_2$ definable, by the following properties: $A$ is not empty;
allempty; all elements of $A$ have the same rank; every element of
$A$ is
not not in OD; every set of rank less than an element of $A$ is in OD;
everyin OD; every set not in $A$, but of the same rank as an element of
$A$,
is is in OD. Each of these properties is either $\Sigma_2$ or
$\Pi_2$, making the set $A$ to be $\Sigma_2\wedge\Pi_2$-definable.
Specifically, the first two requirements are $\Sigma_2$, being
witnessed in a rank-initial segment of the universe; the third is
$\Pi_2$; the fourth and fifth are both $\Sigma_2$, since they are
true just in case there is a large $V_\theta$ which believes them
to be true. Thus, the definition ofI also noted that $A$ has complexity
$\Sigma_2\wedge\Pi_2$is not provably $\Sigma_2$-definable.
Meanwhile, which allowsover at my question Can $V\neq\text{HOD}$ if every
$\Sigma_2$-definable set has an ordinal-definable element?, Emil made a suggestion
leading to the implicationobservation that if ($5\to 1$)$V\neq\text{HOD}$, then there
sinceis a $A$ has$\Pi_2$-definable set with no ordinal-definable memberselements. The
set is simply $U=A\times V_\theta$, where $A$ is as above and $\theta$ is
least such that $V_\theta$ thinks $A$ is the set of minimal-rank
non-OD sets. So I refer the reader to theorem 2 in that answer,
which provides the content of the implication ($5\to 1$).
($1\to 6$) Immediate, since under statement $1$, every set in $M$ is
ordinalis ordinal-definable in $M$.
Conclusion.
Thus Thus, case (1) of the question occurs in exactly the
the models of
$V=\text{HOD}$ that are not pointwise definable. There
There are such
models models, if ZFC is consistent, since one may take any uncountable
modelany uncountable model of $\text{ZFC}+V=\text{HOD}$.
Meanwhile, case (2) of the question — ignoring the issue of
real parameters — does not occur at all, since
if if a set has sets
sets that are not ordinal-definable, then it will
have have a definable set
set with no ordinal-definable members, namely,
the the set of all non
non-OD sets of minimal rank, as in the implication
of of statement 4 to
to statement 1.
Update. I edited to the improved statement 5, which we've now
got down to the case of mere $\Pi_2$-definability, using the
answer to my question Can $V\neq\text{HOD}$ if every
$\Sigma_2$-definable set has an ordinal-definable element?.
