$\newcommand\ZFC{\text{ZFC}}
\newcommand\HOD{\text{HOD}}
\newcommand\P{\mathbb{P}}
\newcommand\Q{\mathbb{Q}}$
It turns out that question 1 has a positive answer, while question
I2 has a negative answer.
The reader may find it useful to know of the characterization of
the $\Sigma_2$ properties as the semi-local properties, those
which are equivalent to an assertion of the form $\exists\theta\
V_\theta\models\psi$, where $\psi$ can have foundany complexity. In
particular, whenever a $\Sigma_2$ property $\varphi(A)$ is true of
a set $A$, it is because there is some $V_\theta$ satisfying
something about $A$. We may therefore preserve that $\Sigma_2$
fact about $A$, while forcing over $V$, provided that we only
force up high and preserve $V_\theta$, the rank-initial-segment of
the universe up to $\theta$.
First, let's consider the positive answer to question 1.
Theorem 1. Every model of $\ZFC$ has a forcing extension
satisfying $V\neq\HOD$, in which every $\Sigma_2$-definable set
has an ordinal-definable element.
Proof idea: perform a forcing iteration, considering each
$\Sigma_2$ formula in turn, where we try to freeze the set defined
by that formula and then code one of its elements (if any) into
the GCH pattern high above the witness to that $\Sigma_2$
property. In the end, every nonempty $\Sigma_2$-definable set will
contain an ordinal-definable element.
Proof. Start inwith $V$ as a ground model. Without loss, by forcing if
necessaryif necessary, we may assume that $V$ satisfies $V=\HOD$, so that
there is a definable well-ordering of the universe. Enumerate the
$\Sigma_2$ formulas $\varphi_0,\varphi_1,\ldots$, and so on. Note that
wethat we may refer to $\Sigma_2$-truth since there is a universal truth
predicatetruth predicate for truth of bounded complexity (so there will be no
issuesno issues with Tarski's theorem on the non-definability of truth). We
defineWe define a full-support forcing iteration $\P$ of length
$\omega$, where the forcing at each stage will become
progressively more highly closed. At the first
stage stage, we consider the
the formula $\varphi_0$. Let $\theta_0$ be any
fixed cardinal, and ask: is there a $\leq\theta_0$-closed forcing extension
extension $V[g_0]$ in which $\varphi_0$ definesholds of a nonempty set
$A_0$? If so
so, we perform such a forcing (choose the least poset forcing
forcing thisthis), and let $\lambda_0$ be the smallest $\beth$-fixed point
point aboveabove the size of that forcing and $\theta_0$ so that also
$\varphi_0$ is witnessed
witnessed in $V_{\lambda_0}^{V[g_0]}$. Next,
perform perform additional
$\leq\lambda_0$-closed forcing over $V[g_0]$ to
an an extension
$V[g_0][h_0]$, where $h_0$ forces to code one of the elements of
$A_0$ into the GCH pattern above $\lambda_0$. This preserves the
definition of $A_0$ by $\varphi_0$, while ensuring that $A_0$ has
an ordinal definable element. Now, let $\theta_1$ be well above this
this coding, and continue.
At stage $n$, we have the partial extension
$V^{(n)}=V[g_0][h_0]\cdots[g_{n-1}][h_{n-1}]$, which performed
forcing below the cardinal $\theta_n$. We ask whether we can perform
perform $\leq\theta_n$-forcing so that $\varphi_n$ definesholds of a nonempty set
nonempty set $A_n$ in the extension. If so, we do that forcing, let
let $\lambda_n$
be be large enough to witness the $\Sigma_2$ property for
for $\varphi_n$, and then perform GCH coding above that so as to make
anmake an element of $A_n$ ordinal-definable, and let $\theta_{n+1}$
larger than all that. (Otherwise, we ignore $\varphi_n$ and let
$\theta_{n+1}=\theta_n$.)
Consider the corresponding extension $V[G]$, where $G\subset\P$ is
$V$-generic, and we use full support in $\P$. Finally, we force to
add add a Cohen subset
$H\subset\delta$, where $\delta$ is a regular
cardinal cardinal above
$\sup_n\theta_n$, since this will force $V\neq\HOD$ in $V[G][H]$. The
The final desired model is $V[G][H]$.
Because Because we used full support
support, it follows that the tail forcing in
$\P$ after stage $n$ is
is $\leq\theta_n$-closed, as is the forcing
to to add $H$, and so preserves
preserves sets of rank below $\lambda_n$. Thus,
if if $\varphi_n$ defines
defines a nonempty set in $V[G][H]$, then itat stage $n$ we would
have done so in someobserved that it was possible to force it to hold of a
nonempty set (with forcing that was sufficiently closed extension of the model at), and so
stagewe would have treated it at stage $n$. That is, and we would have forced to add an ordinal-definable
elementforced to code one of its elements into the GCH pattern,
afterwards always preserving that definition and this coding. So
in the case that $\varphi_n$ does define a nonempty set in
$V[G][H]$, afterwards preservingthen the definition ofstage $n$ forcing exactly ensured that one of
set andthe elements of this set was coded into the GCH pattern of
$V[G][H]$ and was therefore ordinal-definable elementthere. The later
stages of forcing were arranged so as to preserve all these
definitions.
Thus, $V[G][H]$ is a model of $V\neq\HOD$, as the forcing to add
$H$ is weakly homogeneous, such that every $\Sigma_2$-definable
nonempty set has an ordinal-definable element. QEDQED
The argument reminds me of the forcing iteration proof of the
maximality principle, where one forces a given statement to be
true, if it is possible to force it in such a way that it remains
true in all further forcing extensions. The end result is a model
where any statement that could become necessarily true by forcing,
is already true, and this is precisely what the maximality
principle asserts.
Meanwhile, Emil's ideainsightful suggestion in the comments seems to leadleads to a negative answer
tonegative answer to question 2. Perhaps he will post
Theorem 2. If $V\neq\HOD$, then there is a fuller account of itnonempty
$\Pi_2$-definable set (orwith no parameters) containing no
ordinal-definable element.
Proof. Let $A$ be the set of minimal-rank non-OD sets. That is,
$A$ consists of all non-OD sets of rank $\alpha$, where $\alpha$
is minimal such that there is any non-OD set of rank $\alpha$. In
my answer to this related question, I can later if he doesn't carehad proved that $A$ is
characterized by a $\Sigma_2\wedge\Pi_2$ property.
Emil's idea was to) consider not $A$, but a related set, namely the
set $U=A\times V_\theta$, where $\theta$ is the smallest ordinal
such that $A\in V_\theta$ and $V_\theta\models A$ is the set of
minimal-rank non-OD sets.
The set $U$ is defined by the following property: $U$ consists of
the cartesian product $U=A\times B$ of two sets $A$ and $B$ such
that the set $B$ has the form $B=V_\theta$ for some ordinal
$\theta$ such that $A\in V_\theta$ and $V_\theta\models "A$ is the
set of minimal-rank OD sets and there is no $\theta'<\theta$ for
which $V_{\theta'}\models A$ is the set of minimal-rank non-OD
sets"; and finally, the elements of $A$ really are not in OD.
This property altogether has complexity $\Pi_2$, due mainly to the
last clause. The first part, requiring that $U$ has the form
$A\times B$, is $\Delta_0$. The next part, asserting that $B$ has
the form $B=V_\theta$ for some ordinal $\theta$ has complexity
$\Pi_1$, essentially because one need only assert that $B$ is
transitive and satisfies some minimal set theory such that it
thinks it is a $V_\theta$, and such that $B$ contains all subsets
of any of its elements, so that it is using the true power set
operation. The properties asserting that $V_\theta$, that is, $B$,
satisfies certain complication assertions has complexity
$\Delta_0$, since all quantifiers are bounded by $B$ and hence
ultimately by $U$. And finally, asserting that the elements of $A$
are really not ordinal-definable has complexity $\Pi_2$, since
"$x\in\text{OD}$'' has complexity $\Sigma_2$, as any instance of
ordinal-definability reflects to some $V_\theta$ and hence is
locally verifiable; thus, the assertion $\forall x\in A\
x\notin\text{OD}$ has complexity $\Pi_2$.
So altogether, the set $U=A\times V_\theta$ is $\Pi_2$-definable,
but it can have no ordinal-definable elements, since every element
of $U$ has the form $(a,b)$ for some $a\in A, b\in V_\theta$, and
if the pair $(a,b)$ were ordinal-definable, then $a$ would be
ordinal-definable, contradicting $a\in A$ and the fact that every
member of $A$ is not ordinal-definable. QED
Note that the proof is completely uniform, in that the definition
of the set does not depend on the model in any way. Rather, we
have a $\Pi_2$ definition that $\ZFC+V\neq\HOD$ proves is a
nonempty set disjoint from OD.
