Can anyone provide an example of a set S which is definable in ZFC and provable in ZFC to be
denumerably infinite, while at the same time, no set definable in ZFC can be proved in ZFC to
be an element of S? Such examples are easy to find if S is allowed to be uncountable-for instance
S could be the set of all non-measurable sets of real numbers. In most of the examples where S
is uncountable the axiom of choice is needed just to prove that S is non-empty and the cardinal
number of S is greater than the cardinal number of R.



A question about definable non-empty sets containing no definable elements.
Here are a few partial answers.


First, I claim that if ZFC is consistent, then for every
ZFC-definable nonempty set $S$, whether countable or uncountable, it is consistent with ZFC that $S$
contains a definable element. Further, there will be a model of ZFC in which every element of $S$ is definable. 


Suppose that we have defined a nonempty set $S$ in ZFC. What I
mean by this is that we have provided the defining formula $\phi$
of $S$ and proved in ZFC that there is a unique object, which we
call $S$, that satisfies $\phi$, and furthermore that ZFC proves
that this set is nonempty.


Now, it is known that if ZFC is consistent, then there are
pointwise definable models of ZFC, models in which every set is
definable without parameters. The main result of my paper
[Pointwise definable models of set theory](http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/), joint with Jonas Reitz and David Linetsky, is that every countable
model of ZFC and indeed of GBC has an extension obtained via class
forcing that is pointwise definable, in which every set and indeed every class is definable without parameters. My point now is that if $M$ is a pointwise
definable model of ZFC, then every member of $S$ as interpreted in
$M$ will be definable, by some definition.


Although this answer seems relevant to me, it doesn't quite answer
the exact question you asked, since we don't expect that ZFC will
prove that that set is a member of $S$.


Second, I claim that there is an example if you drop the denumerability requirement. That is, there is a definable set $S$, which is provably nonempty,
but which does not provably contain any specific definable object.


(This argument therefore fills in for your proposal with
non-measurable sets, which I don't really see how to make work.)


Define $S$ to be the sets of minimal rank not in $\text{HOD}$, in the case
that $V\neq\text{HOD}$, and otherwise $S=\mathbb{N}$. So ZFC
proves that $S$ is nonempty, but there can be no definable object
$a$ such that ZFC proves $a\in S$, since in this case, in a model
of $V\neq\text{HOD}$, the object $a$ would be a definable set that
is a subset of $\text{HOD}$, by the minimality part of the definition of
$S$, and hence $a$ would be not only definable but hereditarily
ordinal-definable and hence an element of $\text{HOD}$, contrary to the
definition of $S$.


**Further update.** Extending the observation of Emil in the
comments, let's prove that there is a positive solution, if you
change the countability requirement, size $\aleph\_0$, to require
instead $\aleph\_1$.


**Theorem.** There is a ZFC definable set $S$, which ZFC proves
has size exactly $\aleph\_1$, such that if ZFC is consistent, then
for no ZFC-provably definable object $t$ does ZFC prove $t\in S$.


Proof. Let $S$ be the set of reals not in HOD, provided that CH
holds and there are reals not in HOD, and otherwise let $S$ be
$\aleph\_1$ itself. ZFC proves that this set has size $\aleph\_1$, since
either $S$ is explicitly $\aleph\_1$, or else there are reals not
in HOD and CH holds. But if there are reals not in HOD, then it is
not difficult to see that there must be continuum many reals not
in HOD, and so if CH also holds, this has size $\aleph\_1$.


Meanwhile, if ZFC is consistent, then there is a model in which CH
holds and there are reals not in HOD. In this case, every element
of $S$ is not in HOD, and so ZFC cannot prove that any particular
definable object $t$ is in $S$. QED


