Can you define a countable set containing undefinable elements?

Question

Is it possible to construct a countable set that contains an undefinable set as an element?

Consider a set $X$ to be definable if there exists a parameter-free formula $\varphi$ in the language of ZFC $\{ \in \}$ such that $X$ as a class is equal to $\{ x : \varphi(x) \}$. Note that we are working with the class of all sets, not a model of ZFC.

Note that there must not exist a definable bijection with the natural numbers, or any element can be defined using the bijection.

Answer 1

Definability is a very sensitive subject, first of all, note that Tarski's undefinability of the truth tells us that the question "is $X$ undefinable" is not actually a well formed question (but note that saying "$X$ is definable by $φ$" is well formed).

When working with definable objects you need to restrict yourself to set-models to be able to talk about this kind of questions, a variation of your question can be:

Assuming $Con(ZFC)$, Is it possible that there exists a model $(N,ε)⊨ZFC$ such there exists an (internally) countable set $a\in N$ such that for every $φ$ such that $(N,ε)⊨φ(a)$ implies that there exists $a\ne b\in N$ such that $(N, ε)⊨φ(b)$?

Or the stronger

Assuming $Con(ZFC)$, Is it provable that there exists a model $(N,ε)⊨ZFC$ such there exists an (internally) countable set $a\in N$ such that for every $φ$ such that $(N,ε)⊨φ(a)$ implies that there exists $a\ne b\in N$ such that $(N, ε)⊨φ(b)$?

Turns out the answer is "yes" but it is boring, take a model $(N,ε)$ with uncountably many countable sets (note here I mean that externally there are uncountable many sets that are internally countable), we know that there exists only countably many formulaes without parameters, so at least one countable set is undefinable (note that this argument fails in $V$ not because there are somehow more formulaes, but because the question is simply not first order expressable).

A less boring variation will be the dual, is it possible for this property to fail?

Assuming $Con(ZFC)$, Is it possible that there exists a model $(M,ε)⊨ZFC$ such that for every $a\in M$ there exists a formula $φ$ without parameters such that $(M,ε)⊨φ(x)$ if and only if $x=a$?

(note that I removed the condition that $a$ is countable, this is a bit stronger statement than the actual dual)

Or the stronger dual variation, is it provable that this property fails for some model?

Assuming $Con(ZFC)$, Is it provable that there exists a model $(M,ε)⊨ZFC$ such that for every $a\in M$ there exists a formula $φ$ without parameters such that $(M,ε)⊨φ(x)$ if and only if $x=a$?

Turns out that the duals variations are also true, Myhill has proven that if ZFC is consistence than there are $2^{\aleph_0}$ "pointwise-definable" models of ZFC (pointwise definable is the property I stated in the dual variations).

---

Proof of the strong dual form:

Let $\mathscr M=(M,ε)$ be any model of ZFC, we may assume $\mathscr M$ is also a model of $V=HOD$. Models of $ZFC+V=HOD$ are exactly those who have definable (without parameters) well ordering $φ(x,y)$, for each formula $ψ(x;y)$, one has a definable Skolem function $f_ψ(y)=\min_φ(ψ(x,y))$, but then the set $N\subseteq M$ consistent of the definable sets of $\mathscr M$ is it's own Skolem Hull of the set of definable Skolem functions, hence $\mathscr N=(N,ε)\prec \mathscr M$.

So $\mathscr N\models ZFC+V=HOD$ and agree with $\mathscr M$ about what is definable, hence every set in $N$ is definable without parameters by $\mathscr N$. $\square$