TITLE: Is there any countable ordinal number which has a member undefinable?
QUESTION [7 upvotes]: Let $Q\colon=\{\alpha \in Ord \mid \exists \beta \in \alpha(\beta \text{ is undefinable in } \langle\alpha,<\rangle )\}$ denote the set of all ordinal numbers which has an undefinable member.
Since there are only countable many formulae, the number of definable members in $\langle\alpha,<\rangle$ must be at most countable infinite. Thus every uncountable ordinal numbers must belong to $Q$. On the other hand, for each natural number, if it's in $\alpha$, then it must be definable, this conclusion is easy to get by induction.
So $q$, the least member of $Q$ is greater than $\omega$ but at most $\omega_1$.
My question: Is $q$ exactly $\omega_1$, or a countable one?

REPLY [2 votes]: Suppose we have limit ordinals $\alpha < \beta$ and $L_\alpha \prec L_\beta$, then by Theorem 2.4 in Undefinable sets, Rudolf v.B. Rucker, Annals of Mathematical Logic Volume 6, Issues 3–4, March 1974, Pages 395–419 which states

lf x is a model of V=L, then $\alpha$ is strongly inconceivable in x
  iff $L_\alpha\prec x$. Here $\alpha$ is strongly inconceivable in x means for all $\beta\geq \alpha$, $\beta$ is not definable in $\langle x,\in \rangle$ with parameters from $\alpha\cap x$.

Then $\alpha$ is not definable in $\langle L_\beta,\in\rangle$ with finitely many parameters from $\alpha$. Therefore, $\alpha$ is not definable in $\langle \beta,\in\rangle$ since we could define in $L_\beta$ "x is an ordinal" by describing x to be transitive and totally ordered. It is known that $\{\alpha: L_\alpha\prec L_{\omega_1}\}$ is a closed unbounded set so we could always pick two countable ordinals $\alpha,\beta$ whose properties are described as above. However, it does not really give you the exact ordinal though.