{c}

Let $\theta=M\cap\operatorname{Ord}$ be the <ordinal height of a model of set theory>. First $\theta\geq\omega_1$. Indeed, if $\theta$ were countable, the internal <axiom of choice> would give, for every $x\in M$, a bijection in $M$ between $x$ and some ordinal below $\theta$. Such a bijection is also valid externally, making $x$ externally countable. In particular every internal rank $V_\xi^M$, $\xi<\theta$, would be countable. Every element of $M$ lies in one of these internal ranks, so
$$
M=\bigcup_{\xi<\theta}V_\xi^M
$$
would be a <countable union of countable sets>, a contradiction. This is why an <uncountable transitive set model has uncountable ordinal height>.

For $\xi<\theta$, <absoluteness of constructible levels> gives $L_\xi^M=L_\xi$, and that level is an element of $M$. The reason is that satisfaction in a fixed set structure uses the same domain and finite formulas in both universes, so successor definitions agree; <transfinite recursion> then also makes limit stages agree. Transitivity consequently gives
$$
L_{\omega_1}\subseteq L_\theta\subseteq M.
$$
It remains to locate a countability witness. Given an ambient <countable ordinal> $\alpha$, choose an injection $f:\alpha\to\omega$. Since $V=L$, it belongs to the <constructible universe>. Its <transitive closure>, together with $f$ itself, is countable. Choose a countable <elementary substructure> $X\prec L_\eta$ of a sufficiently large limit level containing this closure pointwise. The <condensation lemma for the constructible universe> identifies the collapse with $L_\beta$ for some countable $\beta$. The collapse fixes $f$, since all its hereditary members were included. Thus $f\in L_\beta\subseteq L_{\omega_1}\subseteq M$.

The ordinal $\alpha$ itself is in $M$, since $\alpha<\omega_1\leq\theta$. Being an injection between these fixed sets is a <bounded formula in set theory>, so $M$ recognizes $f$ as a countability witness. This proves <countable-ordinal correctness under constructibility>:
$$
\boxed{\alpha\in M\quad\text{and}\quad M\models\text{“}\alpha\text{ is countable”}\qquad(\alpha<\omega_1).}
$$
The use of condensation supplies a witness below the height of $M$; merely knowing that $f$ belongs somewhere to $L$ would not be sufficient.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 121 / 2 / iii / Solution

Assuming $V=L$, every uncountable <transitive model> of <ZFC> contains all ambient <countable ordinals> and witnesses their countability internally. Its ordinal height is at least $\omega_1$, and <absoluteness of constructible levels> puts $L_{\omega_1}$ inside it. Every countability witness for a countable ordinal can be chosen in $L_{\omega_1}$ by <hereditarily countable constructible sets appear below omega-one>.

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 Incoming links: Countable-ordinal correctness under constructibility