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I know that Joel David Hamkins has constructed a model of set theory where every set and hence every real number is pointwise-definable. But, is there a model of set theory where every real number is definable, but not every set is definable?

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    $\begingroup$ Not sure why this got downvoted... perfectly reasonable question. Note although JDH co-wrote a very nice paper on this topic, this specific result doesn’t originate with that paper but goes back much further. I think to Sheperdson in the 1950s, but not 100% sure. $\endgroup$ Jul 27 at 15:54
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Start with a pointwise definable model satisfying $V=L$, e.g. $L_\alpha$ where $\alpha$ is the least ordinal for which $L_\alpha\models\sf ZFC$.

Over this model force with $\operatorname{Add}(\omega_1,\omega_1)$. Namely, add $\omega_1$ subsets to $\omega_1$ using countable conditions. Let $x_\alpha$ denote the $\alpha$th subset added.

Since we did not add new reals, and since the ground model was $L_\alpha$, all the real numbers are definable still (if not by their original definition, then by relativising it to $L$). However, none of the $x_\alpha$ is definable (without parameters). To see why, simply note that if $p\Vdash\varphi(\dot x_\alpha)$, then there is some $q\leq p$ and $\beta\neq\alpha$ such that $q\Vdash\varphi(\dot x_\beta)$.

Why? Simply take $\beta$ which is not in the support of $p$, let $q$ be the extension of $p$ on which $q(\beta,\xi)=p(\alpha,\xi)$. Then the automorphism of the forcing, $\pi$, given by switching the $\alpha$ and $\beta$ coordinates satisfies that $\pi q=q$, so $\pi q\Vdash\varphi(\pi\dot x_\alpha)$ rewrites itself as $q\Vdash\varphi(\dot x_\beta)$.

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