# Are there models of set theory where every real number is definable but not every set is definable.

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?

• 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. Jul 27 at 15:54

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 $$q\Vdash\varphi(\dot x_\beta)$$.