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?
Are there models of set theory where every real number is definable but not every set is definable.
forcinglogicset-theory
Related Solutions
As Eric Wofsey said, actually building an automorphism of $\mathbb{C}$ which moves a real is difficult: while we can prove that for every transcendental real $r$ there is a field automorphism of $\mathbb{C}$ which moves $r$, this proof is $(i)$ a transfinite recursion argument and $(ii)$ relies on the axiom of choice.
There is, however, an automorphism argument which does work:
We'll show a stronger result - that for any transcendental real number $r$ and any formula $\varphi(x)$ such that $\mathbb{C}\models\varphi(r)$, there is a non-real complex number $s$ such that $\mathbb{C}\models\varphi(s)$. The argument uses automorphisms and elementary submodels, and goes as follows (outlined only - it's a good exercise to fill it all in):
Fix a transcendental real $r$, a transcendental non-real complex number $s$, and a formula $\varphi(x)$ such that $\mathbb{C}\models\varphi(r)$. Let $M$ be a countable elementary submodel of $\mathbb{C}$ which contains $r$ and $s$ - we know such an $M$ exists by the downward Lowenheim-Skolem theorem.
- Note that we've replacing the not-obviously-well-orderable field $\mathbb{C}$ by a countable, and hence well-orderable, field $M$. Intuitively, this should (and does) make the axiom of choice irrelevant here.
Now show that $M$ is a countable algebraically closed field of characteristic zero and infinite transcendence degree.
By a standard back-and-forth argument, there is an automorphism of $M$ swapping $r$ and $s$, so $M\models\varphi(s)$. Back-and-forth arguments are still recursive constructions, but they don't use transfinite recursion; also, this is probably a result you've already proved.
Remembering how $M$ and $\mathbb{C}$ are related, conclude that $\mathbb{C}\models\varphi(s)$ as desired.
Remark $1$. We can avoid elementary submodels entirely by working with Ehrenfeucht-Fraisse games instead; however, elementary submodel juggling is an important technique to learn.
Remark $2$. An even stronger fact is true: $\mathbb{C}$ is strongly minimal, meaning that for any model $N$ of $Th(\mathbb{C})$, every definable-with-parameters subset of $N$ is either finite or cofinite. (First, prove that $\mathbb{C}$ has quantifier elimination, and then show that every quantifier-free formula with parameters defines either a finite or a cofinite set; then lift this to every model of $Th(\mathbb{C})$.)
A Cohen real is a real which is generic for Cohen forcing $\mathbb{C}=(Fin(\omega,2),\supseteq,0)$. Strictly speaking of course this depends on the choice of ground model: if $M$ is a countable transitive model of $\mathsf{ZFC}$ then there will exist (in $V$) a Cohen-over-$M$ real $r$, but $r$ is obviously not Cohen over $M[r]$.
In the usual argument for the failure of $\mathsf{CH}$, for example, the "rows" of the generic function $\omega\times\omega_2^M\rightarrow 2$ are indeed Cohen reals (over the original ground model $M$), and the forcing used has "added $\omega_2^M$-many Cohen reals."
It's worth noting that adding a single Cohen real does not actually add a single Cohen real. Less elliptically, if $r$ is Cohen-over-$M$ then $M[r]$ contains lots of reals that are also Cohen-over-$M$. For example, the real $s(x)=r(2x)$ (basically, "half" of $r$ itself) is a Cohen real and generates a strictly smaller extension $M\subset M[s]\subset M[r]$. Similarly, it is generally a difficult problem to determine whether a forcing notion adds a Cohen real (precisely: given $M$ and $\mathbb{P}$, determine whether for every $\mathbb{P}$-generic-over-$M$ filter $G$ there is a Cohen-over-$M$ real $r\in M[G]$).
Best Answer
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 $q\Vdash\varphi(\dot x_\beta)$.