I found this online (from a Joel David Hamkins's post):

If ZFC is consistent, then there is a model of ZFC in which every set-theoretic object is definable (in the sense of $\emptyset$-definable). This is true in the minimal transitive model of set theory, by observing that the collection of definable objects in that model is closed under the definable Skolem functions of L, and hence by Condensation collapses back to the same model, showing that in fact every object there was definable.

The post was about sets that are definable from ZFC, in the sense of unique solutions of some formulae of the set theory language with a single variable (without parameters). I think that the right way to interpret this is the following: $$\forall M\models ZFC+V=L,\forall x\in M\exists\varphi\big(M\models\varphi(x)\wedge\exists!v_0\varphi(v_0)\big).$$ If it is correct, I can't understand the argument. Can anyone explicit the steps for me?

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    $\begingroup$ Which part of the argument is unclear? $\endgroup$ – Asaf Karagila May 2 at 18:13
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    $\begingroup$ The way you interpreted it is wrong. Here "minimal transitive model" refers to the one of minimal height and not just an arbitrary model of $V=L$. $\endgroup$ – Jonathan May 2 at 20:21
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    $\begingroup$ Jonathan is right. If you start with an arbitrary model $M$ of ZFC + V=L, the argument only shows that the definable elements of $M$ constitute a model of ZFC + V=L in which every element is definable. Minimality is needed to ensure that this submodel is isomorphic to $M$ itself and therefore $M$ has all its elements definable. $\endgroup$ – Andreas Blass May 3 at 3:26
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    $\begingroup$ @Stan Whenever $L \models \varphi(x,\bar a)$ for some $x$ where $\bar a$ was definable, then the $<_L$-least such $x$ is definable. So we can prove elementarity using the Tarski-Vaught criterion. $\endgroup$ – Jonathan May 3 at 16:13
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    $\begingroup$ @Stan I don't understand the first thing you wrote: $\varphi$ was an arbitrary formula and for each of them Tarski-Vaught holds true; that's all we need. To the second part: there are a lot of Skolem-functions and most of them are not definable. The point is that looking at the definable ones suffices. $\endgroup$ – Jonathan May 4 at 11:51

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