Just learned $\mathbb{L}\vDash \mathbb{V}=\mathbb{L}$ and was warned that this property is not obvious with the counterexample mentioned being $HOD$. I can think of a few examples of definable substructures that behave like $\mathbb{L}$. For example given a ring $R$ the center of that ring $Z(R)$ "knows" it's a center since $Z(R)=Z(Z(R))$. So we would have with much notational abuse $Z(R)\vDash R=Z(R)$. Similarly for a monoid $M$ if $U(M)$ is the set of all invertible elements then one would have that $U(M)\vDash M=U(M)$ since $U(M)=U(U(M))$. To avoid abuse of notation both examples are a case where I am given an $L$-structure $M$ satisfying some theory $T$ and there is an formula $\varphi(x)$ and the set $\{x\in M: \varphi(x)\}$ is a substructure satisfying $T$ which I will denote as $\varphi(M)$. What are some classical examples of $\varphi(\varphi(M))\neq \varphi(M)$ or $\varphi(M)\not\vDash\forall x \varphi(x)$ ?

We can call such proposition idempotent as suggested in the comments. It would also be nice to know what properties such sentence shared. For example "x is constructible" is $\Delta_1$

One I thought of right before posting were $(\mathbb{Z},+)$ and $even(x)\equiv \exists y\;\; x=y+y$ then $even(\mathbb{Z})=2\mathbb{Z}$ and $even(even(\mathbb{Z}))=4\mathbb{Z}$.

Edit: Also a suggestion for a better title would really be appreciated.

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    $\begingroup$ hi aldo; just as a terminology suggestion, maybe you could call such formulas "idempotent"? (formulas where $\varphi(M)\models\forall x\varphi(x)$, I mean) $\endgroup$ May 12 at 23:46
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    $\begingroup$ @AtticusStonestrom Yes I will add that $\endgroup$ May 13 at 0:07
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    $\begingroup$ The set of non-isolated points in a topological space sort of counts (the issue being that some work is needed to make this appropriately first-order): the fact that this is not idempotent leads to the Cantor-Bendixson derivative. $\endgroup$ May 13 at 3:30


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