Let $\mathcal{M}$ be a L-model, for any A$\subseteq$M, let dcl(A)={b∈M: b definable in $\mathcal{M}$ over A}, How to show that dcl(dcl(A)=dcl(A)?

Question

two steps

1) suppose $b$ is definable in $\mathcal{M} $ over $A$, then we need to show $b$ is definable over dcl$(A)$.

2) suppose $b$ is definable in $\mathcal{M}$ over dcl$(A)$, then we need to show $b$ is definable over $A$.

I think that we need to construct a formula which satisfy 1) in step 1), and construct a another formula which satisfy 2) in step 2).

But I really don't know how to construct the formulas. So I need your help.

Thanks!

Answer 1

First, it should be obvious that $\mathrm{dcl}(A) \subseteq \mathrm{dcl}(\mathrm{dcl}(A))$. We must now show that $\mathrm{dcl}(\mathrm{dcl}(A)) \subseteq \mathrm{dcl}(A)$.

Assume that $b \in \mathrm{dcl}(\mathrm{dcl}(A))$. Then $\exists \varphi(x,a_1,...,a_n)$ with $\varphi(M,a_1,...,a_n) = b$ and $a_1,...,a_n \in \mathrm{dcl}(A)$. Since $a_1,...,a_n \in \mathrm{dcl}(A) \implies \exists\psi_i(x,\bar{c}_i)$ such that $\bar{c}_i \in A^{i}$ and $\psi_i(M,\bar{c}_i) = a_i$

But then, consider the formula $\theta(x) = (\exists x_1,..,x_n)\bigwedge_{i=1}^n \psi_i(x_i,\bar{c}_i)\wedge\varphi(x,x_1,...,x_n)$. We now have that $\theta(M) = {b}$ and $\theta(x)$ contains parameters only from $A$. Therefore, $b \in \mathrm{dcl}(A)$ and we are done.

Answer 2

For a more geometric proof: notice that if $N\succeq M$, then for any $A\subseteq M$ we have that $\textrm {dcl}(A)$ remains the same in $N$ and $M$.

If you take $N$ to be a sufficiently homogeneous model, then whenever $A$ is a small set, $\operatorname{dcl}(A)$ is just the set of points which are fixed by all automorphisms of $N$ fixing $A$. Thus $\operatorname{dcl}(\operatorname{dcl}(A))$ is simply the set of all points which are fixed by all automorphisms of $N$ fixing $\operatorname{dcl}(A)$. But those are, by definition, exactly the automorphisms fixing $A$.