Definable sets properties that uniquely determine an $\mathscr{L}$-structure

Question

$\newcommand{\Def}{\mathsf{Def}}$ It's known that the set $\Def(\mathfrak{M}, S)$ of $S$-definable sets over an $\mathscr{L}$-structure $\mathfrak{M}$ is the smallest subset of $\bigcup_n M^n$ which

1. is closed under Boolean set operations ($\bigcap$, $\bigcup$, $\cdot^C$), cartesian product, permutation of the coordinates, and linear projections $\pi: \begin{array}{ccc}M^{n+m} &\to& M^n\\(\bar{x}, \bar{y}) &\mapsto& \bar{x}\end{array}$ ;
2. contains $\varnothing$, $M^n$ for $n \in \mathbb{N}$, the diagonals $\Delta^n_{i, j} = \{\bar{a} \in M^n : a_i = a_j\}$, the singletons $\{s\}$ for all $s \in S$, and the sets $\{c^\mathfrak{M}\}$ for all constant symbols $c$, the graph of $f^\mathfrak{M}$ for all function symbols $f$ and $R^\mathfrak{M}$ for all relation symbols $R$.

Now I came across a remark saying that given a language $\mathscr{L}$ and a set $M$, a subset $\mathscr{D} = \bigcup_n \mathscr{D}_n$ of $\bigcup_n M^n$ which satisfying both properties above, determines uniquely an $\mathscr{L}$-structure on $M$. Namely, the interpretation of symbols in $\mathscr{L}$ are specified by $\mathscr{D}$.

Question: How one could "recover" the interpretation of constants, functions and relations symbols from $\mathscr{D}$ ?

Answer 1

I'm not sure where you got this remark from, but it is fairly easy to see that it is not true as written.

For example, if you have a language with two unary predicates $P_1, P_2$ and $X\supseteq Y_1, Y_2$ are any sets with $Y_1\neq Y_2$, then $M_1=(X,Y_1,Y_2)$ and $M_2=(X,Y_2,Y_1)$ give rise to exactly the same $\mathscr D$.

There are two similar observations which are actually true.

One, if you have a $\mathscr D$ and you know which sets correspond to which function or relation symbols, then you can recover the underlying $\mathscr L$-structure. But this is quite trivial, since this is basically the same as handing you the interpretations of symbols, plus some extra data which you don't care about.

The other, slightly deeper one, is that given a family $\mathscr D_0$ of subsets of $\bigcup_n M^n$ satisfying the obvious axioms (being closed under projections, products and Boolean operations, containing each $M^n$ and each diagonal --- although it is enough to include the diagonal in $M^2$, as the others follow), you can endow $M$ with a first order structure such that the resulting $\mathscr D$ is equal to $\mathscr D_0$, and moreover, this structure is unique up to interdefinability.

This is only slightly less trivial: namely, it is enough to consider $\mathscr L=\{R_D\mid D\in \mathscr D_0 \}$, and then you can interpret each $R_D$ as $D$. It is easy to see that this yields $\mathscr D=\mathscr D_0$.

The uniqueness part is actually a separate observation: if $\mathfrak M_1,\mathfrak M_2$ share the same universe and give rise to the same $\mathscr D$, then they are interdefinable, pretty much by definition.