Is definable $\emptyset$-definable?

Question

A set $A$ is $X$-definable in an $L$-structure $\mathcal{M}$ with a domain $M$ iff there are an $L$ formula $\phi(a,\bar x)$ and $\bar x \in X^n$ such that $A=\{a \in M^m | \mathcal{M} \models \phi(a,\bar x)\}$.

Often people say that a set $A$ is just definable. Do they mean that $A$ is $\emptyset$-definable or that $A$ is $M$-definable?

Answer 1

A set $D$ is $A$-definable in $M$ if there is parameter-free formula $\varphi(z,x)$ and a tuple $a\in A^{|z|}$ such that $D=\{b\in M^{|x|}\ :\ M\models\varphi(a,b)\}$. (Were $|\cdot|$ denotes the length of the tuple.)

The following is useful to know (it justifies the notation used above):

[i] My writing $a\in A^{|x|}$ is overscrupulous. Usually it is agreed that $a\in A$ stands for $a\in A^{|a|}$ (it has become very common to use the same notations for tuple and for elements).

[ii] Definable stands for $M$-definable unless the author explicitly says the contrary.

[iii] Sometimes $0$-definable is used for $\varnothing$-definable.

[iv] In 90% of the scientific articles, a structure is denoted with the same symbol used for its domain.

Answer 2

We can see :

• Katrin Tent & Martin Ziegler, A Course in Model Theory (2012), page 4 :

Let $B$ be a subset of $A$ [the domain of the $L$-structure $\mathfrak A$]. By considering every element of $B$ as a new constant, we obtain the new language

$$L(B) = L \cup B$$

Aand the $L(B)$-structure

$$\mathfrak A_B = (\mathfrak A, b)_{b∈B}.$$

Then see page 9 :

[A formula] $\varphi(x_1,\ldots, x_n)$ defines an $n$-ary relation

$$\varphi(\mathfrak A) = \{ \overline a \mid \mathfrak A \vDash \varphi[\overline a] \}$$

on $A$ [the domain of $\mathfrak A$], the realisation set of $\varphi$. Such realisation sets are called $0$-definable subsets of $A^n$, or $0$-definable relations.

Let $B$ be a subset of $A$. A $B$-definable subset of $\mathfrak A$ is a set of the form $\varphi(\mathfrak A)$ for an $L(B)$-formula $\varphi(x)$. We also say that $\varphi$ (and $\varphi(\mathfrak A)$) are defined over $B$ and that the set $\varphi(\mathfrak A)$ is defined by $\varphi$. Often we don’t explicitly specify a parameter set $B$ and just talk about definable subsets. A $0$-definable set is definable over the empty set. structure they define the same set.