Intersection of $L$-structures, substructures The task is to show that the intersection of substructures is a substructure (or empty). Disclaimer: I think it is enough to answer the question in the last orange box, to answer my question... The post got a little longer, but most of the stuff is context and own thoughts. So you do not have to read everything, if you are familiar with the topic.
Definition (substructure):

Let $\mathcal{L}$ be a language and $\mathcal{A},\mathcal{B}$ two $\mathcal{L}$-structures. We say that $\mathcal{B}$ is a substructure of $\mathcal{A}$ if the following holds:
a) $B\subseteq A$
b) For every constant symbol $c\in\mathcal{L}$ is $c^{\mathcal{A}}=c^{\mathcal{B}}$
c) For every n-digit function symbol $f\in\mathcal{L}$ is $f^\mathcal{B}=f^\mathcal{A}|\mathcal{B}^n$ (the restriction of $f^{\mathcal{A}}$ on $\mathcal{B}^n$) and $\mathcal{B}$ is closed with regards to $f^{\mathcal{B}}$
d) For every m-digit relation symbol $R\in\mathcal{L}$ is $R^\mathcal{A}\cap\mathcal{B}^m=R^\mathcal{B}$

So let $L$ be a language, $\mathcal{A}=(A, (z^\mathcal{A})_{z\in L}$ a L-structure, $I$ a set of indices and $\mathcal{A}_i=(A_i, (z^{\mathcal{A}_i})_{z\in L})$ a familiy of substructures of $\mathcal{A}$.
I have to show, that $\mathcal{S}=\bigcap_{i\in I}\mathcal{A}_i$ ('shortend definition' see 'at b)') is a substructure of $\mathcal{A}$, or empty.
Suppose $S\neq\emptyset$.
a): $\bigcap_{i\in I} A_i\subseteq A$ holds true, since $A_i\subseteq A$ for every $i\in I$.
But now I start to struggle with the formalism, I guess.
at b): For every constant $c\in L$ is $c^{\mathcal{A}}=c^{\mathcal{A}_i}$ for every $i\in I$, because $\mathcal{A}_i$ is a substructure of $\mathcal{A}$.
I have to show, that the interpretation of $c$ in $\mathcal{S}$ is the same as in $\mathcal{A}$. Or in other words $c^\mathcal{A}=c^\mathcal{S}$.

How are the constants symbols in $\mathcal{S}$ 'generated'?

Every L-structure has a set of constants symbols. These do not have to be necessarly equal. Here they are, because we are intersecting substructures of $\mathcal{A}$.
Obviously we simply intersect these sets of constant symbols. [Or relation and function symbols]
But how is that formalised?

Do we write simply $\mathcal{S}=(\bigcap_{i\in I} A_i, \bigcap_{i\in I} (z^\mathcal{A_i})_{z\in L})$?

What I find confusing is what we really intersect here?
So can we even make sense out of $\bigcap_{i\in I} (z^\mathcal{A_i})_{i\in I}$. That is if $(z^\mathcal{A_i})_{i\in I}$ is a set (of constant, function and relation symbols) and not, as the notation lets you suspect, a tupel...
So basically (as one would suspect...) $c^\mathcal{S}=c^\mathcal{A}\Leftrightarrow c^{\mathcal{A}_i}=c^\mathcal{A}$ for every $i\in I$.
Thanks in advance.
I will add c), and d) later as this got longer as planned already.

Example: Take a language $L=(0,1,+)$ with the usual interpretations. Now I write down L-structures, where $A,B\subseteq\mathbb{N}$:
$\mathcal{N}=(\mathbb{N},0^\mathcal{N},1^\mathcal{N},+^\mathcal{N})$ (with the usual interpretation)
$\mathcal{A}=(A,0^\mathcal{A},1^\mathcal{A},+^\mathcal{A})$. Where $0^\mathcal{A}=0$, $1^\mathcal{A}=1$ and $+^\mathcal{A}$ is still the usual addition.
Now I take $\mathcal{B}=(B,0^\mathcal{B},1^\mathcal{B},+^\mathcal{B})$, where $+^\mathcal{B}$ still notes the usual addition, but we swap
$1^\mathcal{B}=0$ and $0^\mathcal{B}=1$
Then $\mathcal{B}$ is not a substructure of $N$? Because for $0\in L$ we do not have $0^\mathbb{N}=0^\mathcal{B}$

 A: I will refer to the interpretations of the constant, function and relation symbols in $\mathcal A$ as “the constants, functions and relations.”
A subset $B\subseteq A$ forms a substructure if and only if it is closed under all of the functions and contains all of the constants. Then it makes sense to interpret the constant symbols in $\mathcal B$ as the respective constants and the function symbols as the restrictions of the respective functions. (Restricting the relation always makes sense.)
So you just need to show that the intersection of sets closed under a set of functions is also closed. (And it’s trivial that the intersection of sets that contain some given constants also contains those constants.)
A note on the part about empty/non-empty. Whether to allow empty structures is one of those annoying things where sometimes it is natural and sometimes it isn’t, so the convention of not allowing them can sometimes cause unnecessary complications. Here it is more natural just to let trivial cases be trivial: after all, the empty set is closed under any set of functions. Notice also, if there are any constant symbols that the intersection will not be empty. 
