While studying partial Ebbinghaus-Flum's Mathematical Logic, I came across the partial isomorphism definition, as build upon an $n$-back-and-forth system. Consequently, the question I raise in the title.
Now I quote from the textbook:
«$\mathfrak{A}$ and $\mathfrak{B}$ are said to be finitely isomorphic, written $\mathfrak{A}\cong_{f}\mathfrak{B}$, iff there is a sequence $\left(I_{n}\right)_{n\in\mathbb{N}}$ with the following properties:
• Every $I_{n}$ is a nonempty set of partial isomorphism form $\mathfrak{A}$ to $\mathfrak{B}$.
• (Forth property) For every $p\in I_{n+1}$ and $a\in A$, there is $q\in I_{n}$ s.t. $\left(q\supset p\right)$ and $a\in dom\left(q\right)$.
• (Back property) For every $p\in I_{n+1}$ and $b\in B$, there is $q\in I_{n}$ s.t. $\left(q\supset p\right)$ e $b\in rg\left(q\right)$.»
This definition entails the following one (I quote again):
«$\mathfrak{A}$ and $\mathfrak{B}$ are said to be partially isomorphic, written $\mathfrak{A}\cong_{p}\mathfrak{B}$, iff there is a sequence I such that:
• $I$ is a nonempty set of partial isomorphisms form $\mathfrak{A}$ to $\mathfrak{B}$.
• (Forth property) For every $p\in I$ and $a\in A$, there is $q\in I$ s.t. $\left(q\supset p\right)$ and $a\in dom\left(q\right)$.
• (Back property) For every $p\in I$ and $b\in B$, there is $q\in I$ s.t. $\left(q\supset p\right)$ e $b\in rg\left(q\right)$.»
As far as I can understand, it is a matter of cardinality: two structures $\mathfrak{A}$ and $\mathfrak{B}$ may be uncountable, and yet be partially isomorphic (i.e. countable isomorphic), and in this case I is the set of all partial isomorphisms from $\mathfrak{A}$ to $\mathfrak{B}$; while the chain $\left(I_{n}\right)_{n\in\mathbb{N}}$ indicates the precise $n\in\mathbb{N}$ for which extensions are allowed. However, even if my take on this is correct, I still feel uncomfortable considering the set $I$ as a single set of extended maps.
Thank you in advance for your help.