Concatenation operation on the set of finite sequences in $\{0, 1\}$ Let $A^{\ast} = \bigcup_{I \subset \mathbb{N}} \mathcal{F}(I, \{0, 1\}) = \bigcup_{I \subset \mathbb{N}} (\prod_{i \in I} \{0, 1\})$ be the set of finite sequences in $\{0, 1\}$.
First, if $I = \emptyset$, then we have $\prod_{i \in \emptyset} \{0, 1\} = \{u_{\emptyset}\} \in A^{\ast}$, where $u_{\emptyset} : \emptyset \rightarrow \{0, 1\}$ is the empty mapping, right ?
Second, let $\star$ be the concatenation operation on $A^{\ast}$ defined in the following way :
$$\forall (n, m) \in \mathbb{N}^{\ast} \times \mathbb{N}^{\ast}, \ \forall u = (u_{1}, ..., u_{n}) \in A^{\ast}, \ \forall v = (v_{1}, ..., v_{m}) \in A^{\ast}, u \star v = (u_{1}, ..., u_{n}, v_{1}, ..., v_{m}) \ \text{,}$$
I have seen somewhere that the neutral element for this operation is called the "empty sequence" (denoted here $()$) such that : $\forall u \in A^{\ast}, u \star () = () \star u = u$. The problem is that I don't find a clear definition of what this empty sequence is.
Precisely, my questions are the following : Are the empty sequence and the empty mapping defined above in first point in fact the same element (i.e., $() = u_{\emptyset}$) ? If it is not the case, what is exactly the empty sequence ?
Thank you for your help.
 A: The OP's definition/setup is murky. Although there is no doubt about the intention, it is helpful to frame this with more precision. This is just one way to 'get formal'.
Definition: If $n \in \mathbb N$, then any function $u$ of the form
$\tag 1 u: \{k \in \mathbb N \; | \; 1 \le k \land k \le n\} \to \{0,1\}$
is said to be a finite sequence in {0,1} of length $n$.
If $u$ is a finite sequence in {0,1} of length $n$ and $v$ is a finite sequence in {0,1} of length $m$ we define a finite sequence in {0,1} of length $n +m$, $u*v$, as follows:
$\quad\quad\quad\quad [u*v](k) = u(k) \text{ for } k \le n$
$\quad\quad\quad\quad [u*v](k) = v(k-n) \text{ for } n + 1 \lt k \le n+m$
Informally, we write $u = (u_{1}, ..., u_{n})$.
The definition allows for a finite sequence of length $0$, but there can only be one form of such a sequence, the empty graph $\emptyset$. In the same way that using logic shows that $0! = 1$, you can show that $\emptyset$ serves as an identity.
Now if that sounds 'spooky', you can change the definition so that length $0$ is not allowed. Then, if you want, you can algebraically 'throw in' an identity with our associative binary operation of concatenation.
With our informal notation, using $()$ for $\emptyset$ works great!
