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.


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!

  • $\begingroup$ In my case, the problem is that I need to allow sequences of length $0$ (so I'm agree, there's only one, the one I denoted "$u_{\emptyset}$"), but I also need that when I do concatenation of any sequence $u \ast u_{\emptyset}$, I obtain $u$. So the algebraic identity for $\ast$ can actually be $u_{\emptyset}$ right ? Moreover (maybe I go to far sometimes, sorry), could we actually throw in another algebraic identity that is different from $u_{\emptyset}$ (if it is one) or it is the only identity for $\ast$ ? $\endgroup$ – deeppinkwater Dec 8 '18 at 15:16
  • $\begingroup$ @deeppinkwater You can throw in more than one algebraic identity and still have an associative operation - but why would you? When you put in just one identity, it make sense to say it has length $0$ - the length of a concatenation is the sum of the lengths. $\endgroup$ – CopyPasteIt Dec 8 '18 at 15:32
  • $\begingroup$ I don't want to throw another one. First, I need the existence of a sequence of length $0$ (without talking about concatenation), and then, I need an identity for concatenation and I want this identity to be the sequence of length $0$ I exhibited at first step (also, the problem in my case is that I don't really have an operation since I have a length $N$ that I cannot exceed, so concatenation is not really define for any sequence, but anyway...). $\endgroup$ – deeppinkwater Dec 8 '18 at 15:41
  • $\begingroup$ I'm asking you if I could put another identity because the one we talk about ($\emptyset$) seems obvious to me and I don't realy see which one you could take instead of... $\endgroup$ – deeppinkwater Dec 8 '18 at 15:41
  • $\begingroup$ @deeppinkwater Any identity under multiplication (concatenation) does the same thing - nothing! So they all just look like the $\emptyset$ Graph = Null Sequence. You can name it anything or use any notation you want. So use $\text{ [\0]} $ for the $\emptyset$ Graph, and then $\text{ [\0]} * u = u * \text{ [\0]}$ for all $u$. $\endgroup$ – CopyPasteIt Dec 9 '18 at 1:58

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