My textbook says that the definition of a word is $(s_1,s_2, s_3 , \dots ) $ where $ s_i\in S\cup S^{-1} $ and $s_i =1 \; $ for all $ i $ sufficiently large.
I don't get that why is this "$s_i =1 \; $ for all $ i $ sufficiently large" part is needed.
Is this because elements which are a infinite products of elements are always same with another finite product of elements?
Why only define word only with finite product of elements
free-groupsgroup-theory
Best Answer
So first of all to make that definition formally correct you have to define what $1$ is. Just a special character. Then "$s_i\in S\cup S^{-1}\cup\{1\}$" is what you should have instead of "$s_i\in S\cup S^{-1}$".
So this "sequential" approach should be understood as "infinite multiplication". Of course "infinite multiplication" doesn't make much sense (without some additional structure like metric or topology, but that's offtopic) unless almost all elements in the infinite multiplication are $1$s. For example
$$2\cdot 3$$
can be written as
$$2\cdot 3\cdot 1$$ or $$2\cdot 3\cdot 1\cdot 1$$ or $$2\cdot 3\cdot 1\cdot 1\cdot 1\cdots$$
We can "agree" that the last statement makes sense. We can always ignore $1$s. But this
$$2\cdot 3\cdot 4\cdot 5\cdot 6\cdots$$
doesn't make sense. We don't know how to mulitply infinitely many non-$1$s!
So this is the intuition: a word is a finite sequence filled with infinitely many $1$s.
The other thing is that formally this is necessary because later you define so called concatenation of two words. So first if $w=(w_1,w_2,\ldots)$ is a sequence (not necesarily a word) then you define
$$b(w)=\{n\in\mathbb{N}\ |\ w_i=1\text{ for all }i> n\}$$ $$len(w)=\begin{cases} \min(b(w))&\text{if }b(w)\text{ is nonempty}\\ \infty&\text{otherwise} \end{cases}$$
The $len(w)$ function should be understood as "the length of sequence $w$". With that the condition "$s_i =1 \; $ for all $ i $ sufficiently large" is equivalent to "$len(s)<\infty$".
The importance of "$len(s)<\infty$" shows when we want to concatenate two words. Have you tried joining two infinite sequences? Not an easy task, maybe even impossible, who knows? It is way easier to join finite sequences. :)
Formally if $w,u$ are two words then the concatenation $wu$ is defined as the word $z$ such that:
$$z_i=\begin{cases} w_i&\text{if }i\leq len(w) \\ u_{i-len(w)}&\text{otherwise} \end{cases}$$
note that if $len(w)=\infty$ then this definition simply yields $w$. Not interesting. On the other hand $len(u)$ could be infinite but then we would have a problem that $wu$ makes sense but $uw$ not anymore.
Example. Say you have two words over $S=\{a,b,c\}$: $w=(a,b,c,1,1,1,\ldots)$ and $u=(c,1,b,b,1,1,1,\ldots)$. Then
$$len(w)=3$$ $$len(u)=4$$ $$wu=(a,b,c,c,1,b,b,1,1,1,\ldots)$$ $$uw=(c,1,b,b,a,b,c,1,1,1,\ldots)$$
Note that there's a special "empty" word $(1,1,1,\ldots)$ all $1$s. For it we have $len((1,1,1,\ldots))=0$.
All of that can be approached differently (and in my opinion easier to understand): a word over an alphabet $\mathcal{A}$ is simply a finite sequence $(a_1,\ldots, a_n)\in\mathcal{A}^n$. Note that there's a special empty word $()$ which is the only element of $\mathcal{A}^0$. Many things get simplified with this definition. For example the length:
$$len((w_1,\ldots, w_n))=n$$ $$len(())=0$$
Or concatenation:
$$(w_1,\ldots,w_n)(u_1,\ldots,u_m):=(w_1,\ldots,w_n, u_1,\ldots,u_m)$$
with the special rules for the empty word:
$$(w_1,\ldots,w_n)():=(w_1,\ldots,w_n)$$ $$()(w_1,\ldots,w_n):=(w_1,\ldots,w_n)$$ $$()():=()$$
and so on, and so on. I hope this clarifies things.