Generate a vector space based on a finite set Let $X$ be a finite set and choose a labeling for it, so it will be  $X$= $\{x_1,...,x_n\}$. $\mathbb F$ is a field. Then the vector space over $\mathbb F$ generated by this finite set $X$ is a set containing "formal linear combination" of the elements in $X$. And it should be $a_1x_1+...+a_nx_n$ where$\{a_1,...,a_n\}$ are from field $\mathbb F$. As a result, the basis of this vector space is $(x_1,...,x_n)$
Then I wonder if let  $X$ be a finite set $X$=$\{1,2,3,...,n\}$, and field is $\mathbb C$. Then how can this finite set $X$ and the field together generate a vector space? Specifically,1,2,3,..,n are elements in set $X$, and how can they become linearly independent to become the basis of vector space?
 A: Don't let the names of the elements of $X$ trick you into thinking they have any kind of implicit properties. In your example (for $n\geq2$), a vector might look like
$$
(5+2i){1}+(-3-5i){2}
$$
You could expand the brackets and write
$$
5\cdot{1}+2i\cdot {1}-3\cdot{2}-5i\cdot{2}
$$
but apart from that, not much simplification can be done (I would argue that that's not a simplification, but that's beside the point). Which makes it a bad idea to use that particular $X$, because it really, really feels like those multiplications can be executed in some way. But they can't. That's what the "formal" in "formal linear combinations" means. It is also a bad idea to use that $X$ because it's difficult to tell apart certain elements of $\Bbb C$ that have algebraic properties, from the elements of $X$, that don't.
It can be done, if you're really careful. But most of the time it won't be worth it. It will just confuse your readers, and maybe even yourself.
A: Perhaps a different choice of notation might clarify things. Given any set $X$ and a field $F$ (you can do this with rings, or other stuff as well), let $F^{\oplus X}$ denote the set of all functions $f:X\to F$ such that $\{x\in X\,: f(x)\neq 0\}$ is a finite set.
Then, $F^{\oplus X}$ is a subset of $F^X$, which is the set of all possible functions $f:X\to F$. The latter space can very clearly be given the structure of a vector space over the field $F$, and it is easy to verify that $F^{\oplus X}$ is a subspace of $F^X$.
Very important examples of elements of $F^{\oplus X}\subset F^X$ are: for each $x\in X$, define $\delta_x:X\to F$ by setting
\begin{align}
\delta_x(y)&:=
\begin{cases}
1&\text{if $y=x$}\\
0 & \text{else}
\end{cases}
\end{align}
Then, $\{\delta_x\}_{x\in X}$ forms a (Hamel) basis for the vector space $F^{\oplus X}$, precisely because the definition requires that the support of the functions be finite. Now, given $x_1,\dots, x_n\in X$ and scalars $a_1,\dots, a_n\in F$, it makes perfect sense to consider the linear combination
\begin{align}
a_1\delta_{x_1}+\cdots +a_n\delta_{x_n}
\end{align}
This is just a linear combination of certain functions $X\to F$. So, when speaking of a formal linear combination, you can think of it in this manner.

In the special case that our index set is $X=\{1,\dots, n\}$, then the resulting vector space we get $F^{\oplus \{1,\dots, n\}}$ may set-theoretically be different from $F^n$ (defined as the set of all ordered $n$-tuples), but it's the same idea. So, in terms of my above notation, a basis for the space $F^{\oplus\{1,\dots, n\}}$ is $\{\delta_1,\dots, \delta_n\}$, where $\delta_i:\{1,\dots, n\}\to F$ is the function
\begin{align}
\delta_i(j)&=
\begin{cases}
1&\text{if $j=i$}\\
0&\text{else}
\end{cases}
\end{align}
But if you think about it, this is precisely what everyone writes as $\{e_1,\dots, e_n\}$ being a basis for the vector space $F^n$, where $e_i=(0,\dots, \underbrace{1}_{\text{$i^{th}$ spot}},\dots, 0)$.

Extra ramblings about Polynomials:
If you start with the index set $X=\Bbb{N}_0$ the non-negative integers, then the resulting space you get is the space of polynomials in one variable. It's clear how the vector space structure is defined, because it's a special case of what I've already mentioned above. The multiplication is defined by $\delta_i\cdot \delta_j:=\delta_{i+j}$ for all $i,j\in X$, and then extending bilinearly (again, we can do this because $F^{\oplus X}$ has in its definition the finite support condition).
Of course, when we write the polynomial ring as $F[x]$, to indicate "finite formal sums in the indeterminate $x$", we can either think of $x$ as "a symbolic object to be manipulated according to some rules", or we can think of it as the function $\delta_1$ (which ok you could also argue is a specific symbol and so on), and more generally, $x^i$ as $\delta_i$.
More generally, by taking $X=(\Bbb{N}_0)^k$ for some $k\in\Bbb{N}$, the resulting vector space $F^{\oplus X}$ is what we can think of as the space of polynomials in $k$ variables, with coefficients in the field $F$. Again, the pure vector space structure is clear. The multiplication is defined by extending bilinearly the definition $\delta_{(i_1,\dots, i_k)}\cdot \delta_{(j_1,\dots,j_k)}:=\delta_{(i_1+j_1+\dots, i_k+j_k)}$. In the usual $F[x_1,\dots, x_k]$ notation, what I'm calling $\delta_{(i_1,\dots, i_k)}$ is what would be written as $x_1^{i_1}\cdots x_k^{i_k}$.
