Showing that the indicator function forms a basis 
Let $E$ a non-empty finite set and $F(E)$ the set of all functions $f: E \to \mathbb{R}$. Define for every $f \in E$ the function $X_f : E \to \mathbb{R}$ with $$x \mapsto \begin{cases} 1 \text{, if $x=f$} \\ 0, \text{otherwise}\end{cases}$$ show that $(X_{f_1}, \dots ,X_{f_{n}})$ is a basis for $F(E)$ if the set $E$ has $n$ elements and $\{f_1,\dots, f_n\} = E.$

The notation here is very confusing it seems that the writer wants me to show that the sequence of the functions forms a basis for $F(E).$ Also isn't this function the indicator function or am I missing something here? Any hints on how to start showing something like this would be appreciated. I know that in order to show that some set forms a basis for a vector space I would need to show that it spans the space and that it's elements are linearly independent.
 A: I suggest to rewrite the problem as:

Let $E$ be a non-empty set, and $F(E)$ be the set of all functions from $E$ to $\mathbb R$. For every $e \in E$ define the function $\chi_e : E \to \mathbb R$ (so $\chi_e \in F(E)$) by sending $e$ to $1$ and every other element distinct from $e$ to $0$. Show that if $E = \{e_1,\dots,e_n\}$ for some $n \in \mathbb Z^+$, then the set $\{\chi_{e_1},\dots,\chi_{e_n}\}$ is a basis for $F(E)$.

Now prove that, for every $f \in F(E)$, $f$ is equal to the function $f(e_1)\chi_{e_1} + \cdots + f(e_n)\chi_{e_n}$  (how do we check that two functions are equal?). Also, if $a_1,\dots,a_n$ are $n$ real numbers such that $a_1\chi_{e_1} + \cdots + a_n\chi_{e_n}$ is the zero function, evaluating $a_1\chi_{e_1} + \cdots + a_n\chi_{e_n}$ in $e_j$ (for $j=1,\dots,n$) shows that $a_j=0$ (why?).
Hope this helps you to clarify the things. And let me know if something that I wrote was not clear to you.
A: Recall that a vector space V over a field F is a set where $av_1+bv_2\in V, a,b\in F, v_1,v_2\in V$. A vector space of functions, there is an additional "thing" that is a set X over which the function acts. That is, suppose X is any set, then a vector space V over F is the set of all functions $V=\{f:(f:X\rightarrow F)\}$.
Here, E is the set over which the functions act. Let F be the vector space and R be the field. Then the vector space $F=\{f:(f:E\rightarrow \mathbb R)\}$. Let E have n elements, i.e. $E=\{e_1,...,e_n\}$. The claim is that the set of all $$g_{e_i}(x)=\begin{cases}1&x=e_i\\0&x\ne e_i\end{cases}$$
is a basis of F.
First, show that any vector in F can be written as a linear combination of $g_{e_i}$. Notice that $f(x)=f(e_1)g_{e_1}(x)+...+f(e_n)g_{e_n}(x)$ for any function f. This is because f can take on the finitely many elements in E as inputs, and for each value of e, it has a real valued output. The vector space F consists of infinitely many functions that take the inputs e and outputs a real number for each. Notice the coefficients are real numbers, so they are in $\mathbb R$, the field. You may be thinking that it's weird for the function to depend on its values, but it should be fine by taking a leap of faith.
Next you also have to show that the vectors in the basis are linearly independent, i.e. that $a_1g_{e_1}(x)+...+a_ng_{e_n}(x)=0$ implies that $a=0$. Suppose for the sake of contradiction that $a_j\ne 0$. But then plugging in $e_j$, you get $a_j$, which is not 0. Thus the coefficients must all be 0.
