I won't say anything more than Theo and Eric have already said, but...
As Eric says, every $\mathbb{R}^n$ can be seen as a space of functions $f: T_n \longrightarrow \mathbb{R}$.
That is, the vector $v = (8.2 , \pi , 13) \in \mathbb{R}^3$ is the same as the function $v: \left\{ 1,2,3\right\} \longrightarrow \mathbb{R}$ such that $v(1) = 8.2, v(2) = \pi$ and $v(3) = 13$.
So, the coordinates of $v$ are the same as its values on the set $\left\{ 1,2,3\right\}$, aren't they? Indeed, the coordinates of $v$ are the coefficients that appear in the right-hand side of this equality:
$ (8.2, \pi , 13) = v(1) (1,0,0) + v(2) (0,1,0) + v(3) (0,0,1) \ . $
On the other hand, the coordinates of $v$ are its coordinates in the standard basis of $\mathbb{R}^3$: $e_1 = (1,0,0), e_2 = (0,1,0)$ and $ e_3 = (0,0,1)$ and we can look at these vectors of the standard basis as functions too -like all vectors in $\mathbb{R}^3$. They are the following "functions":
$ e_i (j) = \begin{cases} 1 & \text{if}\quad i=j \\ 0 & \text{if}\quad i \neq j \end{cases} $
This is an odd way to look at old, reliable, $\mathbb{R}^3$ and its standard basis, isn't it?
Well, the point in doing so is to get hold for the following construction: let $X$ be any set (finite or infinite, countable or uncountable) and let's consider the set of all functions $f: X \longrightarrow \mathbb{R}$ (not necessarily continuous: besides, since we didn't ask $X$ to be a topological space, it doesn't make sense to talk about continuity). Call this set
$ \mathbb{R}^X \ . $
Now, you can make $\mathbb{R}^X $ into a real vector space by defining
$ (f + g)(x) = f(x) + g(x) \qquad \text{and} \qquad (\lambda f)(x) = \lambda f(x) $
for every $x \in X$, $f, g \in\mathbb{R}^X $ and $\lambda \in \mathbb{R}$.
And you would have a "standard basis" too in $\mathbb{R}^X$ which would be the set of functions $e_x : X \longrightarrow \mathbb{R}$, one for each point $x \in X$:
$ e_x (y) = \begin{cases} 1 & \text{if}\quad x=y \\ 0 & \text{if}\quad x \neq y \end{cases} \ . $
So, you see $\mathbb{R}^3$ can be seen as a particular example of a space of functions $\mathbb{R}^X$ if you see the number $3$ as the set $\left\{ 1,2,3\right\}$: $\mathbb{R}^3 = \mathbb{R}^\left\{ 1,2,3\right\} = \mathbb{R}^\mathbb{T_3}$ and the "coordinates" of a function $f\in \mathbb{R}^X$ are the same as its values $\left\{ f(x)\right\}_{x \in X}$.
(In fact, a function $f$ is the same as its set of values over all points of $X$, isn't it? -Just in the same way as you identify every vector with its coordinates in a given basis.)
Warning. I've been cheating a little bit here, because, in general, the set $\left\{ e_x\right\}_{x\in X}$ is not a basis for the vector space $\mathbb{R}^X$. If it was, every function $f\in \mathbb{R}^X$ could be written as a finite linear combination of those $e_x$. Indeed you have
$ f = \sum_{x\in X} f(x) e_x \ , $
but the sum on the right needs not to be finite -if $X$ is not so, for instance.
One way to fix this: instead of $\mathbb{R}^X$, consider the subset $S \subset \mathbb{R}^X$ of functions $f: X \longrightarrow \mathbb{R}$ such that $f(x) \neq 0$ just for a finite number of points $x\in X$. Then it is true that $\left\{ e_x\right\}_{x\in X}$ is a basis for $S$.
(Otherwise said, $\mathbb{R}^X = \prod_{x\in X} \mathbb{R}_x$ and $S = \bigoplus_{x\in X} \mathbb{R}_x$, where $\mathbb{R}_x = \mathbb{R}$ for all $x\in X$.)