Proof: infinite parametrisation's of any non-empty subset $\mathbb{R}^n$ Proof: there exists infinite parametrisation's of any non-empty subset $\mathbb{R}^n$
Would this suggestion for a proof be acceptable, if not why and how would I go about proving it. 
let $\alpha \in \mathbb{R}^n$ and $N\in \mathbb{N}$
let $F(x,N)=\sum _{i=i}^N \alpha _i x $
Since there are infinitely many possibilities for N there would be infinitely many possible parameterisation's. 


 A: Your construction does not prove that any set has a parametrization. You don't even start with a fixed set and try to parametrize it.
Your definitions assume no regularity on the functions, so if $Y \subseteq \mathbb{R}^m$ is any non empty infinite set, the inclusion
$$
\begin{align*}
 & f :  Y \subseteq \mathbb{R}^m \longrightarrow \mathbb{R}^m \\
 & \qquad \qquad y \mapsto y
\end{align*}
$$
is a possible parametrization (i.e it sends each point $(y_1, \dots, y_m) \in Y$ to itself) and composing with any bijection $Y \to Y$ gives another parametrization. There are infinitely many such bijections, so there will be infinitely many parametrizations. 
If on the contrary $Y = \{y_1, \dots, y_n\}$ is finite, 
$$
\begin{align*}
& f_n : \mathbb{R}^m \longrightarrow \mathbb{R}^m \\
& \qquad p \mapsto \cases{y_i \text{ if $p = (n+i,n+i)$} \\ y_1 \text{ otherwise}}
\end{align*}
$$
is a family of parametrizations.
It is worth noting that the notion of parametrization here is very weak. Usually more regularity is required, such as $f$ being continuous/differentable and/or the domain being open. For example, a standard definition of a parametrizated curve $\mathcal{C}$ on $\mathbb{R}^n$ is a continuous function $\alpha : [0,1] \to \mathbb{R}^n$ such that $\operatorname{im} \alpha  = \mathcal{C}$.
