Prove that $\mathbb{F}^{\infty}$ is infinite-dimensional I am pretty sure the general idea of my proof is correct, but I am not sure if it is coherent. For example, is something like $i \in I$ , which I mean by for any number $i$ in the index ($I$), common enough notation or is there a better way to state it?

Suppose the span($\mathbb{F}^{\infty}) = \{e_1, e_2, \dots , e_n\}$. Then for any vector $v \in \mathbb{F}^{\infty}$, 
$$v = a_1e_1,  + \dots + a_me_m,~ i \in I,~ a_i \in \mathbb{R}, m > n$$
Then clearly $v$ is not a linear combination of span($\mathbb{F}^{\infty})$ and contradicts the fact that a spanning set must be finite. Thus $\mathbb{F}^{\infty}$ is not finite and is therefore infinite-dimensional.
 A: Well, I think $\mathbb{F}$ is a field and you're considering:
$$
\mathbb{F}^\infty = \{(x_1,x_2,...): x_i \in \mathbb{F}\}
$$
Suppose that $\dim(\mathbb{F^\infty})<+\infty$. Then, exists $n_0 \in \mathbb{N}$ such that $\dim(\mathbb{F^\infty})< n$, $\forall n> n_0$. However, for each $n \in \mathbb{N}$, we considering $T_n: \mathbb{F}^\infty \to \mathbb{F}^n$ defined by:
$$
T_n(x_1,x_2,...) = (x_1,x_2,...,x_n)
$$
Clearly, $T_n$ is a linear transformation and $T_n$ is surjective and not injective. Then, we obtain:
$$
\dim(F^\infty) > n, \forall n \in \mathbb{N}
$$
Thus obtaining a contradiction, therefore $\dim(\mathbb{F^\infty}) = +\infty$. 
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \blacksquare$
A: No, this doesn't work.
Things to address:
1) What is $I$?
2) 'Clearly $v$ is not a linear combination...' This is not clear. In fact it's the opposite: you have written down $v$ as a linear combination of the spanning set, and 
so there is no contradiction.
Try writing down the definition of $\mathbb{F}^\infty$
