Prove $F^\infty$ is infinite dimensional Prove $F^\infty$ is infinite dimensional
I can see that this is trivially true. But how does one mathematically prove it?
It looks true by definition honestly. Do i say that since it is an n-tuple where n is inf that it needs n linearly independent vectors to span it and hence it has infinite dim?
 A: 
Do i say that since it is an n-tuple where n is inf that it needs n linearly independent vectors to span it and hence it has infinite dim?

I think you have the right idea.  But the definition of dimension as the cardinality of the basis is only for the case when the basis is finite.  I believe the definition of infinite dimensional is: not finite dimensional.  That is, there is no finite linearly independent set that spans $F^\infty$.
A: $V = \{(x_1, x_2, \ldots, x_n, 0, 0, ...) \in \mathbb{F}^\infty: x_i \in \mathbb{F}\}$ is a subspace of $\mathbb{F}^\infty$. So we have
$$ n = dim(V) \leq dim(\mathbb{F}^\infty) $$
Since $n$ is arbitrary, $dim(\mathbb{F}^\infty) = \infty$.
A: Similar to the idea/approach of Mr. Leingang,  consider the claim that for any natural $n$,
the set {$\delta_{1j}, \delta_{2j},...,\delta_{nj}$}, with $\delta_{nj}$ identically 1 in the n-th spot and 0 everywhere else, is a linearly-independent set .
Assume
$c_1 \delta_{1j}+c_2\delta_{2j}+ ...+....+c_n\delta_{nj}=(c_1,c_2,....,c_n,0...,0,....0)=(0,0,..,0,0,......) $ clearly implies $0=c_1=c_2=....=c_n$.
This is true for any finite $n$. Then, there are linearly-independent sets of any cardinality in $\mathbb F^{\infty}$
