Prove that $\mathbb{R}^∞$ is infinite-dimensional. 
Prove that $\mathbb{R}^∞$ is infinite-dimensional.

The section that contains this problem deals with the idea of a basis, so the proof probably has something to do with it (since a basis must have a finite set of vectors?).
 A: Assume $\mathbb{R}^\infty$ has dimension $n$. Then it cannot have more than $n$ linearly independent vectors. However, the canonical vectors $e_1,e_2,\ldots\,e_n,e_{n+1}$ (all zeros except the indexed position with one) are linearly independent (easy by definition). This contradicts our assumption.
A: I'm guessing you mean real valued sequences, in which case what can you say about the set 
$$
\left\{\mathbf{a}_i:\mathbf{a}_{ij}=\begin{cases}1\text{ if i=j }\\0\text{ otherwise }\end{cases} \text{ for } i=1,2,3,\dots\right\}
$$
where $\mathbf{a_i}=(\mathbf{a}_{i1},\mathbf{a}_{i2},\mathbf{a}_{i3},\dots)$.
A: If $$\Bbb R^\infty = \Bbb R^{\Bbb N} = \{(a_n)_{n \in \Bbb N} \mid a_n \in \Bbb R,~\forall\,n \in \Bbb N \},$$you can consider the uniform norm $\|\cdot\|_{\infty}$ on the subspace $\ell_\infty(\Bbb N)$ of the bounded sequences. Then ${\bf e}_n = (\delta_{mn})_{m \geq 1}$ all satisfy $\|{\bf e}_n\|_\infty = 1$ and $\|{\bf e}_n-{\bf e}_m\|_\infty = 1$ if $m \neq n$, so the unit ball is not compact. Hence $\infty = \dim \ell_\infty(\Bbb N) \leq \dim \Bbb R^\infty$.
