# 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?).

• How is your $\mathbb{R}^\infty$ defined? What are the vectors? – mvw Jul 13 '15 at 20:13
• $\mathbb{R}^n\subset\mathbb{R}^\infty$ for all $n\in\mathbb{N}$ if one continues $n$-vectors with zeros. – A.Γ. Jul 13 '15 at 20:14

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.
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 $\mathbf{a_i}=(\mathbf{a}_{i1},\mathbf{a}_{i2},\mathbf{a}_{i3},\dots)$.
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 $\infty = \dim \ell_\infty(\Bbb N) \leq \dim \Bbb R^\infty$.