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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?).

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

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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 $\mathbf{a_i}=(\mathbf{a}_{i1},\mathbf{a}_{i2},\mathbf{a}_{i3},\dots)$.

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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 $\infty = \dim \ell_\infty(\Bbb N) \leq \dim \Bbb R^\infty$.

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