Show that $\{ (1, 0, \dots), (0, 1, 0,\dots), (0, 0, 1, 0,\dots), \dots\}$ is not a basis of $\Bbb R^N$

Question

Show that $\beta = \{ (1, 0, \dots), (0, 1, 0,\dots), (0, 0, 1, 0,\dots), \dots\}$ is not a basis for $\Bbb R^N$

We know that for $\beta$ to be a basis we need $(1, 0, \dots), (0, 1, 0,\dots), (0, 0, 1, 0,\dots), \dots$ to be both linearly independent and spanning $\Bbb R^N$. The first condition clearly holds as the only way we get $\sum_i \alpha_i v_i = 0$ is for $\alpha_i = 0$.

However, why doesn't $\Bbb R^N = span((1, 0, \dots), (0, 1, 0,\dots), (0, 0, 1, 0,\dots), \dots)$ hold? Is it because the span of an infinite vector space is ill-defined? Imagine that $N$ is a finite number. Then I guess that $\Bbb R^N = span((1, 0, \dots), (0, 1, 0,\dots), (0, 0, 1, 0,\dots), \dots)$ would not hold either so I suspect that the reason I presented is not the right one.

Do you mean $N$ as in the natural numbers? This is usually denoted as $\mathbb{N}$ or $\mathbf{N}$. Also, the span of any subset of a vector space is well-defined. — shoteyes, Jul 18 '21 at 20:03
The function $f:\mathbb{N}\rightarrow\mathbb{R}$ given by $f(n)=1$ for all $n\in\mathbb{N}$ can't be represented by a finite liner combination of the functions $\mathbf{e}j(n)=\mathbb{1}{{j}}(n)$. — Mittens, Jul 18 '21 at 20:08
Does this answer your question? What is the difference between a Hamel basis and a Schauder basis? — KReiser, Aug 15 '21 at 20:55

score 4 · Accepted Answer · answered Jul 18 '21 at 20:12

Do you mean $\mathbb R^{\mathbb N}$? In any case it depends how you define "basis" and "span". If you mean a Hamel (or algebraic) basis, then by definition this requires that any element can be written as a finite linear combination of the basis vectors. In the context of Hamel basis, $\text{span}(S)$ is always $\text{span}_\text{fin}(S)$, the set of finite linear combinations.

Then $\mathbb R^{\mathbb N} \neq \text{span}_\text{fin}(\{ e_i \mid i=1\ldots n\})$ is quite obvious, since for example $(1,1,1\ldots)$ cannot be expressed a finite linear combination of $(1, 0, \dots), (0, 1, 0,\dots), (0, 0, 1, 0,\dots), \dots$

In fact, no person on this earth can show you an explicit example of how a Hamel Basis of $\mathbb R^{\mathbb N}$ looks like, despite the fact that the existence can be proven via the Axiom of Choice. For this reason, one needs a more sensible notion of basis for infinite dimensional spaces: Schauder basis.

Show that $\{ (1, 0, \dots), (0, 1, 0,\dots), (0, 0, 1, 0,\dots), \dots\}$ is not a basis of $\Bbb R^N$

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