# Do $e_1$, $e_2$, … generate the entire $l_2$ space?

From our textbook goes some statement like this:

... let $X$ be the linear subspace of $l_2$ generated by the vectors $$\left\{e_1,e_2,e_3,...\right\}$$ ...

Which feels strange to me because I thought $e_1=(1,0,0,...)$, $e_2=(0,1,0,...)$, ... are sufficient to generate the entire $l_2$ space, so the $X$ here is actually just $l_2$ itself. Isn't that true?

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The span of the $e_i$'s is a dense subspace in $\ell_2$, namely of all vectors having finitely many non-zero entries. – Lior B-S Nov 28 '12 at 8:06

No. The closed linear subspace they generate is the whole of $\ell^2$, but the generated linear subspace is by defintion the set of linear combinations, that is $$X = \left\{\sum_{i=1}^k \alpha_i e_{k_i} \biggm| \alpha_i \in \mathbb K, k_i, k \in \mathbb N \right\}$$ which equals the set of all sequences that have only finitely many entries different from zero, i. e. $$X = \left\{x \in \ell^2 \mid \exists N \forall n \ge N : x_n = 0 \right\}$$ For example $(\frac 1n) \in \ell^2 \setminus X$.

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Hint: Consider $x=(x_n)$ in $\ell^2$ defined by $x_n=1/n$ for every $n\geqslant1$. Then $x$ is not a linear combination of a finite number of vectors $(e_k)_{k\geqslant1}$.

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The meaning of "generate" depends on context. The set $\{e_1,e_2,\ldots\}$ generates $\ell^2(\mathbb N)$ as Hilbert space (i.e. closed linear span), but it doesn't generate it as a vector space (linear span).

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