Show the subspace $\textrm{span}\{e_n:n\in\mathbb N\}$ is not closed in $\ell^2$

Question

How to show the subspace $\textrm{span}\{e_n:n\in\mathbb N\}$ where $e_n=(\delta_{nk})_{k\in\mathbb N}$ is not closed in $\ell^2$?

Answer 1

Consider $$ x=\sum\limits_{n=1}^\infty 2^{-n}e_n $$

Answer 2

Let $X= \text{span} \{e_n : n \in \mathbb{N} \}$ and $x \in \ell^2$. Then $\displaystyle x= \sum\limits_{i=1}^{+ \infty} x_ie_i$ and $\displaystyle ||x||_2^2= \sum\limits_{i=1}^{+ \infty} |x_i|^2 <+ \infty$. So, for $\epsilon>0$, there exists $N \geq 0$ such that $n \geq N$ implies $\displaystyle \sum\limits_{i \geq n} |x_i|^2 < \epsilon$. Let $\displaystyle y = \sum\limits_{i=1}^N x_ie_i$. Then $y \in X$ and $\displaystyle ||x-y||_2^2= \sum\limits_{i \geq N+1} |x_i|^2 < \epsilon$.

You deduce that $X$ is dense in $\ell^2$. Therefore, $X$ is closed in $\ell^2$ iff $X= \ell^2$. But $\displaystyle z= \sum\limits_{i=1}^{+ \infty} z_ie_i$ belongs to $X$ iff $\{i \geq 1 : z_i \neq 0\}$ is finite; therefore, $X \subsetneq \ell^2$.