Closed and not closed subspaces of $l^{2}$

Question

I am dealing with finding two examples of closed and not closed subspaces of $l^{2}$, where $l^{2}$ is the set of sequences such that: $$\sum_{n=1}^{\infty} |a_n|^{2} < \infty$$

I can think about two closed examples by forcing the components of the sequence of being 0 after some fixed k.

Example. Assuming k = 4, then {$a_k$} = {$a_1$,$a_2$,$a_3$,0,0,...}. A subset like this should be closed, right?

However, I cannot think about a subspace of $l^2$ which is not closed.

Answer 1

$A \subset \ell^2$ is closed if and only if whenever $x^{(n)}$ is a sequence of elements of $A$ and $x^{(n)} \to x$ in the $\ell^2$-norm we have $x \in A$.

Now, if $x^{(n)} \to x$ in $\ell^2$ then we have that for each $k \geq 1$, $x_k^{(n)} \to x_k$ in $\mathbb{R}$ (i.e. convergence in $\ell^2$ implies convergence of the individual coordinates). This makes it easy to see that your example of a closed subset is indeed closed. If $x^{(n)} \to x$ in $\ell^2$ then $x_k = \lim_{n \to \infty} x_k^{(n)} = 0$ for $k \geq 4$ since $x_k^{(n)} = 0$ for all $n \geq 1$ and $k \geq 4$.

The standard example of a subspace of $\ell^2$ which isn't closed is $$c_{00} = \{ x \in \ell^2 : x_k = 0 \text{ for all but finitely many } k\}.$$ It's easy to check that this is a subspace and in fact it is dense in $\ell^2$ since if $x = (x_1, x_2, x_3, \dots) \in \ell^2$ then $x^{(n)} = (x_1, \dots, x_n, 0, 0, 0, \dots) \in c_{00}$ converges to $x$ in $\ell^2$ as $n \to \infty$. So $c_{00}$ is dense in $\ell^2$ and is not all of $\ell^2$ so cannot be closed.

Answer 2

Let $\left\{e_k\right\}_{k\in K}$ be any Hamel basis for $\ell^2$. Since $\ell^2$ is infinite dimensional, $K$ is infinite, and in particular we may pick a countable subset $J=\left\{k_n\right\}_{n\in \mathbb{N}}$. Now define a functional $f:\ell^2\to \mathbb{R}$ by first setting $$f(e_{k_n}):=n,\qquad n\in \mathbb{N} $$ and then taking any linear extension of $f$ to $\ell^2$. Clearly $f$ is linear and unbounded, therefore $\ker f$ is a non-closed subspace (see here).

Alternatively, the vector space $X$ spanned by $\left\{e_k\right\}_{k\in J}$, equipped with the restriction of the $\ell^2$ norm, is a normed space of infinite countable dimension, and hence by a corollary of Baire's category theorem it cannot be complete. Thus $X$ is not closed in $\ell^2$.