Non-closed subspace of a Banach space Let $V$ be a Banach space. Can you give me an example of a subspace $W\subset V$ (sub-vectorspace) that is not closed?
Can't find an example of that yet.
Thanks!
 A: Take the space $L^1 [0,1]$. This space is complete. 
Now take the subspace of continuous functions $C[0,1]$ (with the $L^1$ norm). Then this is not a Banach space (with respect to $\|\cdot\|_{L^1}$). To see this, consider the sequence
$$f_n(x) = 
\begin{cases}
0 & \text{on } \hspace{0.5cm} [0,\frac12 - \frac1n]\\
nx + 1 - \frac{n}{2} & \text{on } \hspace{0.5cm} [\frac12 - \frac1n, \frac12]\\
1 & \text{on } \hspace{0.5cm} [\frac12,1]\\
\end{cases}$$
This is a Cauchy sequence but its limit is not continuous. 
A: There is a regular method to produce a lot of non-closed subspaces in arbitrary infinite dimensional Banach space.
Take any countable linearly independent family of vectors $\{w_i:i\in\mathbb{N}\}\subset V$ and define $W=\mathrm{span}\{w_i:i\in\mathbb{N}\}$. Then, $W$ is not closed.
Indeed, assume that $W$ is closed. Recall that $V$ is a Banach space, then $W$ is also Banach as closed subspace of Banach space. Linear dimension of $W$ is countable, but by corollary of Baire category theorem, Banach space can not have countable linear dimension. Contradiction, so $W$ is not closed.
This general result was demonstrated in Kevin's and J.J.'s answers. I'll show another one.
Consider Banach space $V=(C([0,1]),\Vert\cdot\Vert_\infty)$ of continuous functions with $\sup$ norm. Let $W=(P([0,1]),\Vert\cdot\Vert_\infty)$ be its proper subspace consisting of polynomials. It is of countable dimension because $W=\mathrm{span}\{x^k:k\in\mathbb{Z}_+\}$. From result given above it follows that $W$ is not closed. 
But there is another proof. By Weierstrass theorem $W$ is dense in $V$, i.e. $\overline{W}=V\neq W$. Thus $W$ is not closed, thought it is dense in $V$.
A: A simple example, which gets at the difference between orthonormal and Hamel bases in infinite dimensions, is to take $H$ a separable infinite-dimensional Hilbert space and consider the span of its basis vectors $e_i, i \in \mathbb{N}$. This span, taken algebraically, certainly isn't the entire space because, for instance, $v=\sum_{i\in \mathbb{N}} \frac{1}{2^i} e_i$ isn't in it. Remember that spans are defined via finite linear combinations of the spanning set. But $v$ must be in the Hilbert space, because the partial sums define a Cauchy sequence and Hilbert spaces are complete.
A: Take $V$ to be the space of sequences $(a_1,a_2,\dots)$ of real numbers with norm $\|(a_1,a_2,\dots)\| = \sum_{k=1}^\infty |a_1|$. Consider the subspace of $W \subset V$ consisting of sequences with only finitely many of $a_1,a_2,\dots$ being non-zero.
Then $A_k = (1,\frac{1}{2},\dots,\frac{1}{2^k},0,0,\dots)$, $k=1,2,\dots$, gives a sequence of elements of $W$, but the limit of this sequence is not in $W$.
A: $C_{00}$ with sup norm  is not closed in $l^{\infty}$ 
