Question about normed vector spaces and quotients If you have a norm defined on a vector space $V$ you can define the norm for the quotient by a subspace $W$:
$$ || v + W|| := \inf_{w \in W} || v + w || $$
My question is: why does $W$ have to be a closed subspace?
In $\mathbb{R}^n$, as it happens, any subspace is closed, i.e. if something is a subspace then it's closed (e.g. lines or planes in $R^3$). Is this true for any vector space? I find it difficult to visualise subspaces of e.g. $L^1$.
Many thanks for your help!! (as always ; ) )
 A: Suppose $v$ is in the closure of $W$. Then there exists a sequence $w_n$ of points in $W$ such that $||v-w_n||$ tends to 0. Since $W$ is a subspace, $-w_n$ is in $W$, so $||v+W||=\inf_{w \in W} ||v+w||=0$. If we want our norm on the quotient space to actually be a norm, rather than just a seminorm, then we need $||v +W|| =0$ to imply $v+W=0$, i.e. $v \in W$. So we need everything in the closure of $W$ to be in $W$, i.e. we need $W$ to be closed.
There certainly are natural examples of non-closed subspaces of normed spaces: For example, consider $L^1([0,1])$. The continuous functions form a proper subspace which is dense, and so not closed.
A: For a very simple answer to your second question, let $c_{00}$ be the set of infinite sequences of real numbers having only finitely many non-zero terms. Then $c_{00}$ is a subspace of $\ell^\infty$ that is not closed in $\ell^\infty$. One way to see this is to let $\sigma_n = \langle 1,1/2,1/3,\dots,1/n,0,0,\dots\rangle$ for each $n \in \mathbb{Z}^+$ and let $\sigma = \langle 1,1/2,1/3,\dots\rangle$. Then each $\sigma_n \in c_{00}$, and $\|\sigma-\sigma_n\|_\infty = \frac1{n+1} \to 0$ as $n\to\infty$, so $\sigma_n \to \sigma$ in $\ell^\infty$, but $\sigma \notin c_{00}$.
A: 
why does W have to be a closed subspace?

It is a theorem that the quotient space V/W of topological vector spaces is Hausdorff iff W is closed. For normed spaces this implies, as Chris Eagle says, that the quotient space will be a normed space iff W is closed.

Is this true for any vector space?

Every finite dimensional subspace of a topological vector space is closed, so you can have not closed subspaces in infinite dimensions only. You could try to find one in the complex separable infinite dimensional Hilbert space, as an exercise.

I find it difficult to visualise subspaces of e.g. $L^1$.

Ugh, well, what do you try to visualize? The geometry of infinite dimensional spaces is often counterintuitive. The easiest example is, as I said above, the separable infinite dimensional Hilbert space. You should try to look for examples for stuff that does not happen in finite dimensional geometry there first.
