Is every finite codimensional subspace of a Banach space closed? Is it also complemented? I know how to answer the same questions for finite dimensional subspaces, but couldn't figure out the finite codimension case.



Subspaces of finite codimension in Banach spaces
It's a standard result that a linear functional from a Banach space to the underlying field (real or complex numbers) is continuous if and only if the its kernel is closed. Notice that its kernel is of codimension one. So, use the axiom of choice to find a discontinuous linear functional, and you have found a codimension one subspace which isn't closed. (As I was typing this, rpotrie got the same answer...)


As for complementation: well, this only makes sense for *closed* finite codimension subspaces. But then it's a perfectly reasonable question, and the answer is "yes". If F is of finite codimension in E, then by definition we can find a basis $\{x\_1,\cdots,x\_n\}$ for E/F. For each $k$ let $x\_k^\*$ be the linear functional on $E/F$ dual to $x\_k$, so $x\_k^\*(x\_j) = \delta\_{jk}$. Then let $\mu\_k$ be the composition of $E \rightarrow E/F$ with $x\_k^\*$. Finally, pick $y\_k\in E$ with $y\_k+F=x\_k$. Then the map
$$T:E\rightarrow E; x\mapsto \sum\_k \mu\_k(x) y\_k$$
is a projection of $E$ onto the span of the $y\_k$, and $I-T$ will be a projection onto $F$ (unless I've messed something up, which is possible).


