Banach space modulo a one-dimensional subspace =? My question is the following:
Given an infinite dimensional Banach space $E$ and a one-dimensional linear subspace $F\subset E$. It is well-known that this one-dimensional linear subspace is closed and always complemented. So, we have the following topological isomorphism:
$$E \cong F \times E/F.$$
Now, what is the isomorphism type of $E/F$ ? In particular, is $E/F$ isomorphic to $E$ ? 
I know this is true for all Hilbert spaces since they admit orthonormal bases and I would expect that the existence of some sort of "topological basis" could be related to this question but so far I could not find anything. Furthermore, I did not know where to look for an answer to this question.
By the way: Of course, I believe in the axiom of choice.
EDIT: All isomorphisms are meant to be topological isomorphisms, i.e., linear homeomorphisms. Isometric isomorphism are too restrictive, I would assume (although in the Hilbert space case, one can even get isometric isomorphisms, but I do not need that)
Thank you in advance,
Tom
 A: If you consider the spaces constructed by Gowers, Maurey and others exotic, then it turns out that the answer is also negative in the class of $C(K)$-spaces. Indeed, Koszmider constructed a compact, Hausdorff space $K$ so that $C(K)$ is not isomorphic to its hyperplanes:

P. Koszmider, Banach spaces of continuous functions with few operators. Math. Ann. 
  330, No. 1 (2004), 151–183 .

Of course there is a price you have to pay for this–$C(K)$ is inseparable (i.e. $K$ is non-metrisable) and it is easy to see that a separable $C(K)$-space will never have this property as it contains a complemented copy of $c_0$ which apparently has this property. 
A: Gowers proved in "A solution to Banach’s hyperplane problem" (1994)
An infinite-dimensional Banach space X is constructed which is not isomorphic to X ⊕ R. Equivalently, X is not isomorphic to any of its closed subspaces of codimension one. This gives a negative answer to a question of Banach. In fact, X has the stronger property that it is not isomorphic to any proper subspace. It also happens to have an unconditional basis. 
http://blms.oxfordjournals.org/content/26/6/523.abstract
