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



Banach space modulo a one-dimensional subspace =?
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>


