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

  • $\begingroup$ By isomorphism, you mean isometrically isomorph or linear continuous bijection (whose inverse is hence also going to be continuous) ? $\endgroup$ Commented Dec 1, 2014 at 14:15
  • $\begingroup$ Thanks for your comment. I meant topological isomorphisms and edited the question accordingly. $\endgroup$
    – Tom
    Commented Dec 1, 2014 at 14:41
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    $\begingroup$ $E/F$ is isomorphic to each one-codimensional closed subspace of $E$, and does not need to be isomorphic to $E$. For example, when $E$ is the hereditarily indecomposable space of Gowers and Maurey, $E/F$ is not isomorphic to $E$. $\endgroup$ Commented Dec 1, 2014 at 15:09

2 Answers 2


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.


  • 4
    $\begingroup$ It seems user62448 does not exist! $\endgroup$ Commented Dec 1, 2014 at 15:21
  • 2
    $\begingroup$ The Masked Answerer? $\endgroup$ Commented Dec 1, 2014 at 15:39

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.

  • $\begingroup$ Thank you. It is really intersting to see that this beviour not only accurs for "complicated" Banach spaces but also for "natural" Banach spaces like C(K) (even for complicated and unnatural K). $\endgroup$
    – Tom
    Commented Dec 9, 2014 at 8:49

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