Is it true that, assuming the Axiom of Choice, every infinitedimensional Banach space has an infinitedimensional closed subspace with infinite codimension? Note that this is different from the indecomposability problem, which asks whether every infinitedimensional Banach space has an infinitedimensional closed subspace with a closed infinitedimensional complement (known to be false). This is relevant since with coauthors, and with settheoretic assumptions incompatible with the Countable AC, we have an example of an infinitedimensional Hilbert space in which every closed subspace has either finite dimension or finite codimension.

4$\begingroup$ I should point out that if the Countable AC is assumed, it suffices to answer the question for separable Banach spaces, where no choice may be needed. $\endgroup$– Bruce BlackadarApr 6 at 18:41

1$\begingroup$ Construct an infinite basic sequence $(x_i)$ in the space and take the closed linear span of $(x_{2n})$. The construction is Mazur's argument, and HahnBanach is used. $\endgroup$– Bunyamin SariApr 6 at 18:50

1$\begingroup$ I don't think this works unless it is an unconditional basic sequence. If it did, there couldn't be a hereditarily indecomposable Banach space. $\endgroup$– Bruce BlackadarApr 6 at 19:00

1$\begingroup$ Why does not it work? If this span has a finite codimension, it contains a finite linear combination of $x_1,x_3,\ldots,x_{2m+1}$ for large $m$, say, $\sum_{i=1}^m c_i x_{2i1}$ is arbitrarily close to finite linear combinations of $x_2,x_4,\ldots$. If, say, $c_j\ne 0$, this means that the $(2j1)$st coordinate functional is unbounded (on the span of the basic sequence): there are arbitrarily small elements for which it equals $c_j$. $\endgroup$– Fedor PetrovApr 6 at 19:26

1$\begingroup$ If it worked, in the closed subspace spanned by all the x_n, the closed subspaces spanned by the even and odd ones would be complementary closed subspaces. $\endgroup$– Bruce BlackadarApr 6 at 19:39
2 Answers
Yes, I think this is true. Any infinite dimensional Banach space $V$ contains a basic sequence $(x_n)$. Then $\{x_1, x_3, x_5, \ldots\}$ is linearly independent and therefore its closed span $V_0$ is infinite dimensional. The codimension of $V_0$ must be infinite, otherwise there would be a linear dependence among $\pi(x_2), \pi(x_4), \pi(x_6), \ldots$ where $\pi: V \to V/V_0$ is the natural projection, and this would lift to make some finite linear combination of $x_2, x_4, x_6, \ldots$ with at least one nonzero coefficient belong to $V_0$, which is impossible.
New edit: Bruce has pointed out that it isn't obvious that a nonzero finite linear combination of $x_2, x_4, x_6, \ldots$ cannot belong to $V_0$. Indeed it isn't obvious, but in fact this cannot happen. Let $x = a_2x_2 + \cdots + a_{2n}x_{2n}$ be a finite linear combination with at least one nonzero coefficient. Then its distance to ${\rm span}(x_1, \ldots, x_{2n1})$ is strictly positive, greater than some $\epsilon > 0$. Thus if we fill out $a_2, \ldots, a_{2n}$ to a sequence $(a_n)$ in any way, the norm of $a_1x_1 + a_2x_2 + \cdots + a_{2n}x_{2n}$ must be at least $\epsilon$.
Now we use the fact that since $(x_n)$ is a basic sequence there exists $K > 0$ such that $\left\\sum_{i=1}^{2n} a_ix_i\right\ \leq K\left\\sum_{i=1}^m a_ix_i\right\$ for any $m \geq 2n$ (see the first answer to this question). This shows that the span of any finite subset of $\{x_1, x_3, x_5, \ldots\}$ is at least $\epsilon/K$ away from $x$, and therefore $x$ cannot belong to $V_0$.

1$\begingroup$ No, this does not work. The problem is that, unless the basic sequence is unconditional, or something similar, if a vector is written as an infinite series in the basis vectors, the sum of the even terms does not necessarily converge. Thus there is no "natural projection" onto the closed span of the even basis vectors, or, more precisely, the natural projection from the algebraic span of the basis vectors to the span of the even vectors is unbounded. This is precisely what makes this problem (apparently) difficult. $\endgroup$ Apr 7 at 3:01

$\begingroup$ It is not at all clear that even the odd basis vectors are not in the closed span of the even ones. $\endgroup$ Apr 7 at 3:08

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2$\begingroup$ That means a lot to me, coming from a brilliant guy like you :) $\endgroup$ Apr 7 at 4:27

2$\begingroup$ To clarify one point: while it is true that in general there is no projection onto even subsequence, any subsequence of the basis itself is a basic sequence. $\endgroup$ Apr 7 at 4:44
Let $(x_i)$ be a normalized basic sequence with basis constant 2, and denote by $X$ its closed linear span. Consider $[x_{2i}]$, the closed linear span of the even sequence. Suppose for a contradiction that there exists a finite dimensional subspace $F\subset X$ such that $F\oplus [x_{2i}]=X$. Let $0<\varepsilon <1/2$. By standard approximations there exists even $N$ such that for every $f\in F$ there exists some scalars $(a_i)_{i=1}^N$ such that $\f\sum_{i=1}^N a_i x_i\<\varepsilon$. Consider $x_{N+1}$. By our assumption we must have $\x_{N+1}(\sum_{i=1}^N a_ix_i +\sum_{i=1}^M b_ix_{2i})\<\varepsilon$ for some $(a_i)_{i=1}^N,(b_i)_{i=1}^M$ ($M$ is chosen to approximate a vector in $[x_{2i}]$). But if you write out what is inside the norm explicitly in terms of the basis, you see that it is $$\(a_1, a_2+b_1, \ldots, a_N+b_{N/2}, 1, b_{N/2+1}, \ldots, b_M, 0, \ldots)\$$ which is greater than 1/2 (using basis projection onto the $(N+1)$th coordinate), which is a contradiction.
(I was typing when Nik posted his answer:))