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



Closed subspaces of Banach spaces
Yes, I think this is true. Any infinite dimensional Banach space $V$ contains a [basic sequence](https://matthewdaws.github.io/files/notes/bases.pdf) $(x\_n)$. 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 ${\rm span}(x\_1, \ldots, x\_{2n-1})$ 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](https://math.stackexchange.com/questions/4615388/subset-of-a-basis-for-a-normed-vector-space-cannot-be-cauchy)). 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 $V\_0$.


