Let $V\_0$ be a closed infinite-dimensional subspace of a Banach space $V$ such that the quotient $V/V\_0$ is also infinite-dimensional. Is it always possible to find a closed subspace of $V$ whose sum with $V\_0$ isn't closed?


A positive solution would let me answer [this question](https://mathoverflow.net/questions/431740/under-what-conditions-does-a-continuous-linear-map-map-a-closed-subspace-to-a-cl).



Finding closed subspaces whose sum isn't closed
Probably Spyros has in mind something like the following.


Suppose you have a semi-normalized basic sequence $(x\_n)$ in $V$ with biorthogonal functionals $(f\_n)\_n$
in $V\_0^\perp \subset V^\*$. Take any normalized basic sequence $(y\_n)\_n$ in $V\_0$. If $\epsilon\_n \to 0$ sufficiently quickly with all $\epsilon\_n >0$, then $(y\_n + \epsilon\_n x\_n)\_n$ is a basic sequence that is even equivalent to $(y\_n)\_n$ and $(\epsilon\_n^{-1} f\_n)\_n$ are biorthogonal to $(y\_n + \epsilon\_n x\_n)\_n$. Since $(y\_n + \epsilon\_n x\_n)\_n$ is basic, $(f\_n)\_n$ separates the points in the closed linear span $W$ of $(y\_n + \epsilon\_n x\_n)\_n$, and hence $W \cap V\_0 =\{0\}$. By construction, $W+V\_0$ is not closed. (If it were closed, then its image under the projection from $V$ to $V/V\_0$ would be closed and thus the projection would take $W$ isomorphically onto its image, but $\|y\_n + \epsilon\_n x\_n\| \to 1$ and $\|\epsilon\_n x\_n\| \to 0$.)


To get such $(x\_n)$ and $(f\_n)$, pull back to a semi-normalized sequence $V$ any normalized basic sequence in $V/V\_0$. That sequence in $V$ might not be basic, but you can pass to a subsequence of differences that is basic; see e.g. the early part of the chapter on basic sequences in the book of Albiac and Kalton.


