Kernels of operators on Banach spaces induced by almost disjoint families

Abstract

Let~$\mathcal{A}$ be an almost disjoint family of subsets of an infinite set~$\Gamma$, and denote by~$X_{\mathcal{A}}$ the closed subspace of~$\ell_\infty(\Gamma)$ spanned by the indicator functions of intersections of finitely many sets in~$\mathcal{A}$. We show that if~$\mathcal{A}$ has cardinality greater than~$\Gamma$, then the closed subspace of~$X_{\mathcal{A}}$ spanned by the indicator functions of sets of the form $\bigcap_{j=1}^{n+1}A_j$, where $n\in\N$ and $A_1,\ldots,A_{n+1}\in\mathcal{A}$ are distinct, cannot be the kernel of any bounded operator \mbox{$X_{\mathcal{A}}\rightarrow \ell_{\infty}(\Gamma)$}. As a consequence, we deduce that the subspace \[ \bigl\{ x\in \ell_{\infty}(\Gamma) : \text{the set}\ \{\gamma \in \Gamma : \lvert x(\gamma)\rvert > \varepsilon \}\ \text{has cardinality smaller than}\ \Gamma\ \text{for every}\ \varepsilon>0\bigr\} \] of~$\ell_\infty(\Gamma)$ is not the kernel of any bounded operator on~$\ell_\infty(\Gamma)$; this generalises results of Kalton and of Pe\l{}czy\'{n}ski and Sudakov. The situation is more complex for the Banach space~$\ell_\infty^c(\Gamma)$ of countably supported, bounded functions defined on an uncountable set~$\Gamma$. We show that it is undecidable in \textsf{ZFC} whether every bounded operator on~$\ell_\infty^c(\omega_1)$ which vanishes on~$c_0(\omega_1)$ must vanish on a subspace of the form~$\ell_\infty^c(A)$ for some uncountable subset~$A$ of~$\omega_1$.