Continuum Without Non-Block Points

Abstract

For any composant $E\subset {\mathbb{H}}^{\ast }$ and corresponding near-coherence class $\mathcal{E}\subset {\omega }^{\ast }$ we prove the following are equivalent : (1) $E$ properly contains a dense semicontinuum. (2) Each countable subset of $E$ is contained in a dense proper semicontinuum of $E$. (3) Each countable subset of $E$ is disjoint from some dense proper semicontinuum of $E$. (4) $\mathcal{E}$ has a minimal element in the finite-to-one monotone order of ultrafilters. (5) $\mathcal{E}$ has a $Q$-point. A consequence is that NCF is equivalent to ${\mathbb{H}}^{\ast }$ containing no proper dense semicontinuum and no non-block points. This gives an axiom-contingent answer to a question of the author. Thus every known continuum has either a proper dense semicontinuum at every point or at no points. We examine the structure of indecomposable continua for which this fails, and deduce they contain a maximum semicontinuum with dense interior.