On the reverse mathematics of general topology

by 1978- Mummert, Carl

This thesis presents a formalization of general topology in second-order arithmetic. Topological spaces are represented as spaces of filters on partially ordered sets. If P is a poset, let MF(P ) be the set of maximal filters on P . Let UF(P ) be the set of unbounded filters on P . If X is MF(P ) or UF(P ), the topology on X has a basis {Np | p ? P }, where Np = {F ? X | p ? F }. Spaces of the form MF(P ) are called MF spaces; spaces of the form UF(P ) are called UF spaces. A poset space is either an MF space or a UF space; a poset space formed from a countable poset is said to be countably based. The class of countably based poset spaces includes all complete separable metric spaces and many nonmetrizable spaces including the Gandy–Harrington space. All poset spaces have the strong Choquet property. This formalization is used to explore the Reverse Mathematics of general topology. The following results are obtained. RCA0 proves that countable products of countably based MF spaces are countably based MF spaces. The statement that every G? subspace of a countably based MF space is a countably based MF space is equivalent to ?1 1-CA0 over RCA0. The statement that every regular countably based MF space is metrizable is provable in ?1 2-CA0 and implies ACA0 over RCA0. The statement that every regular MF space is completely metrizable is equivalent to ? 1 2-CA0 over ?1 1-CA0. The corresponding statements for UF spaces are provable in ?1 1-CA0, and each implies ACA0 over RCA0. The statement that every countably based Hausdorff UF space is either countable or has a perfect subset is equivalent to ATR0 over ACA0. ?1 2-CA0 proves that every countably based Hausdorff MF space has either countably many or continuum-many points; this statement implies ATR0 over ACA0. The statement that every closed subset of a countably based Hausdorff MF space is either countable or has a perfect subset is equivalent over ? 1 1-CA0 to the statement that ?L(A) 1 is countable for all A ? N. iii