Interval topology in subsets of totally orderable spaces

Abstract

A topological space is said to be totally orderable if the points of the space can be totally ordered in such a way that the interval topology induced by this ordering coincides with the given topology. Unfortunately, the property of beinga a totally orderable space is not hereditary. Suppose that R is a topological space and that T is a subspace of R and that R is totally orderable. Theorem II of this paper will answer question 1963.4 of Nieuw Archief Voor Wiskunde by giving necessary and sufficient conditions for T to be a totally orderable space. However, the most interesting case is where R is the real line; the solution in this case is quite simple to state and understand and is given in Theorem I. The bibliography lists several papers by I. L. Lynn each of which gives some partial results in the case where R is the real line. If A is a total ordering of a set T, then we will say that I is an interval of A(T) if I c T and no element of TI is between two elements of I in A. We will say that I is an open interval of A(T) if there is a last element of T I preceding every element of I if any element does so, and there is a first element of T I following every element of I if any element does so. If T is a topological space and A a total ordering of the points of T, then we will say that A(T) has interval topology at a point p of T if the open intervals of A(T) form a neighborhood basis for p in T. Then T is totally orderable if and only if there is a total ordering of T with interval topology at every point.