A direct proof that a linearly ordered space is hereditarily collectionwise normal
1970; American Mathematical Society; Volume: 24; Issue: 4 Linguagem: Inglês
10.1090/s0002-9939-1970-0257985-7
ISSN1088-6826
Autores Tópico(s)Rings, Modules, and Algebras
ResumoAlthough it appears well known that a linearly ordered space is completely normal (=hereditarily normal), most available proofs (in, for instance, [1] and [2]) are very indirect. In this paper we present a direct proof of a stronger theorem, namely that the interval topology is hereditarily collectionwise normal.2 If X is linearly ordered, we will call a set SCX convex if a, b E S and a<t<b implies tES. The union of any collection of convex sets with nonempty intersection is convex, so any subset S of X can be uniquely expressed as a union of disjoint maximal convex sets called convex components. Clearly every interval in X is convex but not conversely, and we will, as usual, denote intervals by (a, b), (a, b], [a, b), or [a, b]. In what follows, X will denote a linearly ordered space, i.e., a linearly ordered set endowed with the usual open interval topology. Suppose { A i } is a discrete family of subsets of X. Let
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