From: ikastan@alumni.caltech.edu (Ilias Kastanas) Newsgroups: sci.math Subject: Re: Topological Dimension Date: 19 Jul 1997 15:30:19 GMT In article <869188816.15145@dejanews.com>, Toby Bartels wrote: >What exactly is topological dimension? >I've heard that the empty set has dimension -1 >and every other set has dimension 1 more than its boundary's. >This obviously doesn't make sense in the context of general topology, >where the boundary of any entire space is empty, >even if the space is not what you'd want to call 0 dimensional. >So what is the context of topological dimension, >and how exactly is it defined? The "inductive" dimension ind(X) is defined by ind(0) = -1 and ind(X) <= n+1 iff every x in X has arbitrarily small neighborhoods whose boundaries have ind <= n. So this makes sense in any metric space. An equivalent formulation that also works in arbitrary topological spaces is "for every open U contai- ning x there is an open V, subset of U, containing x with ind(boundary(V)) <= <= n. An alternative is the "great inductive" dimension Ind(X): Ind(X) <= <= n+1 iff for every closed F and open U, F subset of U, there is an open V, F subset V & closure(V) subset U, so that Ind(boundary(V)) <= n. There are other alternatives, but never mind. Now one has some expectations. Dimension should be a topological invariant; it should do the right thing for R^n; it should be monotonic (dimension of a set >= dimension of a subset); the dimension of a countable set should be 0; stuff like that. It turns out that only for separable metric spaces does one get a decent theory (ind and Ind are then equivalent). A quick visit to the Zoo of Topology shows why. Okay, ind() is monotonic... but there is a countable Hausdorff space H that is connected, and therefore has ind(H) > 0. There is also the Tychonoff plank, T = [0, w] x [0, w_1], and the deleted version, T' = T - (w, w_1). T is compact Hausdorff and hence normal; T' is not normal. Easily, ind(T) = 0 (and hence ind(T') = 0). Compactness implies Ind(T) = 0. But Ind(T') > 0, because T' is not normal. So Ind() is not monotonic. That is the way it goes... Ilias