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  <toby@ugcs.caltech.edu> 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