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