# Dimensions and Topology

Back in the early days of Good Math/Bad Math, when it was still at blogger, one of the most widely linked posts was one about the idea of dimension. At the time, I said that the easiest way to describe a dimension was as a direction. If you’ve got a point in a plane, and you want to say where it is, you can do it with two numbers – one for each of the fundamental directions in the plane. If you’ve set an origin, “(5,-2)” is enough to uniquely identify exactly one point. You con reach any point on the plane by moving in two directions: up/down and left/right.

If you’ve got a cube, you can’t uniquely specify a point using its distance in two directions. Up three and left two doesn’t give you one point – there are lots of points that are up three and left two. You need a third direction, forward/back, for depth. That’s the third dimension – a direction that could not be formed by any combination of the two directions you had in the plane.

Topology has its own sense of dimension – in fact, it has several. They’re interesting because, as happens so often in topology, they start with the intuition that we get from simple metric spaces like ℜn, and work it down to its bare essentials by figuring out what it means when you apply it to an arbitrary topological space – that is, an arbitrary structure formed from open sets.

Let’s start by quickly reviewing a couple of important definitions.

1. If we have a topological space $\mathbf{T}$ then an open cover $\mathbf{C}$ of that space is a collection of open subsets $c_i \in \mathbf{C}$ of $\mathbf{T}$ such that the union of all of the subsets is equal to $\mathbf{T}$ (that is, $\mathbf{T}=\cap_{i\in C} c_i = \mathbf{T}$).
2. Given an open cover $\mathbf{C}$, a refinement of $\mathbf{C}$ is another open cover $\mathbf{D}$ such that $\forall d \in \mathbf{D}: \exists c \in \mathbf{C}: d \subset c$; that is, another open cover where every one of its members is a subset of some member of the original cover: none of $\mathbf{D}$s members will span across members of $\mathbf{C}$.

The fundamental concept of dimension in topology comes from open covers. Suppose we have a topological space **T**. **T** has *topological dimension* *d* if and only if for every open cover *X* of **T**, *X* has a refinement *Y*={y1,…,yn} such that there are *no* points in **T** that are members of more than d+1 yis.

That *sounds* horrible. But if we pull back a moment to ℜi, what that really says is that if we’re allowed to shrink open subsets that make up a cover down any way we want, then we can find ways in which for any point *p* we won’t be able to find d+1 open subsets including *p* where none of them are subsets of any of the others.

So, for example, let’s think about a way of doing covers for ℜ2. The most natural one is just the set of all open balls in the plane. Can we cover the plane with a set of open balls (filled circles without their boundary) so that no point is in more that *2* balls?

Let’s try it for a subset of the plane – let’s take a square region.

Now: can we find *any* point in the square where we *can’t* find a refinement where no more than *three* distinct lines include a point *p*? Can we find any point *q* in the square where we can’t find a refinement where *q* is in more than 2 sets?

Try it!

Here’s a perfectly good cover refinement of the open spheres in ℜ2 covering a square. As the arrows point out, there are regions where only one sphere covers, regions where two overlap, and regions where three overlap.

But if we try to do it so that entire plane is covered, but every point is inside of only two circles. No way. Look at this diagram: you the cover is show to the left so that you can see the open circles and the square; to the right, I’ve made the circles a dark opaque so that you can clearly see the gaps. You *can’t* close those gaps without making two circles on opposite sides of the gap overlap on at least one point.

If you think about it, that is, essentially, the same thing as the definition of dimension as direction. The topological definition is a very abstract way of saying something like “If the open set relationships happen to define a metric space, and you want to define every point in the space in terms of its distance from a particular point in different directions, then the maximum number of directions that you can define that *can’t* be formed from any combination of the other directions is *d*.” That’s exactly what we started off with – it’s just pared down.

Of course, that’s not really quite enough. There are some strange cases where the topological dimension just seems *wrong*. Two of the classic examples are the Cantor space and the Sierpinski triangle.

The cantor space is what you get if you take the line segment from 0 to 1, cut it into thirds, and remove the middle third. Then take the two remaining sections, cut *them* into thirds, and remove the middle. And keep going, forever.

The Sierpinski triangle is similar: take an equilateral triangle. Mark the midpoints of the three sides; then *remove* the area of the triangle formed by connecting those sides. Then take each of the three remaining triangles, and do the same thing – and keep repeating. So, for example, here’s the sierpinski triangle after four iterations.

There’s a good chance that these things sound sort of familiar. They’re both very simple examples of *fractals*. And what makes a fractal a fractal is that in some way, for some dimensionality *d*, it’s got *more* than *d* dimensions, but *less* that *d+1* dimensions. The Sierpinski triangle has 1 topological dimension; but it seems to have more than that. The cantor set has topological dimension *0*. But it certainly doesn’t *seem* like it’s got less than one dimension!

Topological dimension is a good idea, an idea which is useful, and reasonably easy to work out. So we like it, and we use it for most spaces without any problem. But sometimes, for what we consider to be sort of pathological spaces, it produces a result that just doesn’t seem to make sense.

Enter *Hausdorff* dimension. The Hausdorff dimension is an attempt to define a sense of dimension that captures the anomalous structure of fractal sets. The complete definition of it is incredibly hairy, so I’m not going to go into the full mess of it; but the application of it to metric spaces is reasonably easy to explain.

So, suppose we’ve got some metric space **M** that’s a subset of ℜn. Given a radius *r*, how many open balls of radius *r* does it take to cover **M**? Call that value N(*r*). Now, look at N(*r*), and see how it changes as *r* gets smaller. Eventually, it will smooth out into a regular curve proportional to *r-d* for some real number *d*. That number *d* is the Hausdorff dimension of **M**.

If you take the Hausdorff dimension of a regular topological space, then it will have the same value as the topological dimension. But if the topological space has anomalous structure to it, so that it’s effective dimensionality is larger than its topological dimension, then the Hausdorff dimension will be larger that the topological dimension. (And in fact, the definition of a fractal is a topological space whose Hausdorff dimension is *larger* than its topological dimension.)

## 0 thoughts on “Dimensions and Topology”

1. James R

You’ve actually defined the self-similarity dimension of M in the second to last paragraph, which is equal to the Hausdorff dimension only for iterated function systems (which are similitudes and satisfy the open set condition)*.
As you’ve said, the actual definition is a bit hairy, but it can be applied to non self-similar objects, whereas the SSD cannot (partly why the box dimension is so useful, since calculating the Hausdorff dimension of a lightning bolt is probably impossible).
*- Hutchinson’s Theorem

2. Mark Chu-Carroll

James:
Yeah, I know that my Hausdorff definition is a bit of a cheat, as I said in the post 🙂 But getting it *right* is really hard, and I think it’s just *too* hard to be written well at the level that I try use when to write this blog.

3. andy.s

In the Cantor set from 0-1, you can cover it with one set of size 1.
If you use sets of size 1/3, you need 2 to cover (0,1/3) and (2/3, 1)
If you use sets of size 1/9, you need 4: (0,1/9), (2/9,1/3)
(2/3,7/9) and (8/9, 1)
In general with a scale of 1/3^n, number of sets = 1/2 ^n
so the Hausdorff dimension is log 1/2^n / log 1/3^n =
-n log 2 / -n log 3 = log 2 / log 3 = a dimension of about 0.63 .
Is that right, teacher?

4. Mark Chu-Carroll

AndyS:
Yup, you got it.
Alon:
I thought I was throwing in too many different intuitions already, and the open-ball intuition that you suggested is recursive, which is a real problem for some readers.

5. Coin

Are there any workalikes of Hausdorff dimension (i.e., which measure a non-integer dimension) which work in non-metric spaces?