For school, I write a kind of ‘paper’ about fractals. I’m currently writing about the definition of fractals, but there’s something I don’t understand.
A shape is a fractal, when its Hausdorff dimension is greater than its topological dimension. For example, the Koch curve has en Hausdorff dimension of $\approx 1.262$, and a topological dimension of 1 (because it is a line). Because the Hausdorff dimension is greater than its topological dimension, the Koch curve is considered a fractal.
However, the Cantor set has a Hausdorff dimension of $\approx 0.631$, and a topological dimension of 1, because it is made out of lines (am I right?). This means that its Hausdorff dimension is smaller than its topological dimension, and, therefore, shouldn’t be considered a fractal. But it is a fractal. What am I missing here? Or does its topological dimension not equal 1?
There are several definitions of dimension :
The small inductive dimension is defined inductively. Let's show that the Cantor set has small inductive dimension $0$: by definition, this means that for every point $x$ and every open $U$ containing $x$ , there exists an open set $V$ containing $x$, whose closure is included in $U$, and such that the small inductive dimension of $\partial V$ is $\leq -1$ (i.e. $\partial V = \emptyset$) .
But the Cantor set is totally disconnected : it has a basis of clopen sets, hence for $x\in U$ such as above, there is a clopen $V\subset U$ containing $x$. For this $V$, $\partial V = \emptyset$ (because $V$ is clopen), and so we find that the Cantor set has small inductive dimension $0$.
The large inductive dimension has the same definition, except that you begin with a closed set $C\subset U$.
But the proof works in the same way, because the Cantor set is compact and Hausdorff, and so $C$ is as well, therefore you can cover it by a finite number of clopen basis sets that are all included in $U$, and their (finite !) union will be clopen as well, so it works too : the large inductive dimension of the Cantor set is $0$ as well.
Then there's also the Lebesgue covering dimension (I think it's the most common notion of topological dimension) : it is defined to be the minimum value of $n$, such that any open cover has a refinement in which no point is included in more than $n+1$ opens, if such $n$ exists. For the Cantor set, this will again yield $0$. Indeed, for metrizable spaces, this dimension is the same as the large inductive dimension: you can also prove it by induction on the size of the open covers if you want.
The intuition for why a totally disconnected space has dimension $0$ is (to me) the following: if you start at a point, you can't draw any lines or anything leaving this point because you're trapped in your clopen neighbourhoods that get closer and closer to you (that's what the first two definitions of dimension try to capture): contrary to what you say, there are no lines in the Cantor set, "only points" in some sense.