# $\dim_\text{topology}(\text{Cantor Sets}) \leq \dim_\text{hausdorff}(\text{Cantor Sets})$?

the Cantor set (a zero-dimensional topological space) is a union of two copies of itself, each copy shrunk by a factor 1/3; this fact can be used to prove that its Hausdorff dimension is $\ln2 / \ln3$, which is approximately 0.63 The Sierpinski triangle is a union of three copies of itself, each copy shrunk by a factor of 1/2; this yields a Hausdorff dimension of $\ln3 / \ln2$, which is approximately 1.58.

Source Wikipedia.

I understand this in a way that topological dimension is a measure of how to discriminate objects, more here, $\dim_\text{topo}(\emptyset) = -1$, $\dim_\text{topo}(\cdot)=0$ because you need nothing to discriminate point, $\dim_\text{topo}(|)=1$ because you need a point to discriminate a line. Similarly for:

• $\dim_\text{topo}(\#)= 1$ because you need four points to discriminate it and the supremum of the local dimesion is 1.
• $\dim_\text{topo}(\text{keyboard}) = 3$ because I need a plane to discriminate it.

But what about them, how can I use similar logic as above to discriminate them?

• $\dim_\text{topo}(\text{Cantor sets}) = ?$
• $\dim_\text{Hausdorff}(\text{Cantor sets}) =?$
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The Cantor set is totally disconnected; you need nothing to separate parts of it, so the topological dimension will be $0$. –  Arturo Magidin Sep 22 '11 at 4:03

The Hausdorff dimension measures the growth rate of the following function as $R$ tends to $0$: How many balls of radius $R$ are needed to cover the whole space?
The topological dimension is a topological concept and is independent of the metric on your space (independent in the sense that you get the same dimension when switch to a different metric that gives the same open sets). But observe that there are various topological notions of dimension around. There is the Lebesgue covering dimension, the small inductive dimension, and so on. These different notions agree on non-pathological spaces such as $\mathbb R^n$.