$\dim_\text{topology}(\text{Cantor Sets}) \leq \dim_\text{hausdorff}(\text{Cantor Sets})$? Please, explain:

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?


*

*$\dim_\text{topo}(\text{Cantor sets}) = ?$

*$\dim_\text{Hausdorff}(\text{Cantor sets}) =?$ 

 A: First observe that the source that you linked to talks about separating spaces, not discriminating things.
Remove a point from a line (a point that is not an end point) and you get two pieces (which are open in the subspace topology of the line).
But the Cantor space is already totally disconnected:  every subset with at least two points is disconnected.  No need to remove anything to separate it.
Hence it is zero dimensional.
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?
This is a metric concept.  You can construct variants of the Cantor set
that have different Hausdorff dimensions.  These variants are topologically the same as the classical middle-thirds Cantor set, but not metrically.
You can even construct a Cantor set of Hausdorff dimension 0, by taken out not middle thirds, but larger and larger chunks of the remaining intervals.
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 $\mathbb R^n$.  
