Covering number definition, general metric space question. While reading "On the mathematical foundations of learning" by F.Cucker and S.Smale I came across this definition:
Let $S$ be a metric space and $s>0$. We define the covering number $\mathcal{N}(S,s)$ to be the minimal $ l \in \mathbb{N}$ such that exist $l$ disks in $S$ with radius $s$ covering $S$. 
My questions are kind of silly. The first one is: are those disks are to be understood as open balls?
Second one: I do not know how to interpret that those disks are "in $S$". They cannot be subsets of S, because if they were, the covering number would be infinite if S wasn't a disk (open ball?) itself. 
 A: Let $X$ be a metric space. When the authors are talking about 'open discs' in $X$, they mean sets of the form $B(x_0, r) = \{x \in X : d(x,x_0) < r\}$. In Euclidean space $\Bbb R^n$ these actually look like discs; in other spaces, they don't need to. (Consider, for instance, the torus with the metric it inherits from $\Bbb R^3$; if you pick a ball around the point closest to us with radius just big enough to contain every point on the 'meridian circle' it belongs to, this neighborhood will look like $S^1 \times (-1,1)$, roughly.) 
So we define $\mathcal N(X,r)$ to be the minimal number of balls of radius $r$ we need to cover our whole space. Equivalently, this is the minimal number of points we need to pick such that every other point of our space is within $r$ of one of them. On the unit circle with the metric it inherits from $\Bbb R^2$, if we set $r=1.5$, this is two; the top and bottom of the circle suffice for our two points, but because the diameter of the circle is $2$, picking one point does not suffice.
For what's called a compact space, this $\mathcal N(X,r)$ is always finite, for any $r>0$. To be compact means that for any collection of open sets $\{U_\alpha\}$ such that $U_\alpha$ cover $X$ (i.e., every point is contained in some $U_\alpha$), there is a finite subcollection $U_1, \dots, U_n$ that also covers $X$. Some familiar examples of compact spaces are spheres $S^n$, the torus, boxes $[-1,1]^n$...
Why is $\mathcal N(X,r)$ automatically finite? Consider the collection $\{U_x\}$, where $U_x = B(x,r)$ - so we have one open set for every point. By the assumption that $X$ is compact, we can pick a finite subcollection of this that covers all of $X$. So we're done!
