I am confused by the terminology concerning $n$-dimensional holes in algebraic topology. A circle is said to have a one-dimensional hole, and a sphere a two-dimensional hole for example. However I cannot see why the circle should be described to have a one-dimensional hole $-$ surely if drawn in two dimensions the 'gap' left in the middle of the circle is two-dimensional? If we think of the circle as a one-dimensional space only, then there is nowhere to have a 'hole' in the space?
[Math] $n$-dimensional holes
algebraic-topologydefinitiongeneral-topologyhomology-cohomologyterminology
Related Solutions
I think Alexander Duality is what you are looking for. I gather that you are a non-expert, so I will attempt to describe in fairly informal terms how Alexander duality deals with the questions that you are interested in. Consequently, I'll suppress the inevitable technicalities, since they don't enter into the very geometric situations that you are interested in.
Alexander duality deals with the following situation. Let $S^n$ denote the $n$-dimensional sphere. Note that you can think of the $n$-sphere as being standard $n$-dimensional space $\mathbb{R}^n$ with an extra point "at infinity" added in (take a look at stereographic projection if this is unfamiliar to you). The upshot of this is that working with the $n$-sphere is not too far away from the situation you are interested in. Now take some subspace $X$, like the $m$-balls you are removing. Or take anything else; a solid torus of whatever genus you like, higher-dimensional manifolds, etc. Let $Y$ denote the complement of $X$ inside $S^n$. Then, in very informal terms, Alexander duality asserts that homologically, $Y$ is exactly as complicated as $X$. Somewhat more technically, Alexander duality asserts that for all $q$, there is an isomorphism $$ \tilde H _q(Y) \cong \tilde H^{n-q-1}(X) $$ between the reduced homology of $Y$ and the reduced cohomology of $X$ (for whatever coefficient group we choose). If the meaning of this is somewhat unfamiliar to you, you might be more interested to know that this says that the Betti numbers of $Y$ can be computed from those of $X$. In the range $1 \le q \le n-2$, a consequence of Alexander duality is that $$ B_q(Y) = B_{n-q-1}(X), $$ where $B_k(Z)$ indicates the $k^{th}$ Betti number of a space $Z$. If the piece $X$ that you are removing has $p$ components, this also implies that $B_{n-1}(Y) = p-1$.
As I remarked before, Alexander duality says that the topology of a space obtained by cutting a piece out is, from the point of view of homology (which encapsulates Betti numbers) exactly as complicated as the piece being removed. It is interesting to note that from the point of view of homotopy theory, this is incredibly far from being true. A basic technique in knot theory is to study a knot by studying the topology of its complement in $\mathbb R^3$ (or $S^3$). Homologically, Alexander duality says this is very boring, but from the perspective of homotopy theory, the story is very interesting indeed.
$|\;\;|$ cylinder (understood as having no "cap" nor "base"; surely that's your issue)
$(\; \;)$ sphere with two holes.
Note that the surface around a cube (drawn by the faces) is homeomorphic to the (empty) cylinder with a cap and base, and it's homeomorphic to the sphere (without holes).
Best Answer
You got ample feedback on the notion of a "hole" in the comments. Here is my take.
Topologists do not use the "hole" terminology with two exceptions:
Kreck, Matthias, Differential algebraic topology. From stratifolds to exotic spheres, Graduate Studies in Mathematics 110. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4898-2/hbk). xii, 218 p. (2010). ZBL1420.57002.
Then Kreck, of course, gives rigorous definitions, none of which requires the "hole" terminology.
In order to appreciate how Kreck's notion of an extrinsic hole works, take your example of the unit circle $S^1$. The unit circle $S^1$ in the plane bounds the unit disk $$ X=D^2=\{(x,y): x^2+y^2\le 1\}. $$ The subset $L$ is the open unit disk $$ L=\{(x,y): x^2+y^2< 1\}. $$ Then $Y=S^1 = X - L$. Of course, you are right (and Kreck makes this point too): The "extrinsic hole" is not a part of the space $Y$ you are actually interested in. The next step in Kreck's informal definition involves "throwing a net" (mapping $S^1$ to $Y$ by, say, the identity map $f: S^1\to S^1$). The fact that one "cannot shrink the net to a point" is formally described by saying that the continuous map $f: S^1\to Y$ cannot be extended to a continuous map $F: D^2\to Y$.
This discussion leads to the notion of the fundamental group of a space.
To conclude: There is no mathematical definition of a hole here. Formal definitions define something else.
The precise meaning (there are some minor variations) of this notion is the following:
Suppose that $X$ is a compact connected (without boundary) surface without boundary. Consider a finite subset $P\subset X$ of cardinality $n\ge 1$. Then the surface $Y=X-P$ is said to be "a surface with $n$ holes." In other words, one says that a surface $Y$ has $n$ holes if there is a compact surface $X$ and a finite subset $P\subset X$ of cardinality $n\ge 1$ such that $Y$ is homeomorphic to $X-P$.
Another commonly used terminology here is an "$n$ times punctured surface."