Closures and boundaries

Today, I was working on some technical report (an internal publication for the university) for my PhD. I was dealing with topology of sets in metric spaces, which is a special case of general topology, well suited just for my needs, a computer graphics field.
I had minor troubles with a definition of a set boundary, but I think I finally got it right and actually some things surprised me a bit - that's why I'm sharing it here!
At first I was saying about boundary of the set A without mentioning about the metric space X that contains it. Usually I'm operating in R^n euclidean space, and all of my sets was subsets of R^n. But this time, I was dealing with a space constructed from a subset of R^n, specifically, a unitary box: [0..1]^n, further called I^n space, I={x|0<=x<=1}, where x is a real number. Let Y be a continuously deformed ball that is a subset of I^n . I was "intuitively" (by analogy to R^n space) thinking that the boundary of Y should be always connected (or even simply-connected). You can not imagine a ball that has disconnected boundary, right ?
As usual, it all depends on the definition on the boundary.
In topology, the boundary can be defined several ways that are equivalent for metric spaces. We can for example say, that a boundary of a set Y included by a metric space X is a set of points in X for which any open ball (with radius > 0) contains points both inside Y and outside Y.
To define an open ball in X you surely need to define a metrics of X. If it is euclidean R^n space, a ball "looks" like a ball = it is a sphere, if not, it can be almost "anything". But stick to the euclidean first. The I^n set has a boundary in R^n. It is an "empty" box. But any open ball in euclidean I^n space will not always be an open sphere - near the boundary of I^n in R^n, it will be just a part of a sphere cut by the walls of I^n box. Now, if we take any connected subset Y of I^n that touches the walls of I^n box, the set of points at the walls of I^n box will be not a part of boundary of Y! This is why we can construct Y that will have a disconnected boundary in I^n, an example provided below:


In the figure above, a boundary is denoted as a thicker line. Note that any open ball co-centered with "a" and smaller than "a" will contain only points inside Y, thus the center of "a" is not a part of the boundary of Y. We can also clearly see that the center of "b" lies on the boundary of Y (any ball around it contains point both inside and outside Y). Obviously the boundary of Y' in R^n is connected.