What is a saturation of a set?

So, I encountered in my topology book the saturation of a set and In my first language the translated word is rarely used and those papers I found who use it don't explain it since it seems to be something very basic. Unfortunately the part where the saturation is mentioned doesn't really give many information about what it could mean.

My Question: When do we call a set saturated? (I guess I understand it to a certain extend) What is the saturation of a set (which is not saturated)?

I hope someone can explain it to me. The corresponding wikipedia entry http://en.wikipedia.org/wiki/Saturated_set only explains when you call a subset of a topoligical space saturated but doesn't explain how to achieve the saturation of a set which isn't saturated.

-
I am not familiar with the first definition given on Wikipedia, but the second one reminds of the saturation w.r.t. partition (or a function that induces that partition). Let $f:X\to Y$ be a function - then it has certain level sets, i.e. $f^{-1}(y)$ may not be a single point of $X$ but a collection thereof - think of $f(x) = x_1^2 + x_2^2+x_3^2$ which level sets are spheres. The set $A\subset X$ is saturated w.r.t. $f$ if it is a union of level sets. E.g. in the example that I gave every ball and ring is a saturated set, but boxes are not. – S.D. Nov 29 '12 at 17:06
From what I know a subset $A$ of a set $X$ is saturated with respect to a map $f:X\to Y$ if $A=f^{-1}(f(A))$. They are of relevance in topology for example when dealing with quotient maps. If $f:X\to Y$ is a surjection from the topological space $X$ to a set $Y$, then we can equip $Y$ with a topology by defining a set to be open if it is the image of an open saturated set. A nice property is that if $A$ is saturated and $B$ is arbitrary, then $f(A)\cap f(B)=f(A\cap B)$. From this it follows that the restriction of an open (closed) map to a saturated set is open (closed). – Stefan Hamcke Nov 29 '12 at 18:52
@StefanHamcke Another nice property is that if $B$ is saturated, then $f(A\setminus B) = f(A) \setminus f(B)$ – Eric Auld Feb 15 '14 at 5:49

In my experience the usual context in which saturated sets appear in topology is the one metioned by Stefan H. in the comments: you have a map $f:X\to Y$, and you say that a set $A\subseteq X$ is saturated with respect to $f$ iff $A=f^{-1}\big[f[A]\big]$. More generally, if $\mathscr{P}$ is a partition of $X$, a set $A\subseteq X$ is saturated with respect to $\mathscr{P}$ iff $A=\bigcup\{P\in\mathscr{P}:P\cap A\ne\varnothing\}$. (This really is a generalization: in the case of the map $f$, the associated partition of $X$ is $\{f^{-1}[\{y\}]:y\in Y\}$, the set of fibres of the map $f$.)
If $\mathscr{P}$ is a partition of $X$ and $A$ is an arbitrary subset of $X$, there are two saturated sets naturally associated with $A$. One, which we might call the outer saturation of $A$, is $$\bigcup\{P\in\mathscr{P}:P\cap A\ne\varnothing\}\;;$$ if $\mathscr{P}$ is generated as above by a map $f:X\to Y$, the outer saturation of $A$ is $A=f^{-1}\big[f[A]\big]$. The other, which we might call the inner saturation of $A$, is $$\bigcup\{P\in\mathscr{P}:P\subseteq A\}\;;$$ it’s the complement of the outer saturation of $X\setminus A$, so if $\mathscr{P}$ is generated as above by a map $f:X\to Y$, the inner saturation of $A$ is $X\setminus f^{-1}\big[f[X\setminus A]\big]$.
My best guess without having seen the actual context is that by the saturation of a set $A$ they mean what I’ve called here the outer saturation of $A$.