I am currently listening to my online lectures on quotient topology, and my professor first builds up the term saturated using set theory. He defines as
For $X$ and an equivalence class $P$ induced by a partition $p$ of $X$, subset $A \subset X$ is a saturated set if $A$ is a union of elements in $P$.
So for example just to clear up the definition, suppose we have $X=[0, 1]$ and an equivalence relation $\sim$ such that $0\sim1$ and $x\sim x$, the equivalence classes are $P=\{\{0, 1\}, A_t\}$ where $A_t=\{t\}$ for $^\forall t \in (0, 1)$. Then, we can say that the set $\{0, 0.5, 1\}$ is a saturated set in $X$ since it is a union of elements of $P$, but $\{0, 0.5\}$ is not a saturated set.
In Munkres, though, the author defines saturated set somewhat differently:
A subset $A$ of $X$ is saturated(with respect to surjective map $p:X \to Y$ if $A$ contains every set $p^{-1}(\{y\})$ it intersects. This, $C$ is saturated if it equals the complete inverse image of a subset of $Y$.
I don't think these definitions are different, they obviously have the same name, so there must be a reason. In my opinion, the first definition of saturated set is a specific case of the second definition, in particular when the map is defined as $p:X \to X/P$ such that $p(x)=Q$ where $Q\in X/P$ and $x \in Q$; so the map where the element is mapped to its partition belonging to. Is my understanding correct?
Best Answer
An equivalent (and simpler, in my opinion) definition is that $A$ is saturated by a surjective map $p:X \to Y$ when $A = p^{-1}(p(A))$. This is equivalent to Munkres definition, which states that $A$ is saturated if there exists a subset $S$ of $Y$ such that $A = p^{-1}(S)$. The trick is that, since $p$ is surjective, $p(A) = p(p^{-1}(S)) = S$.
It is easy to pass from the Munkres definition to the partition definition, as explained by Alessandro Codenotti. In the opposite direction, just define $p$ to be the quotient map from $X$ onto $X/{\sim}$, where $\sim$ is the equivalence relation defined by the partition.