Varying definitions of disconnectedness As I've taken courses in real analysis and topology, I've worked with connectedness and disconnectedness a fair bit and I have an ample understanding of the concept. Unfortunately, however, in my browsing, I've encountered two varying definitions of connectedness and disconnectedness.
Generally speaking, a subset of a topological space is disconnected if it can be written as the union of two disjoint open sets. That's the first definition of disconnectedness I've seen. The second definition (which is more scarcely encountered) necessitates all of the same things as the first definition but also includes that the closures of the open sets be disjoint as well, i.e. $\overline{U} \cap V = \varnothing = U \cap \overline{V}$.
Now, I understand that in some topologies (like $\mathbb{R}$ for example), this detail won't make much of a difference. But I'm sure there's an entire collection of topological spaces in which this minute change in definition makes a world of a difference. So, which definition is the "correct" or "most utilizable" definition? Does the distinction matter or does it all just depend on what you're trying to show?
 A: The second definition does not say that the closures of the two open sets that witness disconnectedness are disjoint. It says that the closure of either of the sets is disjoint from the other one. (You wrote this correctly in symbols.) The two definitions are equivalent. If $U$ and $V$ are disjoint open sets, then $\overline{U} \cap V = U \cap \overline{V} = \emptyset$, because the complement of $V$ (resp. $U$) is a closed set that contains $U$ (resp. $V$) and hence contains $\overline{U}$ (resp. $\overline{V}$).
As an example, consider $X = \Bbb{R} \setminus \{0\}$ and take $U = (-\infty, 0)$ and $V = (0, \infty)$. Then $U$ and $V$ disconnect $X$ under either of the definitions.
A: If $A,B$ form a partition of a space $X$ (i.e. $A\cup B = X, A \cap B = \emptyset$) the following statements are equivalent:

*

*$A$ and $B$ are separated (i.e. $A \cap \overline{B} = \emptyset = \overline{A} \cap B$; the notion that Rudin uses).


*$A$ and $B$ are both closed in $X$.


*$A$ and $B$ are both open in $X$.
Proof: if 1. holds then $A = X\setminus \overline{B}$ so $A$ is open as the complement of a closed set. $B$ is open in the same way, as $B = X\setminus \overline{A}$. And as $A$ and $B$ are each other's complement both are closed too, so both 2 and 3 hold. If 2 holds, 1 is trivial as $\overline{A}=A, B = \overline{B}$ and if 3 holds 1 is also trivial, as is easy to see.
So it's always equivalent to define connectedness of $X$ as having no such non-trivial (not the $\{\emptyset,X\}$ partition) partition. If you have one type of partition of $X$ it's also of the other type...
For a non-partition the notions do differ: $(0,1]$ and $[2,3)$ are clearly separated in the reals but neither set is open or closed. If we can separate disjoint closed sets in $X$ we call a space $X$ "normal", and if we can separate separated sets (sounds weird...) a space is called "completely normal", which is stronger.
