Is the notion of separation and separated sets interchangeable? I was looking at the following problem:
let $A,B$ be separated subsets of a space $X$. If $C$ is a connected subset of $A \cup B$, then either $C \subset A$ or $C \subset B$.
The definition of separated sets given to me is that $A,B$ are separated sets if $A,B$ are disjoint nonempty and $A \cap \overline{B}=\overline{A} \cap B=\varnothing$.
To solve the problem in question I think it would be easier to assume $A,B$ are disjoint nonempty open sets whose union is $X$. Then if $C$ is not entirely contained in $A$ or $B$, $A \cap C$ and $B \cap C$ are a pair of disjoint nonempty open sets with union $C$. This contradicts the assumption that $C$ is connected.
So it seems easier to use the nonempty open sets to prove this, not two separated sets. My question is, is it valid to interchange the notion of disjoint nonempty open sets constituting a "separation" of a space with separable sets in a space? If not how would this proof go using the definition of two separated sets?
Best Answer
First of all, separable (for a space $X$ or a subspace $A$) means that it has a countable dense subset, and has nothing to do with connectedness properties. So keep the term separable out of it, please.
$A$ and $B$ are separated in $X$ iff $\overline{A} \cap B = \emptyset = A \cap \overline{B}$ (i..e no point of $A$ is "close to" $B$ and vice versa). For disjoint sets $A$ and $B$ this means the same as saying that $A$ and $B$ are closed-and-open (clopen) in $C=A \cup B$ (in the subspace topology); this is easy to see. A space being disconnected is defined as being able to write $X$ as the union of two non-empty separated sets; and connectedness is defined as "not disconnected".
So if $S$ is connected and $S \subseteq A \cup B$, $S$ cannot contain points from $A$ and $B$ both, because that would indeed mean that $S \cap A$ and $S \cap B$ are relatively clopen non-empty subsets of $S$ partioning $S$, contradicting connectedness of $S$. So indeed $S \subseteq A$ or $S \subseteq B$ holds.