# Thread: An open set is disconnected iff it is the union of two non-empty disjoint open sets

1. ## An open set is disconnected iff it is the union of two non-empty disjoint open sets

How do I proof this is the definition of disconnectedness that I am using that $S$ is disconnected if there exists two sets $S_{1}, S_{2}$ such that they are both non-emtpy, $S_{1} \cup S_{2} = S$ and $cl(S_{1})\cap S_{2} = cl(S_{2})\cap S_{1} = \emptyset$

Attempting to prove the first condition, that disconnected implies the open set union condition, I tried to show that if $(S_{1},S_{2})$ form a disconnection, then so does $(S_{1} - \partial S_{1}, S_{2}-\partial S_{2})$ as well, but I am not even sure if this is the correct approach because I am not nsure how to show that those two sets would be non-empty and that their union would be $S$. And I am not sure at all how to show the other direction of the proof (that the open set union condition implies disconnectedness). Any help would be appreciated.

2. This is an ideal case for using RL Moore’s way of defining connected sets.
Two non-empty sets are separated if neither contains a point or a limit point of the other.
Then A set is connected if and only if it is not the union of two separated sets.

3. Then would I use that to assume to the contrary that a set is disconnected yet it is not the union of two non-empty disjoint open sets, so the disconnection must be of closed sets...? I am not sure how that definition follows to the proposition that I want to prove.

Edit: So here's my attempt at a proof:

Suppose $S_{1}, S_{2}$ are two non-empty disjoin open sets such that $S = S_{1} \cup S_{2}$ but suppose to the contrary that they do not form a disconnection of $S$. Then, without loss of generality, $cl(S_{1}) \cap S_{2} \ne \emptyset$. Since $S_{1} \cap S_{2} = \emptyset$, it follows that there exists $x\in \partial S_{1} \cap S_{2}$. So there exists $\delta > 0$ such that $B(\delta, x) \subset S_{2}$ and this ball contains elements in $S_{1}$ and $(S_{1})^{c}$ since x is a boundary point of $S_{1}$. This means there exists $y \in S_{1} \cap S_{2}$, which is a contradiction, so $cl(S_{1}) \cap S_{2} = \emptyset$ and since the index of the sets was arbitrarily fixed, it follows $(S_{1}, S_{2})$ is in fact a disconnection.

But now I'm still having trouble proving the converse...