# $(X,d)$ m.e., with $Y \subset X$: $Y$ is open, $Y$ is connected it's equivalent to another property.

Let $(X,d)$ be a metric space and let $Y \subset X$: $Y$ is open. Prove that $Y$ is connected if and only if there aren't $A,B \subset X$ non-empty such that $Y=A \cup B$ and $A \cap \overline B=\emptyset$ and $\overline A \cap B=\emptyset$.

My attempt at a solution:

I could prove that the second statement implies $Y$ is connected (here I've used that $Y$ is connected if and only if the only clopen sets are $Y$ and $\emptyset$:

Suppose the second condition holds but $Y$ is disconnected. Let $S \subset Y$ such that $S$ is clopen and suppose $S\neq Y$ and $S \neq \emptyset$. Clearly, $Y=S \cup S^c$. $S=\overline S$ and $S^c=\overline S^c$, so $S \cap \overline S^c=S \cap S^c=\emptyset$, analogously, $\overline S \cap S^c=\emptyset$. But this is absurd by the hypothesis, this means that the only clopen sets are $Y$ and $\emptyset$, it follows that $Y$ is connected.

I couldn't do much with the other implication: My hypothesis is that $Y$ is connected and I want to prove that this implies the second condition holds. I've tried to prove it by the absurd, i.e, I suppose there exist $A,B \subset X$ non-empty such that $Y=A \cup B$ and $A \cap \overline B=\emptyset$ and $\overline A \cap B=\emptyset$. I should conclude that $Y$ is disconnected. In some part of the proof I have to use the fact that $Y$ is open in $X$.

Hint: Since $A\cap\overline B=\emptyset,$ then $X\setminus\overline B$ is an open superset of $A,$ so since $B\subseteq\overline B$ and $Y=A\cup B,$ then $A=Y\cap(X\setminus\overline B),$ so $A$ is open in $Y.$ Similarly, $B$ is open in $Y.$ What can we then conclude? (Can you justify these claims?)
• If we restric ourselves to $(Y,d)$, then $A^c=B$ and $B^c=A$ , we can conclude that both $A$ and $B$ are clopen proper subsets of $Y$, then $Y$ is disconnected. Now I'll try to prove all the things you've stated. Your answer was clear and extremely helpful, thanks! Nov 25 '13 at 22:17