0
$\begingroup$

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 $X$.

$\endgroup$
1
$\begingroup$

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?)

$\endgroup$
2
  • $\begingroup$ 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! $\endgroup$
    – user100106
    Nov 25 '13 at 22:17
  • $\begingroup$ Bingo! Well done. $\endgroup$ Nov 25 '13 at 22:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.