If $C$ is connected subset of a disconnected metric space $X=A\cup B$ then either $C\subseteq A $ or $C \subseteq B$

Question

If $C$ is a connected subset of a disconnected metric space $X=A\cup B$ where $\overline{A}\cap B = A\cap \overline{B} = \phi$ then either $C\subseteq A $ or $C \subseteq B$

If $C\subseteq A $ or $C \subseteq B$ then we're done. Assume neither $C\subseteq A $ or $C \subseteq B$

$\Rightarrow \; \; \exists \; x,y \in X $ such that

$x\in C\cap A$ and $y \in C\cap B$, obviously $x\neq y $ since $A\cap B= \phi$

We have $C\cap A \subseteq C$ and $C\cap B \subseteq C$ and both are non empty.

C=$C\cap X = C\cap (A\cup B) = (C\cap A)\cup (C\cap B)$

Also,

$C\cap A \subseteq A$ and $C\cap B \subseteq B$ and since $A\cap B = \phi $

$(C\cap A)\cap (C\cap B) = \phi$

Here is where I am having problem :-

Can I directly say that $(C\cap A)$ and $(C\cap B)$ are open in C?

Because that will immediately show that C is disconnected and we'd have the required contradiction.

Answer 1

Since $A\cap \overline B=\emptyset$ and $X=A\cup B$, then $B=\overline B$ is closed and $A$ is open. Therefore $C\cap A$ is open in $C$ (because open subsets of $C$ are just intersection of $C$ with open subset of $X$.)

The same reasoning shows that $C\cap B$ is open.

Answer 2

Observe that $B$ is the complement of closed set $\overline A$ hence is open.

Similarly $A$ is open.

Then $C\cap A$ and $C\cap B$ are open in $C$.

You can now finish your proof by stating that one of them must be empty (because $C$ is connected) so that $C\subseteq A$ or $C\subseteq B$.