I'm trying to prove or disprove the following statement:
If $X$ is connected, $A \subset X$ is connected, and $C$ a component of $X\backslash A$ then $\overline A \cap \overline C \ne \emptyset$.
I know that $C$ is closed in $X\backslash A$, so $C = cl_{X \backslash A}C = \overline C \cap (X \backslash A)$ - But I don't know how to continue.
I also couldn't find a counter example, and I'm not even sure where to look.
Any suggestions?
Just a remark: Since both $A$ and $C$ are connected, if $A\cup C$ is disconnected, then $A$ and $C$ form a separation of their union. So there are closed sets $B$ and $D$ in $X$ such that $A\subseteq B\subseteq X\setminus C$ and $C\subseteq D\subseteq X\setminus A$. Since $C$ is closed in $X\setminus A$, there is a closed set $E$ such that $C=E\setminus A$. Thus $C=D\cap E$ is closed in $X$. It follows that $\overline C\cap\overline A=\emptyset$. On the other hand, this property implies that $A\cup C$ is disconnected.
So $\overline A\cap\overline C=\emptyset$ is equivalent to $A\cup C$ being disconnected.
Now here's a counterexample: Let $X=\{(0,0)\}\cup \bigcup_{n\in\Bbb N}L_n$, with the topology from $\Bbb R^2$, where $L_n=\left\{\left(t+\frac{1-t}n,\ t\right)\mid 0\le t\le1\right\}$. This space is connected. Let $A=\{(x,y)\in X\mid y\ge 1/2\}$. Then $A$ is connected, too. Now $C=\{(0,0)\}$ is a component (even a quasicomponent) of $X\setminus A$, but $C$ and $A$ are disjoint closed sets.
By the way, a counterexample is not possible if we require $C$ to be clopen in $X\setminus A$, as this thread shows