Let $A$ be a connected subset of a connected topological space $X$ and let $B$ be a connected component of $ A^c$. Prove that $ B^c$ is connected.Let $A$ be a connected subset of a connected topological space $X$ and let $B$ be a connected component of $ A^c$. Prove that $ B^c$ is connected.
We know that a set is connected if and only if it is not disconnected, and a connected component is a maximal subset of a topological space that cannot be covered by the union of two disjoint open sets. We also know by a proposition that a connected component is a closed set, which would mean that $B^c$ is an open set. We have that $B$ is a closed connected subset of $ A^c$ and $ A^c$ is disconnected. Now, $ B^c$ is an open subset of a connected space and $ B^c$ is path connected which implies that $ B^c$ is connected. Is this correct?
I'm afraid very little is correct. Firstly, $B^c$ need not be open. $B$ is closed in $A^c$, but that does not imply it is closed in $X$. Secondly, an open subset of a connected space need not be connected. Thirdly, path-connectedness has not been specified, but an open subset of path-connected space may also be not connected or path-connected.
