$A,B\subset \mathbb R^2$ s.t $A\cap B=\Phi$ and $A\cup B$ is open in $\mathbb R^2.$ Given that $A$ is open and $A\cup B$ is connected then is $B$ closed or open ?
Now if , $B$ is open too then $A\cup B$ cannot be connected so $B$ must be closed.
Now one possibility of this is that $A$ is any open set and $B=A^c$ the closed set.So the union is the whole set.
If $A$ is an open disc and $B$ is just any closed disc then the union is not open.
So my question is whether there is such a thing that this case is meaningful iff $A\cup B$ is allowed to be the whole set and otherwise not.Can I prove that union of an open set A and a closed set B can never be open in $\mathbb R^2$ unless $B=A^c$ and $A\cup B=\mathbb R\ ?$ Thanks.