If $A \subset X$ and $X$ is a topological space, I endow $A$ with the subspace topology.
Let $n \geq 2$.
If $A$,$B$ are connected sets in $\mathbb{R}^n$ such that $A \cap B = \varnothing$, $A \cup B$ can be connected. Indeed, we can have in this case $A \cup B = \mathbb{R}^n$ : $A \subset \mathbb{R^n}$, $n \geq 2$, such that $A$ is homeomorphic to $\mathbb{R^n} \setminus A$, and $A$ is connected .
If $A$ and $B$ are convex sets in $\mathbb{R}^n$ such that $A \cap B = \varnothing$, $A \cup B$ can also be connected. Indeed, we can take $A = B((-1,0,\cdots,0),1) \cup \{(0,\cdots,0) \}$ and $B = B((1,0,\cdots,0),1)$, where $B(x,r)$ means the open ball of center $x \in \mathbb{R}^n$ and radius r.
But if $A$,$B$ are closed sets in $\mathbb{R}^n$ such that $A \cap B = \varnothing$, $A \cup B$ can't be connected. This is a consequence of the fact that $\mathbb{R}^n$ is normal : we can find open sets $U$ and $V$ in $\mathbb{R}^n$ such that $A \subset U$ and $B \subset V$, with $U \cap V = \varnothing$.
My questions are :
(1) What condition $C_1$ weaker than (or different from) "closedness" would force the union of two connected subsets of $\mathbb{R}^n$ verifying $C_1$ and such that $A \cap B = \varnothing$, to be not connected ? And in this case, does it force $A \cup B$ to have exactly 2 connected components ?
(2) What condition $C_2$ weaker than (or different from) "closedness" would force the union of two convex subsets of $\mathbb{R}^n$ verifying $C_2$ and such that $A \cap B = \varnothing$, to be not connected ? And in this case, does it force $A \cup B$ to have exactly 2 connected components ?
For example, if $n = 1$, $C_1$ and $C_2$ seem to be : $\overline{A} \cap B = \varnothing$ and $A \cap \overline{B} = \varnothing$.
