If $A \cup B$ is open and disconnected in $\mathbb{R}^2$, does it follows both $A$ and $B$ are open

Question

Let $A$ and $B$ be two disjoint subset of $\mathbb{R}^2$ such that $A \cup B$ is open and disconnected in $\mathbb{R}^2$. does it follows both $A$ and $B$ are open.

If $A$ and $B$ are both open, since $A$ and $B$ are disjoint by hypothesis. then by definition $A \cup B$ is disconnected. But for the reverse case i guess it is not true. But i dont get any counter example for this.

So i will be happy if someone help me to get this.

since the same question already exist here. But there is no proper counter example for this

Answer 1

You've got multiple counterexamples under the hypotheses of the question.

If you assume $A$ and $B$ are connected, however, then "$A \cup B$ open and disconnected" implies $A$ and $B$ are open.

To prove this, note that $A \cap B = \varnothing$: If a family of connected sets has an element in common, then the union is connected.

Let $\{U, V\}$ be a separation of $A \cup B$. Each of the sets $U_{0} := (A \cup B) \cap U$ and $V_{0} := (A \cup B) \cap V$ is open and non-empty as an intersection of finitely many open sets.

Since $A$ is connected, one of the sets $A \cap U$ or $A \cap V$ is empty, say $A \cap V$ without loss of generality. This means $A \subset A \cap U$, so $A \subset U$ and therefore $A = A \cap U = (A \cup B) \cap U = U_{0}$ is open. A similar argument shows $B$ is open.

Answer 2

No, since $A=(0,1)$ and $B=[1,2)\cup(3,\infty)$ provides a counterexample.

Answer 3

No. Let $$A:=((0,1)\times (2,3))\cup ((0,1)\times (0,1)) \cup (I\times \{0\})$$ and $$B:=((0,1)\times (-1,0)) \cup (J\times \{0\})$$ where $I=\mathbb{Q}\cap (0,1)$ and $J=(0,1)\setminus I$.

Then $A\cup B=((0,1)\times (-1,1))\cup ((0,1)\times (2,3))$ is open and disconnected in $\mathbb{R}^2$.

On the other hand $A$ and $B$ are not open and they are disjoint in $\mathbb{R}^2$.