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

Hint in $\mathbb R$: $A=(0,1]\cup (3,4)$ and $B=(1,2) \cup (2,3]$. — user251257, Nov 25 '16 at 12:08

Robert Z · Answer 1 · 2016-11-25T12:22:31.500

2

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 $\mathbb{R}^2$.

edited Nov 25 '16 at 12:22

answered Nov 25 '16 at 12:17

Robert Z

145,942

Andrew D. Hwang · Accepted Answer · 2016-11-25T12:56:08.387

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.

But $A$ and $B$ are not disjoint... – 5xum Nov 25 '16 at 12:37 — 5xum, Nov 25 '16 at 12:37
@5xum: Thanks, edited. – Andrew D. Hwang Nov 25 '16 at 12:56 — Andrew D. Hwang, Nov 25 '16 at 12:56

score 1 · Answer 3 · edited Jul 08 '20 at 18:12

1

For counterxample , take $A=[(0,1)\times (0,1)] \cap [\mathbb Q \times \mathbb R] $ and $B=[(0,1)\times (0,1)] \cap [\mathbb Q^c \times \mathbb R] $

edited Jul 08 '20 at 18:12

vidyarthi

7,028

answered Jun 17 '20 at 06:38

Unknown x

703

5xum · Answer 4 · 2016-11-25T12:36:18.563

0

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

edited Nov 25 '16 at 12:36

answered Nov 25 '16 at 12:10

5xum

123,496
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@user251257 They are now after I fixed my typo :) – 5xum Nov 25 '16 at 12:36
That $A$ and $B$ are subset of $\mathbb{R}^2$. So this is not a counterexample for given problem. – vijayanand Nov 28 '16 at 05:26
@vijayanand yeah, op totally changed the question after I gave this answer – 5xum Nov 28 '16 at 05:28

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

4 Answers4