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Question is :

If $A$ and $B$ are two disjoint non empty subsets of $\mathbb{R}^2$ such that $A\cup B$ is open in $\mathbb{R}^2$ . Then which of the following is true?

  • If $A$ is open and $A\cup B$ is connected then $B$ must be closed in $\mathbb{R}^2$.
  • If $A$ is closed then $B$ must be open in $\mathbb{R}^2$.
  • If both $A$ and $B$ are connected, then $A\cup B$ must be disconnected.
  • If $A\cup B$ is disconnected then both $A$ and $B$ are open.

I am sure that first and second bullet are correct but do not know how to make it more clear.

For first bullet,

Suppose $B$ is open then there is no chance of $A\cup B$ being connected though $A$ and $B$ are connected. (As $A$ and $B$ are disjoint) But $B$ being Not open does not imply $B$ being closed. So, I would not say this is a proof but i would see this can be made to a proof.(I guess it can)

For second bullet,

Suppose $B$ is also closed then $A\cup B$ is closed but we have given that $A\cup B$ is open. But $B$ being Not closed does not imply $B$ being open. So, I would not say this is a proof but i would see this can be made to a proof.(I guess it can)

For third bullet,

I can surely say that this is false. Suppose $A\cup B$ is connected then $A$ and $B$ should have a common point But $A$ and $B$ are disconnected.Thus third option is false.

For fourth bullet,

I think it is true.In general, $A\cup B$ being disconnected does not imply both $A$ and $B$ are open. but we have a condition $A\cup B$ is open. I feel this would force both $A$ and $B $ to be open. I could not make this more clear.

I would be thankful if someone can help me to clear this.

Thank you :)

  • Why dont you write your thoughts as proper proofs? I think the first one is false – Lost1 Dec 26 '13 at 02:00
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    I don't write my thoughts as proper proofs because they are neither proper nor proofs... –  Dec 26 '13 at 02:09

1 Answers1

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The problem with your approach is that a set can be neither open nor closed. So you cannot prove the contrapositive by turning open into closed.

For the first bullet, consider $A = (0, 1) \times (0, 1)$, $B = [1, 2) \times (0, 1)$. Then $A \cup B = (0, 2) \times (0, 1)$.

For the second bullet, consider $(A \cup B) \cap A^\complement$. What is it equal to? Why is it open?

For the third and fourth bullets, use examples similar to the first bullet.

Ayman Hourieh
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  • I would be so thankful if you can say how did that examples was obvious choice of yours... I mean i am expecting some natural proof/natural counter example.... Please take some more pain to clear my problem.... –  Dec 26 '13 at 02:12
  • @PraphullaKoushik I started with an open square for $A$ (the simplest possible open set). I tried to add a non-closed square to it so that the result was open and connected. From that, the choice of $B$ came up naturally. – Ayman Hourieh Dec 26 '13 at 02:24
  • Yes.. I got it... Thank you :) –  Dec 26 '13 at 02:30
  • @Praphulla Koushik : what is the counter example for 4. – Struggler Jun 12 '15 at 04:40
  • Have you proved option 4? – MathBS Jul 23 '20 at 19:06