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Let $A\subseteq\mathbb{R}$ be open. If $A\cup (0,1)$ is connected then

  1. A must be connected.

  2. A must have one or two component.

  3. $A\setminus(0,1)$ has at most two component.

  4. $A$ must be a cantor set.

Take $A=(0,1/2)\cup(1/2,1)$. Then $A\cup (0,1)=(0,1)$ is connected but $A$ is not, so $1$ is false. $2$ is also false as I can take 3 or more components. I am not sure about 3 and 4, thank you for help.

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2  
In the title you have intersection, but in the question you have union - which one is correct? –  Old John Jul 11 '12 at 20:31
    
I am extremely sorry. –  Bunuelian Trick Jul 11 '12 at 20:34
    
@Patience What is your definition of "A cantor set"? –  Alex Becker Jul 11 '12 at 20:38
    
I dont know, actually I am solving one question paper of past year, I have copied the question. I must confess I dont know what does it mean by "A must be a cantor set" –  Bunuelian Trick Jul 11 '12 at 20:41

2 Answers 2

up vote 4 down vote accepted

HINT Can you characterize the connected subsets of $\mathbb{R}$?

EDIT If you just want to answer the question, you can easily rule out $1$, $2$ and $4$, by taking $$A = (-1/2,1/4) \cup (1/2,3/4) \cup (8/9,2)$$

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only intervals are connected thats what you want to say Marvis? –  Bunuelian Trick Jul 11 '12 at 20:44
    
@Patience Yes. Precisely. –  user17762 Jul 11 '12 at 20:45

Hint: As mentioned in another answer, you could ask what are the connected components of $\mathbb{R}$. Indeed if $A\cup (0,1)$ is connected, you would at most have two connected components if you remove a connected subset. So $3$ sounds fine.

For (4), what would you think about $A = (\frac{1}{2},2)$?

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then $A\cup (0,1)=(0,2)$ –  Bunuelian Trick Jul 11 '12 at 20:45
    
@Patience: And that is not the Cantor set (en.wikipedia.org/wiki/Cantor_set). –  Thomas Jul 11 '12 at 20:46
    
Thomas, thank you :) –  Bunuelian Trick Jul 11 '12 at 20:51

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