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The following is an exercise in my textbook.

Fill in the missing details in the following proof that the Least Upper Bound property implies that $\mathbb{R}$ is connected.

(a) Let $\mathbb{R}=A \cup B$ where $A$ and $B$ are mutually separated. Then, $A$ and $B$ are both open.

In my class notes, I have the following written down.

$\mathbb{R} = A \cup B$ where $A$ and $B$ are mutually separated. Mutually separated implies that $A \cap \overline{B} = \emptyset$.

I wanted to know if someone can give a more detailed explanation as to how mutually separated implies that $A \cap \overline{B} = \emptyset$?

Also in my notes.

$\mathbb{R} = A \cup B$

$A = \overline{B^c} \rightarrow$ $A$ is open. Similarly, $B$ is open.

So, I understand that $\overline{B^c} \cap \overline{B} = \emptyset$. But, why does that automatically mean that we can set $A = \overline{B^c}$?

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    $\begingroup$ What does mutually separated mean in this context? $\endgroup$
    – ε-δ
    Sep 27, 2018 at 20:40
  • $\begingroup$ neither set contains a boundary point of the other. $\endgroup$
    – Skm
    Sep 27, 2018 at 20:44
  • $\begingroup$ So $A \cap (\overline{B} \setminus int(B)) = B \cap (\overline{A} \setminus int(A)) = \emptyset$? $\endgroup$
    – Paul Frost
    Sep 27, 2018 at 22:05
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    $\begingroup$ Is there a missing condition here? Because if $A = B = \mathbb{R}$ then we have that $bd(A) = bd(B) = \emptyset$, so neither $A$ nor $B$ contain a boundary point of the other, but clearly $A \cap \overline{B} \neq \emptyset$ $\endgroup$
    – David Lui
    Sep 27, 2018 at 22:54
  • $\begingroup$ @Paul Frost: yes, that's an equivalent way of saying that neither set contains a boundary point of the other $\endgroup$
    – Skm
    Sep 28, 2018 at 4:20

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