My teacher works with two definitions of disconnected sets in $\mathbb{R^n}$:

1) $\Omega \subset \mathbb{R^n}$ is disconnected iff there exists open sets $A,B$ such that: 1) $A\cap \Omega \neq \varnothing$ ; $B\cap \Omega \neq \varnothing$ 2) $A\cap B = \varnothing$ ; 3)$\Omega \subseteq A\cup B$

2)$\Omega \subset \mathbb{R^n}$ is disconnected iff there exists relatively open sets $A´,B´$ to $\Omega$ such that: 1) $A´ \neq \varnothing$ ; $B´\neq \varnothing$ 2) $A´\cap B´ = \varnothing$ ; 3) $\Omega = A´\cup B´$

I was trying to prove that this definitions are equivalent, but I was only able to prove that Definition 1) implies definition 2) But I´m having a hard time proving that definition 2) implies definition 1)

My attempt was like this: We know that $A´$ and $B´$ are relatively open to $\Omega$ so $\exists$ open sets $A$ and $B$ such that:

$A´=A\cap \Omega \neq \varnothing$ and $B´=B\cap \Omega\neq \varnothing$ but I don´t know how to prove that this sets $A$ and $B$ are disjoint and $\Omega \subseteq A\cup B$

Any suggestions would be highly appreciated

  • 1
    $\begingroup$ Consider using A' - closure B' and B' - closure A' . $\endgroup$ – William Elliot Sep 18 '17 at 7:04

Theorem. Within a metric space (S,d), if A is a disconnected subset,
then there are open disjoint U,V with A subset U $\cup$ V and not empty
U $\cap$ A, V $\cap$ A. Proof upon request.


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