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The definition of a disconnected set seems a bit ambiguous in the book I am reading :

$1.$ A subset $D$ of $\mathbb R^p$ is said to be disconnected if there exist two open sets $A$ and $B$ such that $A \bigcap D $ and $B \bigcap D$ are disjoint non empty sets whose union is $D$.

Whose union is $D \implies A \bigcup B = D$ or does it mean : $(A \bigcap D)~~ \bigcup ~~(B \bigcap D)=D $

Also, I am encountering a difficulty in the proof of the theorem :

$2.$ The closed unit interval is a connected set . Incidentally, a similar kind of problem has been asked here . But, please note that my query with respect to this proof is a little different.

The proof goes like this :

Suppose that $A,B$ are open sets forming a disconnection of $\mathbb{I}$. Thus $A\cap \mathbb{I}$ and $B\cap \mathbb{I}$ are non-empty bounded disjoint sets whose union is $\mathbb{I}$. Since $A$ and $B$ are open, the sets $A\cap \mathbb{I}$ and $B\cap \mathbb{I}$ cannot consist of only one point. (Why?) For the sake of definiteness, we suppose that there exist points $a\in A$, $b\in B$ such that $0<a<b<1$. Applying the supremum property, we let $c=sup\{x\in A:x<b\}$ so that $0<c<1$; hence $c\in A\cup B$. If $c\in A$ then $c\ne b$ and since $A$ is open there is a point $a_1\in A$, $c<a_1$, such that the interval $[c,a_1]$ is contained in $\{x\in A: x<b\}$ contrary to the definition of $c$.

Query: Why does $c \in A \bigcup B$ where $c = \sup ( A \bigcap \mathbb I)$ .

Since, it's possible that a supermum need not belong to it's set : $A \bigcap \mathbb I$ .Hence, isn't it possible that c does not belong to neither $A$ nor $B$ and hence, does not belong to $A \bigcup B$ either?

Thank you for your help.

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    $\begingroup$ For the second, $c$ belongs to the closed unit interval. That is the union of $A$ and $B$ by assumption. Hence $c\in A\cup B$. $\endgroup$ – Daniel Fischer Aug 31 '14 at 23:38
  • $\begingroup$ Ohkay .. that means in the original definition : $D = A \bigcup B$ ? $\endgroup$ – MathMan Aug 31 '14 at 23:39
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    $\begingroup$ D'oh, parse error. No, as written, $A$ and $B$ are open in $\mathbb{R}^p$, and $(A\cap D)$ and $(B\cap D)$ are nonempty, and $D = (A\cap D) \cup (B\cap D)$. Sorry. $\endgroup$ – Daniel Fischer Aug 31 '14 at 23:42
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    $\begingroup$ Part of the assumption is that $\mathbb{I} \subset A\cup B$. Now $c$ is the supremum of a nonempty subset of $\mathbb{I}$, hence $c\in \mathbb{I}$. Therefore, $c\in A\cup B$. $\endgroup$ – Daniel Fischer Aug 31 '14 at 23:48
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    $\begingroup$ "Suppose that $A,B$ are open sets forming a disconnection of $\mathbb{I}$." That means $(A\cap\mathbb{I})\cap (B\cap\mathbb{I}) = \varnothing$, $(A\cap\mathbb{I}) \neq \varnothing \neq (B\cap\mathbb{I})$, and $\mathbb{I} = (A\cap\mathbb{I})\cup (B\cap\mathbb{I})$ [the latter can equivalently be stated $\mathbb{I}\subset A\cup B$]. $\endgroup$ – Daniel Fischer Aug 31 '14 at 23:54

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