The closed unit interval $\mathbb{I}=[0,1]$ is a connected subset of
$\mathbb{R}$.
I am having difficulty understanding the proof in my book, which goes:
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 $c$.
Why is $0<c<1$? They define $c=\sup\{x\in A:x<b\}$ so it is the supremum of $A$ which is less than $b$ so doesn't that mean $c$ must be less than $b$ ? Also, why is $c\in A\cup B$ ? Lastly, since $c$ is the supremum of $A$ how is there a point in $A$ such that $c<a_1$ and why is it contrary that it is contained in $\{x\in A: x<b\}$ ?
Best Answer
$A$ meets $\mathbb{I}$ non-trivially so $c$ is bigger than $0$ necessarily. Since $b<1$ and $b$ is an upper bound for $\{ x \in A : x < b \} $, $c \leq b < 1$. So, $0<c<1$.
It follows that $c \in (0,1) \subseteq [0,1] = A \cup B$ by supposition.
Now we'll assume that $c \in A$.
Necessarily $c < b$.
If $c=b$ then $c \in A \cap B$ however $A \cap B = \varnothing$ and also $c$ cannot exceed $b$.
We can pick some $\varepsilon_1$ neighborhood around $c$ that is contained in $A$ since it is an open set. Pick some $c< a_1< c+\varepsilon_1$ in that neighborhood. $A$ and $B$ are disjoint so this neighborhood will not contain $b$ so this $a_1 < c+\varepsilon_1 < b$ so $a_1 \in \{ x \in A : x < b \}$ but then $a_1 \leq c$ since it is the supremum but also we have $a_1 > c$, a contradiction.