to a point, the demonstration is clear for me. I'll sum it up like this:
So firstly I'll try to demonstrate that I=[0,1] is not connected to get to a contradiction, which will conduct to the fact that [0,1] is actually connected.
In case [0,1] is disconnected, that means there are two open sets A and B, so that $A\cap I$, respectively $B\cap I$ that are non-empty, disjoint, and their union is I. (from the definition of disconnected sets)
Without loss of generality, let's say 1 E B we define c=sup($A\cap I$) which means c>=0 and c<=1 which conducts to the fact hat c E [0,1], so c E I which means c E ($A\cap I$)U($B\cap I$) because I is the union of $A\cap I$ and $B\cap I$ from the definition of disconnected sets this means that c E $(AUB)\cap I$ which implies that c E (AUB)
then, there are two cases case 1) c E A case II) c E B
for case I) it says that in this case cE (O,1) * Because A is open and it contains c, then there are points in $A\cap I$ which are higher than c, which contradicts the definition of supremum**
case II) is similar and conducts to a contradiciton as well
*which is where my confusion starts, how do you know for sure that it is (0,1) and not just [0,1) ? **you can know for sure that there is such a point in A(because it's an open set) but how do I know that there is such a point in $A\cap I$ ?
in my book i have this definition of connectedness
I is connected if I is not disconnected
I is disconnected if there are two open sets A & B, so that :
i) $A\cap I$ is non empty set
$B\cap I$ is non empty set
ii) ($A\cap I$) U ($A\cap I$) = I
iii) $$A\cap I)\cap (B\cap I$$ is empty set
Best Answer
Your question is a little convoluted, so I'll just answer the question in the title. I hope this helps anyway.
Suppose $[0,1]$ can be separated into $A,B$. We can suppose $1 \in B$ without loss of generality.
Let $c:=\sup A$.
Now, $c$ can't belong to $A$, because if it did it would not be $1$ and, since $A$ is open, it would imply that a greater element than $c$ would belong to $A$. But $c$ can't belong to $B$ either, because this would imply either that $c=0$ (from where $A$ would not be open) or that there would exist someone smaller than $c$ in $B$ which would still be an upper bound for $A$, since $B$ is open.
Therefore, $c$ is not in $A$ nor $B$, a contradiction.