First, the definition of a connected set:
Definition: A topological space is connected if, and only if, it cannot be divided in two
nonempty, open and disjoint subsets, or, similarly, if the empty set and the whole set are the
only subsets that are open and closed at the same time.
I don't understand some points in the following proof, that every interval $I \subset \mathbb{R}$ is connected.
Suppose $I = A \cup B$ and $A \cap B = \emptyset$, $A$ and $B$ are both non-empty and open in the subspace-topology of $I \subset \mathbb{R}$. Choose $a\in A$ and $b\in B$ and suppose $a < b$. Let $s := \mathrm{inf}\{ x \in B ~|~ a < x \}$. Then in every neighborhood of $s$ there are points of $B$ (because of the definition of the infimum), but also of $A$, then if not $s = a$, then $a < s$ and the open intervall $(a,s)$ lies entirely in $A$. And so $s$ cannot be an inner point of $A$ nor $B$, but this is a contradiction to the property that both $A$ and $B$ be open and $s \in A \cup B$.
With the bold part, I have a problem, why it follows that $(a,s)$ lies entirely in $A$ and it that case it must be that $A = (a,s)$, or not?
Best Answer
If not $s = a$, then $a<s$ because $s$ is the infimum of a set of elements all larger than $a$.
The entire interval $(a, s)$ must be in $A$ because of how $s$ was chosen. The number $s$ has been chosen so that it is smaller than any element in $B$ greater than $a$, so there can't be any elements of $B$ in the interval $(a, s)$, and since $A$ and $B$ between them contain all of $I$, $(a, s)$ must be completely within $A$.
The crux of the proof is that with this $s$ as chosen (and exisiting, by Archimedes' lemma), it can't be in neither $A$ nor $B$, but at the same time it has to be in one of them, since $A$ and $B$ together cover the whole of $I$. Thus we have a contradiction, and we know that our assumption must be wrong, namely that $I$ can be written as the union of two nonempty, disjoint open sets.