I struggled to understand a part of the following proof.
Topological Proof that every Interval $I \subset \mathbb{R}$ is connected
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.
Proof. 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 interval $(a,s)$ 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 $s \in A \cup B$.
My Problems
I am struggled with the bold parts. Probably because I'm more familiar with metric spaces than with topological spaces. I know that $\forall \epsilon>0, \exists y\in B:s+\epsilon>y$ and this implies that every open ball $\beta (s,\epsilon)$ contains a point of $B$. Furthermore, if $a=s\in A$ clearly $\beta (s,\epsilon)$ contains a point of $A$, otherwise, if $a<s$ then $(a,s)\subset A$ which also means that $\beta (s,\epsilon)$ contain a point of $A$. If this argument is correct, then my problem turns out to be the translation of these ideas of open balls and distances to the topological framework. I have the intuition that he is trying to show that $s$ is a boundary point of $B$ in $I$. As $A$ and $B$ separates $I$, $B^c=A$. So, $s$ can never be (an interior point) either in $A$ or in $B$.
I'm trying to furnish a proof in the context of a topological space. Can someone clarify my doubts?
Thanks in advance!
Best Answer
The first two bolded statements are less about using ideas from abstract topological spaces and more about using the definition of infimum, properties of real intervals and manipulating inequalities:
In what follows let us denote $B_a = \{r \in B\ :\ a < r\}$.
Suppose $(p, q)$ is a basic neighborhood of $s$ in $\mathbb{R}$; so $ p < s < q$. Now since $s$ is the largest lower bound of $B_a$ and $q$ is larger than $s$, $q$ cannot be a lower bound of $B_a$.
So there must be a point $x$ of $B_a$ (and thus of $B$) such that $x < q$. Also since $s \leq x$ by the lower bound property of $s$, we have $p < s \leq x < q$ i.e. $x \in (p, q)$. Thus every basic neighborhood $(p, q) \cap I$ of $s$ in $I$ with the subspace topology, still contains that point $x$ of $B$ and we are done.
Again, consider our neighborhood $(p, q)$ of $s$ from earlier. If $s = a$, we are done because then $s$ is a common element of $A$ and the neighborhood $(p, q) \cap I$ of $s$ as required.
Otherwise, if $s \neq a$, then $a < s$. Again this is because of the largest lower bound property of $s$. $a$ is a lower bound on $B_a$ by definition. So since $s$ is the largest lower bound of $B_a$, $a \leq s$. Along with $s \neq a$ this implies $a < s$.
Now analyze the number $$x = \frac{\max(a, p) + s}{2}$$ Since both $p < s$ and $a < s$ in this case, we have $a, p \leq \max(a, p) < s$. So $p < x < s$ which implies $x \in (p, q)$. And $a < x < s \leq b$ which implies $x \in I$ because $I$ is an interval. Finally as $s$ is a lower bound on $B_a = \{r \in B\ :\ a < r\}$, any real $r$ which is less than $s$ but still $a < r$ cannot be in $B$. So $a < x < s$ implies $x \not\in B$. And this gives $x \in A$ as $A$ and $B$ partition $I$.
I understand why this compact statement might lead you to think of $s$ being a boundary point of either $A$ or $B$. That is certainly a correct intuition which can be used to complete the proof. But the direct intention of this statement was much simpler: it was intending to derive a contradiction by showing that either $A$ is not open or $B$ is not open like so:
Note that $a \leq s \leq b$ which gives $s \in I = A \cup B$ as $I$ is an interval. Now:
if $s \in A$, then by definition of $A$ being open, there is a basic neighborhood $U = (p, q) \cap I$ of $s$ entirely contained in $A$. So since $A$ is disjoint from $B$, $U$ cannot contain any point of $B$. But this contradicts what we just showed above: every such $U$ contains a point of $A$ and a point of $B$.
and if $s \in B$, we get a completely analogous contradiction.