Open subset of $\mathbb{R}$ can be written as countable union of disjoint open intervals, intuition?

Question

Here is the proof of this proposition from my Real Analysis textbook.

Proposition 1.5. Suppose $G \subset \mathbb{R}$ is open. Then $G$ can be written as the countable union of disjoint open intervals.

Proof. Let $G$ be an open subset of the real line and for each $x \in G$, let$$A_x = \inf\{a : \text{there exists }b\text{ such that }x \in (a, b) \subset G\}$$and$$B_x = \sup\{d : \text{there exists }c \text{ such that }x \in (c, d) \subset G\}.$$Let $I_x = (A_x, B_x)$.

We prove that $x \in I_x \subset G$. If $y \in I_x$, then $y > A_x$, and so there exist $a$ and $b$ such that $A_x < a < y$ and $x \in (a, b) \subset G$. Because $y < B_x$ there exist $c$ and $d$ such that $y < d < B_x$ and $x \in (c, d) \subset G$. The point $x$ is in both $(a, b)$ and $(c, d)$, hence their union $J = (\min(a, c), \max(b, d))$ is an open interval. $J$ will be a subset of $G$ because both $(a, b)$ and $(c, d)$ are. Both $x$ and $y$ are greater than $a > A_x$ and less than $d < B_x$, so $x \in I_x$ and $y \in J \subset G$.

We next argue that if $x \neq y$, then either $I_x \cap I_y = \emptyset$ or else $I_x = I_y$. Suppose $I_x \cap I_y \neq \emptyset$. Then $H = I_x \cup I_y$ is the union of two open intervals that intersect, hence is an open interval, and moreover $H \subset G$. We see that $H = (\min(A_x, A_y), \max(B_x, B_y))$. Now $x \in I_x \subset J \subset G$. It follows form the definition of $A_x$ that $A_x \le \min(A_x, A_y)$, which implies that $B_x \ge B_y$. Hence $I_y \subset I_x$. Reversing the roles of $x$ and $y$ shows that $I_x \subset I_y$, hence $I_x = I_y$.

We therefore have established that $G$ is the union of a collection of open intervals $\{I_x\}$, and any two are either disjoint or equal. It remains to prove that there are only countably many of them. Each open interval contians a rational number and the rational numbers corresponding to disjoint open intervals must be different. Since there are only countably many rationals, the number of disjoint open intervals making up $G$ must be countable.

I can follow this proof line by line, but I'm curious. What is the intuition behind this proof? What are the one to three key ideas this proof boils down to?

Answer 1

Connected sets of $\mathbb{R}$ are intervals.

You are decomposing an open set in its connected components. The proof as stated is a way to avoid this terminology, but the idea is the same. You only have the technicality to prove that the intervals (which are the connected components) are indeed open intervals, but it is easy to see why they must be (if they weren't, since the set is open, you would be able to find a larger connected set containing the endpoint of the supposed not-open interval).

The countability of such family is a result from the fact that you can choose a distinct rational number on each one.

Answer 2

@fleablood pretty much has it in his comment.

Given an open set $G \subset \mathbb{R}$, for any point $p \in G$ there must be an open ball $U_p = (p - \epsilon, p + \epsilon)$ around $p$ where $U_p \subset G$.

Thus $G$ can be written as a union of open intervals:

$$G = \bigcup_{p \in G} U_p$$

We just need to make this into a countable union somehow.

To do this, we basically union all overlapping $U_p$ intervals (using their suprema and infima as detailed in the given proof) into a set of disjoint open intervals $\{I_x\}$ ($G$'s connected components).

Now each interval $I_x$ can be indexed by a chosen rational number that it contains (since rationals are dense in $\mathbb{R}$). We also know the rationals are countable, so the collection $\{I_x\}$ is also countable.