Any subset of $\mathbb{R}$ is a countable union of pairwise disjoint intervals.

Question

Let $A\subseteq\mathbb{R}$ be non-empty and bounded. Also, $\forall x\in A$, $\exists\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subseteq A$. Prove that $A$ is a finite or countable union of pairwise disjoint open intervals.

I'm aware that this question has many different proofs, but I'm interested in knowing how one would go about proving it in this particular way.

For $q\in\mathbb{Q}\cap A$, define

$\alpha(q):= \text{inf}\{x\in\mathbb{R}:(x,q]\subseteq A\}$; $\beta(q):= \text{sup}\{x\in\mathbb{R}:[q,x)\subseteq A\}$

First, why do $\alpha$ and $\beta$ exist and $\alpha<\beta$? Then, how can one show that $I_q:=(\alpha(q),\beta(q))\subseteq A$? Lastly, how can one prove that

• $\bigcup_{q\in\mathbb{Q}\cap A}I_q=A$
• $\forall q, s\in\mathbb{Q}\cap A$, either $I_q=I_s$ or $I_q\cap I_s=\emptyset$

Thanks in advance.

Answer 1

Okay, so:

Firsly, let's take any $q\in Q\cap A$. We know that $ q \in A$, so there exists $\epsilon_q$, such: $(q-\epsilon_q;q+\epsilon_q) \subseteq A$.

Now, we're ready to look at $\alpha(q),\beta(q)$. Why do they exist? A is bounded, $[q-\frac{\epsilon_q}{2}; q] \subseteq A $, and $[q;q + \frac{\epsilon_q}{2}] \subseteq A$. It is clear now, that both $\alpha(q),\beta(q)$ exists and $\alpha(q) < \beta(q)$.

I suppose u denote $I_q = (\alpha(q),\beta(q)) $.

$I_q \subseteq A$ because $\alpha(q) \ (\beta(q))$ by definition are such that $(\alpha(q),q] \subseteq A$ and $[q,\beta(q)) \subseteq A \ \ \ \ (*)$

To prove 1 and 2:

1) I'll show two inclusions. Let's pick any $x \in \bigcup_{q\in Q \cap A} I_q $. I'll show that $x \in A$. Clearly for particular $q_x$, $x \in I_{q_x}$, that means $ x \in (\alpha(q_x),\beta(q_x)) $ and by $(*)$ we get $x \in A$

Now let's pick any $y \in A$. There exists $ \epsilon_y $ such that $(y-\epsilon_y;y+\epsilon_y) \subseteq A $. In any interval, there is $ p \in Q $, $p \in (y-\epsilon_y;y+\epsilon_y), p>y $. Let's look at $I_p = (\alpha(p),\beta(p)) $. I'll show that $\alpha(p)<y$, which is obvious because by definition of $\alpha(p)$, there must holds inequality: $\alpha(p) \leq y-\epsilon_y < y $.

So $ y \in (\alpha(p),p) \subseteq I_p \in \bigcup_{q \in Q \cap A} I_q $

2) If $I_q = I_s$ everything's good, if no, let's assume that $I_q \cap I_s \neq \emptyset$

Just to fix something, say $q<s$ ($s<q$ case is really the same).

Clearly then $\beta(q) < \beta(s) $, but $\beta(q) \in (\alpha(s),\beta(s))$ that means $\beta(q)$ cannot be $ \sup\{x\in R: [q,x) \in A\} $ (because $[q,\beta(q)) \in A , (\alpha(s), \beta(s)) \in A , \alpha(s) < \beta(q) < \beta(s)$ (otherwise it really would be $I_q \cap I_s = \emptyset $)

So,there we have a problem, cause whole $[q,\beta(s)) \subseteq A$, and that means $\beta(q)$ wasn't supremum. Contrary