So, I have proven already that every open set in $\mathbb{R}$ can be written as a countable union of disjoint open intervals; i.e., that $\mathcal{O} = \bigcup_{i=1}^\infty (a_{x_i}, b_{x_i})$.
Now, consider the closed interval $\left [ a + \frac{1}{n}, b – \frac{1}{n}\right ] \subset (a, b)$.
Since the midpoint of $(a,b)$ is $\frac{b-a}{2}$, we need $\displaystyle n > \frac{2}{b-a}$.
Otherwise, $\displaystyle a + \frac{b-a}{2} = a + \frac{1}{ \frac{2}{b-a}} = b – \frac{1}{ \frac{2}{b-a}} = b – \frac{b-a}{2}$, and our interval will be degenerate.
Note that $\forall (a,b) \in \mathcal{O}$,
$\displaystyle \bigcup_{n = \frac{b-a}{2}+1}^{\infty} \left[ a + \frac{1}{n}, b – \frac{1}{n} \right] = (a,b)$.
So, we have $\displaystyle \mathcal{O} = \bigcup_{i=1}^\infty \left ( \bigcup_{n = \frac{b-a}{2} + 1}^\infty \left[ a_{x_i}+\frac{1}{n}, b_{x_i}-\frac{1}{n} \right] \right)$.
But, I don't like this. It looks like I'm still just taking a union of open sets. Please help me get this to look like the union of closed sets.
(For unbounded intervals, I know what to do – it should follow from this relatively easily).
Best Answer
This property actually holds in any metric space:
Proof. Let $F$ be a closed set of the metric space $(E, d)$. Set, for each $n > 0$, $$ U_n = \bigcup_{x \in F}\ \bigl\{y \in E \mid d(x,y)< {1 \over n} \bigr\} $$ Then each $U_n$ is open and it is not difficult to see that $F = \bigcap_{n > 0} U_n$. This proves the first part of the statement. The second part follows by taking complements.