TITLE: why is each open set an $F_\sigma$?
QUESTION [13 upvotes]: In Royden's Real Analysis:

On $\mathbb{R}$, a set which is a countable union of closed sets is called an $F_\sigma$. 
  Thus every countable set is an $F_\sigma$, as is, of course, every closed set. 
  A countable union of sets in Fa is again in $F_\sigma$. 
  Since
  $$(a, b)= \cup_{n=1}^\infty [a + 1/n, b - 1/n],$$
  each open interval is an $F_\sigma$, and hence each open set is an $F_\sigma$.

Each open interval is an $F_\sigma$, but why is each open set an $F_\sigma$? An open set can be uncountable union of open intervals. 
If we are considering a general Borel sigma algebra instead of $B(\mathbb R)$, will each open set be an  $F_\sigma$?
Thanks!

REPLY [3 votes]: As a slight expansion of Brian Scott's argument every open set in $\mathbb R$ is the disjoint union of a countable number of open intervals. In particular if $U$ is an open subset of $\mathbb R$ then for each $x \in U$ we can consider the largest interval $x \in I_x \subset U$. Then the set $\mathcal I=\{I_x : x \in U\}$ is a disjoint open cover of $U$ whose union is precisely $U$. We see that it's countable from observing that each interval contains a rational so there's an injection from $\mathcal I$ to the rationals.