Show that any open subsets of the real line are $F_\sigma$-sets.

This is an exercise from a topological book. It is this:

Show that any open subsets of the real line are $F_\sigma$-sets.

Could anybody help to solve it?

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You could do it for metric spaces in general. Let $F$ be a non-empty closed set, then $x \mapsto d(x,F)$ is a continuous function vanishing exactly on $F$. It follows that $F$ is a $G_\delta$, being the intersection of countably many open sets: $F = \bigcap_{n=1}^\infty \{x\,:\,d(x,F) \lt 1/n\}$. Thus $F^c$ is an $F_\sigma$. – t.b. Aug 7 '12 at 7:57
@t.b. Very nice! – Paul Aug 7 '12 at 8:00

First note that open subsets of the real line are countable unions of (pairwise disjoint) open intervals. Next show that every open interval is an F$_\sigma$-set. As countable unions of F$_\sigma$-sets are also F$_\sigma$, this gives the result.
My idea is this: for an open interval $[a,b]$, there are countably open intervals such that their union is $[a,b]$, and hence we may get countably closed set which are the closues of the open intervals and their union is also $[a,b]$. Am I right? – Paul Aug 7 '12 at 8:10
@John: In you comment, is $[a,b]$ an open interval or a closed interval? – arjafi Aug 7 '12 at 19:11
Sorry. $[a,b]$ should be $(a,b)$. – Paul Aug 8 '12 at 12:05
@John: Don't worry about "closures of open intervals," just worry about closed intervals: if $(a,b)$ is any open interval, then it is a countable union of (non-degenerate) closed intervals (which happen to be the closures of their interiors, but that is not too important). – arjafi Aug 8 '12 at 13:45