Every open set in $\mathbb{R}$ is $F_\sigma$ sets I was given this assertion in class and told that the proof is really short.  I understand that for an open interval that is bounded such as $(a,b)$ we can define a closed set $[a-\frac{1}{n},b+\frac{1}{n}]$ with $n \in \mathbb{N}$.  However, I'm not sure how to extend this to an unbounded interval.
 A: For unbounded intervals such as $( a , + \infty )$ you just have to look at the bounded side: for each $n$ let $F_n = [ a + \frac{1}{n} , + \infty )$, which is closed, and show that $( a , + \infty ) = \bigcup_{n=1}^\infty F_n$.  (And analogously for $( - \infty , b )$.)
To get that every open set in $\mathbb{R}$ is F$_\sigma$, note the following:


*

*Every open set in $\mathbb{R}$ is the (disjoint) union of countably many open intervals.

*Every open interval in $\mathbb{R}$ is F$_\sigma$.

*It is not too hard to prove that countable unions of F$_\sigma$ sets are also F$_\sigma$.

A: Sets of the forms $[a,\to)$ and $(\leftarrow,a]$ are closed, and
$$(a,\to)=\bigcup_{n\in\Bbb Z^+}\left[a+\frac1n,\to\right)\;.$$
Don’t forget, though, that it’s not enough just to show that every open interval is an $F_\sigma$; you’ve got to show it for all open sets. You’ll need to use the fact that every non-empty open set in $\Bbb R$ can be written as the union of at most countably many pairwise disjoint intervals. (I’m including the possibility of unbounded open intervals.) If you already know this fact, there’s very little left to do; if not, you should prove it, which does take a bit of work. 
In case you do have to prove it, here’s a pretty big hint. Let $U$ be a non-empty open set in $\Bbb R$, and define a relation $\sim$ on $U$ by $x\sim y$ if $x\le y$ and $[x,y]\subseteq U$, or $y\le x$ and $[y,x]\subseteq U$. Show that $\sim$ is an equivalence relation and that its equivalence classes are pairwise disjoint open intervals whose union is $U$. Then use the fact that $\Bbb R$ is separable to show that there are at most countably many equivalence classes.
