Is every Borel set finitely decomposable into open sets? Given a topological space $\langle X, T \rangle $, consider the Borel algebra $B$ generated by $T$. The question is, can you write any $b \in B$ using only finitely many open (or closed) sets using set operations $\cup$ and $\sim$ (and $\cap$) ?
For closed and open sets, this is obviously the case. So the characterization is only interesting for sets which are neither open nor closed.
For example, consider $\mathbb{R}$ with the standard topology and the following set: $$A = \{\frac{1}{n} : n \in \mathbb{N}\}.$$ 
$A$ is a Borel set and obviously, $A$ is neither open nor closed. But $A' = A \cup \{ 0 \}$ is closed and one can rewrite $A$ as $A' \setminus \{ 0 \}$. Since both sets are closed, $A$ is "finitely decomposable into open (or closed) sets."
If not, can you give a counter example?
 A: Here is an elementary proof that $\mathbb{Q} \subset \mathbb{R}$ can not be constructed as you describe.
Consider $\mathbb{R}$ with the standard topology.  We will prove that all sets $A$ generated in the way you describe have the following property (let's call it property P):
Property P: For any open interval $I$, either $I \cap A$ or $I - A$ contains an open interval.
This is clearly true of closed and open sets: Either $I \cap A$ or $I - A$ is open, thus either contains an open interval or is empty, but if $I \cap A$ is empty, $I - A = I$ and vice versa.
Thus we need only show this property is preserved under complements and finite unions.  Complements are clear, since they merely switch $I \cap A$ and $I - A$.  
So we now need only show that if $A_1,A_2 \subset R$ have property $P$, then so does $A_1 \cup A_2$.  So fix an $I$.  Clearly if $I \cap A_1$ or $I \cap A_2$ contains an open interval then we're done, so we can assume $I - A_1 $ and $I - A_2$ both contain open subintervals of $I$, call them $I_1$ and $I_2$, respectively.
By property $P$, either $I_1 \cap A_2$ or $I_1 - A_2$ contains an open interval $I'$.  If the former, then $I' \subset I \cap (A_1 \cup A_2)$.  If the latter, then $I'$ is disjoint from both $A_1$ and $A_2$, thus $I' \subset I - (A_1 \cup A_2)$.  This shows that $A_1 \cup A_2$ has property P.
Since $\mathbb{Q}$ is dense in $\mathbb{R}$ with dense complement, it clearly does not satisfy property $P$.  But it is Borel (as it is a countable union of closed points), and thus is a counterexample.
