Given a subset $S$ of the powerset of $R$, define the following operations on it:
$f(S)=$ All countable unions of elements in $S$. $g(S)=$ The union of $S$ and complements of sets in $S$.
Say we start with the set of all open sets, $U$.
Can we reach the Borel set by applying finitely many $f,g$?
The OP asks (clarified in the comments below this answer): "Is there a finite $n$ such that the sets constructed in $n$ steps from open sets via countable union and complementation, are exactly the Borel sets?"
The answer is no, (see pages 23-24 of these notes (but beware of typos!), or Theorem 22.4 in Kechris' book Classical Descriptive Set Theory).
The meat of the proof is the construction of universal sets. Let $\mathcal{C}$ be Cantor space with the usual topology, and write $\Sigma^0_n(X)$ (etc.) for the class of subsets of $X$ which are $\Sigma^0_n$ in the sense of $X$ (etc.). Fix $n\in\mathbb{N}$ and $X$ a Polish space; then a set $U\subseteq \mathcal{C}\times X$ is $\Sigma^0_n$-universal if $U$ is $\Sigma^0_n(\mathcal{C}\times X)$ and for all $\Sigma^0_n(X)$ $V$ we have $$V=\{x: (c, x)\in U\}$$ for some $c$; $\Pi^0_n$ universality is defined analogously.
Intuitively, a $\Gamma$-universal set is a $\Gamma$-set which contains every $\Gamma$-set as a "row"; if you know some computability theory, think about how the complete c.e. set has every c.e. set "coded" into it appropriately (this analogy can in fact be made precise and proved).
The existence of universal sets (for any Polish space $X$!) is not at all obvious, and is proved by induction on $n$. Rather than post it here, I'll direct you to the sources above, since it's somewhat messy.
Now, suppose we have a $\Sigma^0_n$-universal set $U\subseteq\mathcal{C}\times\mathcal{C}$. Then if we take $Y=\{x: (x, x)\not\in U\}$, $Y$ is a $\Pi^0_n$ set, but $Y$ can't be $\Sigma^0_n$: otherwise by universality of $U$ there would be an $x\in X$ such that $y\in Y$ iff $(x, y)\in U$, but then $$x\in Y\iff (x, x)\in U\iff x\not\in Y.$$ Now, since every $\Pi^0_n$ set is $\Sigma^0_{n+1}$, this means that the Borel hierarchy doesn't stop after finitely many levels . . .
. . . on Cantor space, that is! What about $\mathbb{R}$? Well, here we run into the fact that, basically, all the Polish spaces (other than $\mathbb{N}$ and its variants) have the same Borel structure. If $X$ is an uncountable Polish space, $X$ contains a perfect $P\subseteq X$ which is homeomorphic to $\mathcal{C}$, and we can run the above argument on that perfect set.
What about if we weaken the requirement a little - does every specific Borel set arise after finitely many stages?
No, finitely many steps are not enough - you need to continue through the countable ordinals (so each specific Borel set can be built in countably many stages, but to build them all we in fact need uncountably many stages).
Why? Well, here's a rough explanation. Let $\Sigma^0_1$ be the class of open sets, $\Pi^0_n$ be the class of complements of $\Sigma^0_n$ sets, and $\Sigma^0_{n+1}$ be the class of countable unions of $\Pi^0_n$ sets. Then the class $$\Sigma^0_\omega=\bigcup_{n\in\mathbb{N}} \Sigma^0_n$$ is the collection of all Borel sets built in finitely many "steps".
Now for each $n\in\mathbb{N}$, let $X_n\subseteq (n, n+1)$ be $\Sigma^0_{n+1}$ but not $\Sigma^0_n$. (The existence of such $X_n$ is a good exercise.) Then $$Y=\bigcup X_n$$ is a Borel set, being the union of countably many Borel sets; but $Y$ can't be $\Sigma^0_n$ for any $n\in\mathbb{N}$.
Why can't $Y$ be $\Sigma^0_n$? Well, suppose it were. Then look at $Y\cap (n, n+1)$. As the intersection of an open set and a $\Sigma^0_n$ set, this is $\Sigma^0_n$ (this is a good exercise - induction on $n$); but $Y\cap (n, n+1)=X_n$ is not $\Sigma^0_n$.
Your answer gives a set which is not in any finite steps, which is interesting, but it uses my question which is that finitely many steps isn't everything.
– Andy Nov 16 '16 at 20:07