Assuming $\mbox{BPP}\subseteq \Pi_2$ -What conclusions can we make?

Question

I'd like your help with the following question:

Assume we proved that $\mbox{BPP}\subseteq \Pi_2$ -What conclusions can you make?

BPP is the class of decision problems solvable by a probabilistic Turing machine in polynomial time, with an error probability of at most 1/3 for all instances,

$\Pi_2$ is the class of all languages $L$ such that there's a polynomial algorithm $M$ and a polynom $p$ so that $\forall x.x\in L\Leftrightarrow \forall u\in \{ 0,1 \}^*.\exists v \in \{ 0,1 \}^*.M(x,u,v)=1$.

We already know that $\mbox{BPP}\subseteq \Sigma_2$, so $\mbox{BPP}\subseteq \Pi_2\cap \Sigma_2$.

Answer 1

It is known. As your final statement says, $\mbox{BPP} \subseteq \Pi_2 \cap \Sigma_2$. It is called the Sipser–Gács–Lautemann theorem. (All though your "so" is somewhat misleading.)

It can be strengthened to $\mbox{BPP} \subseteq \mbox{MA} \subseteq \mbox{S}^P_2 \subseteq \Pi_2 \cap \Sigma_2 \subseteq \Pi_2$ (see also Arthur-Merlin protocol).

Answer 2

Just to add a minor point to Pal GD's answer, $BPP$ is closed under complementation. So, once you prove that $BPP \subseteq \Sigma_2$, you can easily deduce $BPP \subseteq \Sigma_2 \cap \Pi_2$. Also, (Just to improve the inclusions) even better $BPP \subseteq NP^{BPP} \subseteq MA \subseteq S_{2}^{P} \cdots$ is known.