# A simple clarification on polynomial hierarchy

$$P^{NP}\subseteq BPP^{NP}$$ holds. According to current knowledge $$BPP$$ is in $$\Sigma_2^P\cap\Pi_2^P$$ holds. So according to current knowledge is following true?

1. $$P^{\Sigma_2^P\cup\Pi_2^P}\subseteq BPP^{NP}\subseteq(\Sigma_2^P\cap\Pi_2^P)^{NP}\subseteq\Sigma_3^P\cap\Pi_3^P$$?

2. Is $$P^{\Sigma_2^P\cup\Pi_2^P}$$ the largest standard polynomial hierarchy class in $$BPP^{NP}$$ and is $$(\Sigma_2^P\cap\Pi_2^P)^{NP}$$ the smallest standard polynomial hierarchy class containing $$BPP^{NP}$$?

• How is $(\Sigma_2^\mathrm{P} \cap \Pi_2^\mathrm{P})^{\mathbf{NP}}$ defined? Note relativized classes only make sense if you start out with a class which has a machine characterization (and AFAIK $\Sigma_2^\mathrm{P} \cap \Pi_2^\mathrm{P}$ is not one such class). – dkaeae May 27 '19 at 7:59
• @dkaeae I thought $NP\cap coNP$ had a machine charaterization and likewise.. no? – T.... May 27 '19 at 8:05
• Well, we have $\mathbf{NP} \cap \mathbf{coNP} = \mathbf{P}^{\mathbf{NP} \cap \mathbf{coNP}}$, but I'm unsure how you'd make the oracles "stack". – dkaeae May 27 '19 at 8:10
• @dkaeae I see.... – T.... May 27 '19 at 8:12
• @dkaee we know $BPP$ is in $\Sigma_2^P\cap\Pi_2^P$. What is the best way to say where $BPP^{NP}$ is properly best contained in (I thought it was $(\Sigma_2^P\cap\Pi_2^P)^{NP}$ and you have shot it down then)? – T.... May 27 '19 at 8:17

The NP-machine hypothesis implies that $$BPP^{NP}=P^{NP}$$.
So combining with the hypothesis that $$PH$$ does not collapse, this will falsify the proposition that $$P^{\Sigma_2^p\cup\Pi_2^p}\subseteq BPP^{NP}$$.