The polynomial hierarchy is a hierarchy. Specifically: $$ \Sigma^p_i\cup\Pi^p_i\subseteq\Delta^p_{i+1}\subseteq\Sigma^p_{i+1}\cap\Pi^p_{i+1} $$

Proof

To see that $\Sigma^p_i\cup\Pi^p_i\subseteq\Delta^p_{i+1}=\mathcal{P}^{\Sigma^p_i}$ , observe that the machine which checks its input against its oracle and accepts or rejects when the oracle accepts or rejects (respectively) is easily in $\mathcal{P}$ , as is the machine which rejects or accepts when the oracle accepts or rejects (respectively). These easily emulate $\Sigma^p_i$ and $\Pi^p_i$ respectively.

Since $\mathcal{P}\subseteq\mathcal{NP}$ , it is clear that $\Delta^p_i\subseteq\Sigma^p_i$ . Since $\mathcal{P}^{\mathcal{C}}$ is closed under complementation for any complexity class $\mathcal{C}$ (the associated machines are deterministic and always halt, so the complementary machine just reverses which states are accepting), if $L\in\mathcal{P}^{\Sigma^p_i}\subseteq\Sigma^p_i$ then so is $\overline{L}$ , and therefore $L\in\Pi^p_i$ .

Unlike the arithmetical hierarchy, the polynomial hierarchy is not known to be proper. Indeed, if $\mathcal{P}=\mathcal{NP}$ then $\mathcal{P}=\mathcal{PH}$ , so a proof that the hierarchy is proper would be quite significant.