It seems to me that there are two different situations which get called ``PH collapse",
(1) That $\exists i \geq 1$ s.t $\Sigma_i ^p = \Sigma_{i+1}^p$
(2) That $\exists i \geq 1$ s.t $\Sigma_i^p = \Pi_i^p$
- Are these two different independent situations?
AFAIK the only natural relation is that $\Sigma_i^p \subseteq \Pi_{i+1}^p \subseteq \Sigma_{i+2}^p$ and I don't think this is enough to make the two above scenarios be equivalent.
- Is there any natural relation even between $\Sigma_i^p$ and $\Sigma_{i+1}^p$ or between $\Pi_i^p$ and $\Pi_{i+1}^p$? For some $k > i$, if one shows that $\Sigma_k^p \subseteq \Sigma_i^p$ does that say anything about $\Pi_k^p$ vs $\Sigma_i^p$ ?
- Do both these scenarios lead to the conclusion that $PH = \cup _{j \geq 1} \Sigma_j^p = \Sigma_i ^p$ ? (If yes, then can someone kindly help with the proof?)
Roughly I guess the proof has to go like this :
That inductively assume that for some $k > i$ one has shown that $\Sigma_j^p \subseteq \Sigma_i^p$ $\forall j$ s.t $i\leq j \leq k$. Now sit at the threshold of the inductive hypothesis — and take some $L \in \Sigma_{k+1}^p$ and by truncating the first quantifier construct a language $L'$ such that $\langle x,u_1\rangle \in L'$ iff $x \in \Sigma_{k+1}^p$ with its first existential quantifier fixed to $u_1$. This makes $L' \in \Pi_{k}^p$
Now I don't know... something needs to be claimed about $L' \in \Pi_{k}^p$ to show that $L \in \Sigma_i^p$. But I don't see what goes here.