We know $NP=\bigcup_{k\in\Bbb N}NTIME(n^k)$ and $\Sigma_2^P=NP^{NP}$.

1. Does $\Sigma_2^P=\bigcup_{k\in\Bbb N}NTIME(n^k)$ also hold (we can do $O(n^k)$ queries to $NP$ oracle which runs in non-deterministic $O((n^k)^c)$ time which is non-deterministic $O(poly(n))$ time)?

2. Is there a reason we cannot conclude $NP=\Sigma_2^P$ from this?

From this $NP=PH$ should hold.

We know $NP=\bigcup_{k\in\Bbb N}NTIME(n^k)$ and $\Sigma_2^P=NP^{NP}$.

1. Does $\Sigma_2^P\subseteq\bigcup_{k\in\Bbb N}NTIME(n^k)$ also hold (we can do $O(n^k)$ queries to $NP$ oracle which runs in non-deterministic $O((n^k)^c)$ time which is non-deterministic $O(poly(n))$ time)?

2. Is there a reason we cannot conclude $NP=\Sigma_2^P$ from this?

From this $NP=PH$ should hold.

We know $NP=\bigcup_{k\in\Bbb N}NTIME(n^k)$ and $\Sigma_2^P=NP^{NP}$.

1. Does $\Sigma_2^P=\bigcup_{k\in\Bbb N}NTIME(n^k)$ also hold?

2. Is there a reason we cannot conclude $NP=\Sigma_2^P$ from this?

From this $NP=PH$ should hold.

We know $NP=\bigcup_{k\in\Bbb N}NTIME(n^k)$ and $\Sigma_2^P=NP^{NP}$.

1. Does $\Sigma_2^P=\bigcup_{k\in\Bbb N}NTIME(n^k)$ also hold (we can do $O(n^k)$ queries to $NP$ oracle which runs in non-deterministic $O((n^k)^c)$ time which is non-deterministic $O(poly(n))$ time)?

2. Is there a reason we cannot conclude $NP=\Sigma_2^P$ from this?

From this $NP=PH$ should hold.