NP-Complete Problem and Polynomial Hierarchy I have tried to search the internet to check if the following is correct:
If $\sum_{2}$ contains a NP-Complete problem then PH collapses to NP: $PH=NP$
For example if $SAT\epsilon\sum_{2}$ than: $PH=NP$
$\mathrm{SAT}$ and all $\mathrm{NP}$-complete problems are certainly in $\Sigma_2^P$, as $\mathrm{NP} \subseteq \Sigma_2^p$. It is true however that if $\mathrm{SAT}$ (or any $\mathrm{NP}$-complete problem) ever gets proven to be $\Sigma_2^p$-hard, then $\mathrm{NP} = \Sigma_2^p$, and consequently the polynomial hierarchy collapses to $\mathrm{NP}$.