Why does PCP theorem imply that NP problems are hard to approximate? What I only got currently from PCP theorem is that it needs at most $O(\log n)$ randomness and $O(1)$ query of proof to approximate. So how does this result relate to the fact that solution to NP problems are hard to approximate?
Take as an example SAT. The PCP theorem (in one formulation) shows that there is a PCP for SAT which uses $O(\log n)$ bits of randomness, queries $O(1)$ places, always accepts on YES instances, and rejects on NO instances with probability at least $\alpha &gt; 0$. The PCP queries at most $2^{O(\log n)} = n^{O(1)}$ different locations in the proof, so the entire proof can be encoded in polynomial size. For each of the $2^{O(\log n)}$ possible queries, we can write a formula of constant size that is true if and only if the verifier accepts. Taking all these formulas together in CNF form, we get a CNF $\psi$ (which depends on the instance of SAT) on polynomially many variables (representing the bits of the proof) with polynomially many clauses (each possible query gives rise to $O(1)$ many clauses).

By the definition of PCP, for each $\phi \in \mathrm{SAT}$, there is a proof that satisfies the verifier on all possible queries. In other words, $\psi$ is satisfiable. On the other hand, for each $\phi \notin \mathrm{SAT}$, every proof is rejected with probability at least $\alpha$. This means that any assignment satisfies at most $1-\alpha/O(1)$ of the clauses of $\psi$, the $O(1)$ factor coming from the fact that the result of each query is encoded using $O(1)$ different clauses, one of which must be falsified if the query leads to rejection by the verifier.