Define $io$-$SUBEXP$ to be the class of languages $L$ such that there is a language $L' \in \cap_{\varepsilon > 0} TIME(2^{n^{\varepsilon}})$ and for infinitely many $n$, $L$ and $L'$ agree on all instances of length $n$. (That is, this is the class of languages which can be solved infinitely often, in subexponential time.)Is there an oracle $A$ such that $NP^A \not\subset io$-$SUBEXP^A$? If we equip SAT with the oracle $A$ in the usual way, can we say that $SAT^A$ is not in this class? (I'm asking separate questions here, because we have to be careful with infinitely-often time classes: just because you have a reduction from problem $B$ to problem $C$ and $C$ is solvable infinitely often, you may not actually get that $B$ is solvable infinitely often without further assumptions on the reduction: what if your reduction from $B$ misses the input lengths that you can solve $C$ on?)
Is there an oracle such that SAT is not infinitely often in sub-exponential time?
cc.complexity theory;sat;oracles
You can just take the oracle A s.t. NP$^A$=EXP$^A$ since EXP is not in i.o.-subexp. For SAT$^A$ it depends on the encoding, for example if the only valid SAT instances have even length then it is easy to solve SAT on odd-length strings. But if you use a language like $L=\{\phi 01^*\ |\ \phi\in SAT^A\}$ then you should be fine.