Prove that \(o(g(n)) \cap \omega(g(n))\) is the empty set.

\(o(g(n))\) implies that for any positive constant \(c > 0\), there exists a constant \(n_a > 0\) such that \(0 \leq f(n) < cg(n) \ \forall \ n \geq n_a\). Likewise, \(\omega(g(n))\) implies that for any positive constant \(c > 0\), there exists a constant \(n_b > 0\) such that \(0 \leq cg(n) < f(n) \ \forall \ n \geq n_b\).

In order to find \(o(g(n)) \cap \omega(g(n))\) we must combine these two inequalities, so we have:

\[0 \leq cg(n) < f(n) < cg(n) \ \forall \ n > \max(n_a, n_b)\]

This inequality has no solution, and therefore \(o(g(n)) \cap \omega(g(n))\) is the empty set.