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

#### Using Classical Definitions

By definition, $$o(g(n))$$ is the set of functions $$f(n)$$ such that $$0 \le f(n) < c_1g(n)$$ for any positive constant $$c_1 > 0$$ and all $$n \ge n_0$$.

And, $$\omega(g(n))$$ is the set of functions $$f(n)$$ such that $$0 \le c_2g(n) < f(n)$$ for any positive constant $$c_2 > 0$$ and all $$n \ge n_0$$.

So, $$o(g(n)) \cap \omega(g(n))$$ is the set of functions $$f(n)$$ such that,

$0 \le c_2g(n) < f(n) < c_1g(n)$

Now, the above inequality cannot be true asymptotically as $$n$$ becomes very large, $$f(n)$$ cannot be simultaneously greater than $$c_2g(n)$$ and less than $$c_1g(n)$$ for any constants $$c_1, c_2 > 0$$

Hence, no such $$f(n)$$ exists, i.e. the intersection is indeed the empty set.

#### Using Limit Definitions

We can use the limit definitions of $$o(n)$$ and $$\omega(n)$$ to draw same conclusion.

$o(g(n)) = \lim_{n \to \infty} \frac {f(n)} {g(n)} = 0$

and

$\omega(g(n)) = \lim_{n \to \infty} \frac {f(n)} {g(n)} = \infty$

Both of these cannot hold true as $$n$$ approaches $$\infty$$.

Hence, no such $$f(n)$$ exists, i.e. the intersection is indeed the empty set.