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

This question is from CLRS (3rd Edition).

So, I know that this is true, but I can't seem to prove it. My approach was the following:

Let $$f(n) = o(g(n)) \Rightarrow 0 \le f(n) < c_2g(n)$$ for some $$c_2 > 0$$ and for all $$n \ge n_1 \ge 0$$

Similarly, $$f(n) = \omega(g(n)) \Rightarrow 0 \le c_1g(n) < f(n)$$ for some $$c_1 > 0$$ and for all $$n \ge n_2 \ge 0$$

Therefore, $$0 \le c_1g(n) < f(n) < c_2 g(n)$$

But this can be true for $$f(n) = 2n$$ and $$g(n) = n$$ and $$c_1 = 1$$, $$c_2 = 3$$, and $$n_1 = n_2 = 1$$

So what exactly am I not picking up?

What is wrong is your definition of $$o$$ and $$\omega$$.
$$f\in o(g)$$ if and only if $$f(n) = h(n)g(n)$$ with $$h(n) \underset{n\rightarrow +\infty}{\longrightarrow}0$$.
$$f\in \omega(g)$$ if and only if $$g\in o(f)$$.
What you used as definition are $$\mathcal{O}$$ and $$\Omega$$ (which are different).
• In this definition, the property $f(n) < c g(n)$ must be true for ANY $c>0$, not for SOME $c>0$. The quantification is universal, not existential. Mar 7 at 14:11