# Is $\Omega(n) = O(n \log n)$? [closed]

problem $$T(n) = 3T(n/4) + n \log n$$

$$f(n) = n \log n$$ and $$g(n) = n$$

Why is $$\Omega(g(n)) = O(n \log n)$$? Is it because $$\Omega$$ means at some $$n$$ and constant, $$\Omega(g(n)) = O(n \log n)$$?

• Why do you think that $\Omega(g(n)) = O(n\log n)$? Some context seems to be missing here. May 14 at 11:14

No. Consider $$n^2 \in \Omega(g(n))$$, obviously, $$n^2\notin O(n\log n)$$ and therefore $$\Omega(g(n)) \neq O(n \log n)$$.
Intuitively: $$\Omega(\cdot)$$ gives a lower bound, $$O(\cdot)$$ gives an upper bound. If I tell you $$x > 1$$, you can't deduce an upper bound for $$x$$ from it.