given $f(n)=O(g(n))$ and $f(n)=\Omega(g(n))$ $\rightarrow$ $f(n)=\theta(g(n))$
consider $f(n)=n$ and $g(n)=2n$
$f(n)=O(g(n))$ and $f(n)=\Omega(g(n))$
$f(n)=\theta(g(n))$
$f(n)\not=\omega(g(n))$
$f(n)\not=o(g(n))$
$f(n)\not=o(g(n))$, by definition for this to be true $f(n)$ should be less for all values of c in $cg(n)$ which is not the case in the above example
$f(n)\not=\omega(g(n))$, by definition for this to be true $f(n)$ should be greater for all of c in $cg(n)$ values which is not the case in the above example
