I want to prove the following
$$\Omega(f(n)) \subset \Omega(g(n)), iff : g(n)\in \mathcal{O}(f(n)) \wedge f(n) \not\in \mathcal{O}(g(n))$$
What I did so far is:
$$t(n)\in\Omega(f(n)) \rightarrow \exists c_{1} \exists n_{0} \forall n>n_{0} , t(n)\geq c_{1}\cdot f(n)$$
Now I need to use the given details:
$$g(n)\in \mathcal{O}(f(n)) \rightarrow \exists c_{2} \exists n_{1} \forall n>n_{1} , g(n)\leq c_{2}\cdot f(n)$$
$$\frac{1}{c1}\cdot t(n) \geq f(n)$$
$$\frac{1}{c2}\cdot g(n) \leq f(n)$$
$$\rightarrow \frac{1}{c2}\cdot g(n) \leq f(n) \leq \frac{1}{c1}\cdot t(n) $$
$$\rightarrow t(n) \in \Omega(g(n))$$
I dont know how to continue and I dont know if its right
any suggestions?
thanks.
Let's start with $f(n) \in \Omega(f(n))$. Since $\Omega(f(n)) \subset \Omega(g(n))$, we have $g(n) \in O(f(n))$ by partial ordering. You showed this correctly and more formally above. If you have already proven partial ordering of Big-O and Big-Omega, it's easier to use that.
Now, since $\Omega(f(n)) \subset \Omega(g(n))$, we let $j(n) \in \Omega(g(n))$ with $j(n) \not \in \Omega(f(n))$ by definition of a proper subset. And so since $j(n) \not \geq C * f(n)$ (for all $n \geq n_{0}$), we have $j(n) < C * f(n)$, which implies $j(n) \in O(f(n))$. And so we have this strict inequality that $c_{1} g(n) \leq j(n) < c_{2} f(n)$. Thus, $g(n) \not \in O(f(n))$.
I leave the other side for you.