Big O comparison

Question

If $T(n) = \mathcal{O}(n^3)$ Then $T(n) = \mathcal{O}(n^2)$, is this statement right?

Same question for omega, if $T(n) = \Omega(n)$ does that mean $T(n)= \Omega(n^2)$?

Answer 1

No for both. Let $T(n) = n^3$ for the first. Then $T(n) = \mathcal{O}(n^3)$ but $T(n) \neq \mathcal{O}(n^2)$. Let $T(n) = n$ for the second. Then $T(n) = \Omega(n)$ but $T(n) \neq \Omega(n^2)$.

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Let $T(n) = n^3$. Claim that $T(n) \neq \mathcal{O}(n^2)$. Since $\displaystyle \lim_{n\rightarrow\infty} \dfrac{n^3}{n^2} = \lim_{n\rightarrow\infty} n = \infty \not< \infty$, $n^3 \neq \mathcal{O}(n^2)$. Similarly, $\displaystyle \lim_{n\rightarrow\infty} \dfrac{n}{n^2} = \lim_{n\rightarrow\infty} \dfrac{1}{n} = 0 \not> 0$. So $n \neq \Omega(n^2)$.