Using time hierarchy theorem to show $Time(n^7)$ strictly contained in P

Question

I'm relatively new to computational complexity and am trying to use the time hierarchy theorem to show that $Time(n^7)$ is strictly contained in P. I understand that the time hierarchy theorem says that if the limit as n tends to infinity of $\frac{T_1(n)}{T_2(n)}=0$ then $Time(T_1(n))$ is strictly contained in $Time(T_2(n))^2$. It seems obvious to me that this implies the desired result because for any number $x$ greater than 7, $Time(n^7)$ will be strictly contained in $Time(n^{2x})$ and then P is the union over all such time complexity classes. Could anyone help with how I might write this more formally? Is it necessary to assume the existence of some language in $Time(n^{16})$ for example? I feel like I might be missing something because it just seems clear to me, any help is much appreciated.

Answer 1

By the time hierarchy theorem, if $f$ and $g$ are time-constructible functions and $f(n) \log f(n) = o(g(n))$ then $\mathsf{DTIME}( f(n) ) \subsetneq \mathsf{DTIME}(g(n))$.

From your question you seem to be using the following weaker for of the theorem: if $f$ and $h$ are time-constructible and $f(n) = o(h(n))$ then $\mathsf{DTIME}( f(n) ) \subsetneq \mathsf{DTIME}( h(n)^2 )$. This follows from the previous statement by choosing $g(n)=h(n)^2$ and noticing that $f(n) \log f(n) = O(f(n)^2) = o(h(n)^2)$. I will therefore use this latter version.

Choosing $f(n) = n^7$ and $h(n) = n^8$: $$ \mathsf{DTIME}( n^7 ) \subsetneq \mathsf{DTIME}( n^{16} ) \subseteq \bigcup_{k \in \mathbb{N}} \mathsf{DTIME}(n^k)=\mathsf{P}. $$