On asymptotic complexity class notation? Is the class of problems with complexity $O(n^{n^\epsilon})$ at every $\epsilon&gt;0$ same as class of problems with complexity $O(n^{f(n)})$ at every $f(n)\in\omega(1)$ and hence both classes are equivalent to $\mathsf P$ ($n$ is number of input bits)?

(This is my view. Denote $C(n^{n^\epsilon})$ to be class of problems solvable in $O(n^{n^\epsilon})$ at every $\epsilon&gt;0$ and let $D(n^{f(n)})$ be class of problems solvable in $O(n^{f(n)})$ at every $f(n)$ in $\omega(1)$. Does $\cap_{f(n)\in\omega(1)}D(n^{f(n)})=\cap_{\epsilon&gt;0}C(n^{n^\epsilon})$ hold?  $\forall n\in\Bbb N\wedge\forall\delta&gt;0 \exists\epsilon&gt;0:n^\epsilon&lt;1+\delta$ and so $C(n^{n^\epsilon})$ is smaller than $D(n^{f(n)})$ at every $f(n)$ in $\omega(1)$. However for every $\epsilon&gt;0$ there is an $f(n)\in\omega(1)$ such that $f(n)&lt;n^\epsilon$ at large enough $n\in\Bbb N$ and so $D(n^{f(n)})$ at any $f(n)$ in $\omega(1)$ is smaller than $C(n^{n^\epsilon})$. We then conclude both $O(n^{n^\epsilon})$ and $O(n^{f(n)})$ at any $f(n)\in\omega(1)$ are morally equal. Since $O(n^{f(n)})$ at any $f(n)\in\omega(1)$ is same as $\mathsf P$ we have the notation.).
The time hierarchy theorem shows that there are problems which can be solved in time $O(n^{\log n})$ but not in time $O(n^{\log\log n})$. These problems can be solved in time $O(n^{n^\epsilon})$ for any $\epsilon &gt; 0$, but not in polynomial time.