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Is the class of problems with complexity $O(n^{n^\epsilon})$ at every $\epsilon>0$ 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 $\epsilon>0$ and let $D(n^{f(n)})$ be class of problems solvable in $O(n^{f(n)})$ at every $f(n)$ in $\omega(1)$.  $\forall n\in\Bbb N\wedge\forall\delta>0 \exists\epsilon>0:n^\epsilon<1+\delta$ and so $C(n^{n^\epsilon})$ is smaller than $D(n^{f(n)})$ at every $f(n)$ in $\omega(1)$. However for every $\epsilon>0$ there is an $f(n)\in\omega(1)$ such that $f(n)<n^\epsilon$ 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.).

Is the class of problems with complexity $O(n^{n^\epsilon})$ at every $\epsilon>0$ 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 $\epsilon>0$ and let $D(n^{f(n)})$ be class of problems solvable in $O(n^{f(n)})$ at every $f(n)$ in $\omega(1)$. Does $\cap_{f(n)\in\omega(1)}D(n^{f(n)})=\cap_{\epsilon>0}C(n^{n^\epsilon})$ hold? $\forall n\in\Bbb N\wedge\forall\delta>0 \exists\epsilon>0:n^\epsilon<1+\delta$ and so $C(n^{n^\epsilon})$ is smaller than $D(n^{f(n)})$ at every $f(n)$ in $\omega(1)$. However for every $\epsilon>0$ there is an $f(n)\in\omega(1)$ such that $f(n)<n^\epsilon$ 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.).

If you say a problem has complexity $O(n^{n^\epsilon})$ at every $\epsilon>0$ then can we denote the complexity as $O(n^{f(n)})$ at any $f(n)\in\omega(1)$ or equivalently in $\mathsf P$ ($n$ is number of input bits)?

(This is my view. $\forall n\in\Bbb N\wedge\forall\delta>0 \exists\epsilon>0:n^\epsilon<1+\delta$ and so $O(n^{n^\epsilon})$ is smaller than $O(n^{f(n)})$ at any $f(n)$ in $\omega(1)$. However for every $\epsilon>0$ there is an $f(n)\in\omega(1)$ such that $f(n)<n^\epsilon$ at large enough $n\in\Bbb N$ and so $O(n^{f(n)})$ at any $f(n)$ in $\omega(1)$ is smaller than $O(n^{n^\epsilon})$. 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.).

Is the class of problems with complexity $O(n^{n^\epsilon})$ at every $\epsilon>0$ 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 $\epsilon>0$ and let $D(n^{f(n)})$ be class of problems solvable in $O(n^{f(n)})$ at every $f(n)$ in $\omega(1)$. $\forall n\in\Bbb N\wedge\forall\delta>0 \exists\epsilon>0:n^\epsilon<1+\delta$ and so $C(n^{n^\epsilon})$ is smaller than $D(n^{f(n)})$ at every $f(n)$ in $\omega(1)$. However for every $\epsilon>0$ there is an $f(n)\in\omega(1)$ such that $f(n)<n^\epsilon$ 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.).

If you say a problem has complexity $O(n^{n^\epsilon})$ at every $\epsilon>0$ then can we denote the complexity as $O(n^{f(n)})$ at any $f(n)\in\omega(1)$ or equivalently in $\mathsf P$ ($n$ is number of input bits)?

(This is my view. $\forall n\in\Bbb N\wedge\forall\delta>0 \exists\epsilon>0:n^\epsilon<1+\delta$ and so $O(n^{n^\epsilon})$ is smaller than $O(n^{f(n)})$ at any $f(n)$ in $\omega(1)$).

If you say a problem has complexity $O(n^{n^\epsilon})$ at every $\epsilon>0$ then can we denote the complexity as $O(n^{f(n)})$ at any $f(n)\in\omega(1)$ or equivalently in $\mathsf P$ ($n$ is number of input bits)?

(This is my view. $\forall n\in\Bbb N\wedge\forall\delta>0 \exists\epsilon>0:n^\epsilon<1+\delta$ and so $O(n^{n^\epsilon})$ is smaller than $O(n^{f(n)})$ at any $f(n)$ in $\omega(1)$. However for every $\epsilon>0$ there is an $f(n)\in\omega(1)$ such that $f(n)<n^\epsilon$ at large enough $n\in\Bbb N$ and so $O(n^{f(n)})$ at any $f(n)$ in $\omega(1)$ is smaller than $O(n^{n^\epsilon})$. 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.).