On different characterizations of $\mathsf P$
I will consider the following simpler version of your question: is there a function $g(n,\epsilon)$ such that the following are equivalent, for a function $f(n)$:


$f$ is bounded.
For every $\epsilon &gt; 0$ there exists $N$ such that for $n \geq N$, $f(n) \leq g(n,\epsilon)$.


Consider first the case in which $g(n,\epsilon_1)$ eventually exceeds $g(n,\epsilon_2)$ whenever $\epsilon_1 &gt; \epsilon_2$ (like your example $g(n,\epsilon) = n^\epsilon$). Then the second condition can be rewritten equivalently as


For every integer $M$ there exists $N$ such that for $n \geq N$, $f(n) \leq g(n,1/M)$.


Define the function
$$ f(n) = \max_{M \leq n} g(n,1/M). $$

This function satisfies condition (3), and so is bounded, say $f \leq C$. This implies $g(n,1) \leq C$. But then the bounded function $C+1$ doesn't satisfy condition (3) for $M=1$. This contradiction shows that under the stated constraint, no function $g$ fits the bill.

On the other hand, as you mention, $f$ is bounded iff it is eventually dominated by every $\omega(1)$ integer function. Indeed, suppose that $f$ is unbounded. Then for every integer $M$ there exist infinitely many $n$ such that $f(n) \geq M$. In particular, we can find an increasing sequence $n_1,n_2,\ldots$ such that $f(n_M) \geq M$. Define a function $g$ by $g(n) = M-1$ for $n_{M-1} &lt; n \leq n_M$. Then $g = \omega(1)$ but $f$ is not eventually dominated by $g$.

The functions from $\mathbb{N}$ to $\mathbb{N}$ have the cardinality of the continuum, and so they can be put (constructively!) in one-to-one correspondence with the real interval $(0,\infty)$. This immediately gives a function $g(n,\epsilon)$ which does fit your bill, though is not particularly natural.