Is there a function $t$ fulfilling $\mathcal{O}(0) \subset \mathcal{O}(t) \subset \mathcal{O}(1)$?
Yes, for example: $f(n)=\cases {0&amp;$n\in\mathbb{N}_{even}$\\1&amp;$n\in\mathbb{N}_{odd}$}$
This function is not in $O(0)$, since it has infinitely many $1$'s, but also $1$ is not $O(f)$ since it has infinitely many $0$'s.
To see more clearly why $1$ is not $O(f)$, consider any constant $c&gt;0$ and any constant $n_0\in \mathbb{N}$. Then, if $1 \le c\cdot f(n)$ for any $n&gt;n_0$, we will have that for any $n_0&lt;n\in \mathbb{N}_{even}$, $1 \le c\cdot f(n) = c\cdot 0 = 0$ which is a contradiction. Hecne $1\notin O(f)$.
Therefore $O(0)\subsetneq O(f)\subsetneq O(1)$.
However, for computer science usages, as Pal GD said, we consider only monotonically increasing functions - in which case there is no such $f$.