I need to help with mathematical problem. Let's have function $g(n) : \mathbb{N} \to \mathbb{R}^+$ and $2$ sets defined: $$o(g(n)) = \{f(n) : \mathbb{N} \to \mathbb{R}^+\;|\; \forall c \in \mathbb{R}^ +\; \exists n_0 \in \mathbb{N}\; ∀n \geq n_0 : f(n) < c · g(n)\} \\ and \\ \bar{o}(g(n)) = \{f(n) : \mathbb{N} \to \mathbb{R}^ +\;|\;\forall c \in \mathbb{R}^ +\; \exists n_0 \in \mathbb{N}\; \forall n \geq n_0 : f(n) \leq c · g(n)\}$$ How can I prove that $o(g(n))$ is equal to $\bar{o}(g(n))$? Which consequence does it have to definition of little o ($o$)?
I know that $o(g(n)) \implies \bar{o}(g(n))$ side of equivalence should be trivial.
Let $f\in\bar o(g)$. To prove $f\in o(g)$, let $c$ be any positive number. Since $f\in \bar o(g)$, there exists a natural number $n_0$ such that, for all $n\ge n_0$, we have $$f(n)\le \frac c2g(n)<cg(n).$$