Showing equivalence of 2 Big O notations I have the task to show that the following two big O definitions are equivalent. My first thought was that it has to be equivalent because the set of natural numbers excludes all broken rational numbers.
$$
O (g) := \big\{f: \mathbb{N{_0}} \rightarrow \mathbb{R^+} | \space \exists c \in  \, \mathbb{R^+} : \forall n \in \mathbb{N{_0}} : f(n) \leq c \space  \cdot g(n)\big\}
$$
$$
O (g) := \big\{f: \mathbb{N{_0}} \rightarrow \mathbb{R^+} | \space \exists c \in  \, \mathbb{R^+} : \exists n{_0} \in \mathbb{N{_0}} : \forall n \geq n{_0} :  f(n) \leq c \space  \cdot g(n)\big\}
$$
$$
g: \mathbb{N{_0}} \rightarrow \mathbb{R^+} 
$$
Please give me hint. I would really appreciate it.
Thank you very much.

Let me temporarily call these sets $O_1(g)$ and $O_2(g)$. It should be clear that $O_1(g)\subseteq O_2(g)$: if $f(n)\le cg(n)$ for all $n\in\Bbb N_0$, then we can take $n_0$ to be any natural number and have $f(n)\le cg(n)$ for all $n\ge n_0$. Thus, we need only show that $O_2(g)\subseteq O_1(g)$ in order to prove that the two are equal.
Suppose that $f\in O_2(g)$; then there are $c\in\Bbb R^+$ and $n_0\in\Bbb N_0$ such that $f(n)\le cg(n)$ for all $n\ge n_0$. For $k<n_0$ let
$$c_k=\frac{f(k)}{g(k)}\;,$$
let $c'=\max_{k<n}c_k$, and let $c''=\max\{c,c'\}$; now just check that $f(n)\le c''g(n)$ for all $n\in\Bbb N_0$ to see that $f\in O_1(g)$.
