# Equivalene of big O definitions (Limit Definition $\Longleftrightarrow$ Quantifier Definition)

I need to proof, that both definitions of the Big 0 notation are equiavlent, but I am not sure if my proof works both ways of the equivalence.

Definitions:
Let f,g be functions.

1. $$f(n)\in \mathcal{O}(g)\Longleftrightarrow \exists c>0\exists n_0 \forall n\geq n_0 :f(n)\leq cg(n)$$
2. $$f(n)\in \mathcal{O}(g)\Longleftrightarrow \limsup\limits_{n\to\infty}\frac{f(n)}{g(n)}<\infty$$

My proof:
$$f(n)\in \mathcal{O}(g)\Longleftrightarrow \limsup\limits_{n\to\infty}\frac{f(n)}{g(n)}<\infty$$
$$\Longleftrightarrow \exists c>0 : \limsup\limits_{n\to\infty}\frac{f(n)}{g(n)}\leq c$$
$$\Longleftrightarrow \exists c>0\exists n_0 \forall n\geq n_0 : \frac{f(n)}{g(n)}\leq c$$
$$\Longleftrightarrow \exists c>0 \exists n_0 \forall n\geq n_0 : \frac{f(n)}{cg(n)}\leq 1$$
$$\Longleftrightarrow \exists c>0 \exists n_0 \forall n\geq n_0 :f(n)\leq cg(n)$$
$$\Longleftrightarrow f(n)\in \mathcal{O}(g)$$

I would like to know, if the conversion is correct or if I made any mistakes. Thanks in advance!

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Commented Nov 9, 2020 at 22:22

## 1 Answer

This 2 definitions, of course, are not equivalent, when $$g$$ have $$0$$ subseuquence i.e. when $$\exists n_k \in \mathbb{N}, n_{k+1}>n_k , k\in \mathbb{N}$$ such that $$g(n_k)=0$$. In this case, assuming, that $$\frac{0}{0}$$ have no sense as well as $$\frac{a}{0}, a \ne 0$$, then $$\frac{f(n_k)}{g(n_k)}$$ is not defined, while definition with quantifiers have no problem.

This is why, imho, first should be taken as base definition and second considered only in appropriate situations and why I count your question important.

If such subsequence does not exist, then your proof is correct.

• Yes you are correct. I forgot to mention that I assume that I pressuposed $g(n)>0$. Thanks a lot! Commented Nov 10, 2020 at 9:06