Jesus Lopez wrote:

> One partial order more, on the set of functions of the natural numbers on themselves. One function is bigger than another, when it is bigger in all natural numbers of the domain except at finite quantity at worst.

You mean except on a finite set of natural numbers? ("At finite quantity" isn't quite how English-speaking mathematicians talk.)

> The order is not total.

Right: if we take the characteristic function of the even numbers versus the characteristic function of the odd numbers, neither is \\(\le\\) the other.

> Plotkin has translated the German papers of Hausdorff on order theory in a beautiful book and this order appears there nontrivially.

Hmm, interesting! Gordon Plotkin? I know him, he's a great guy...but I have trouble imagining him translating the German papers of Hausdorff.

There's another great partial order on the set of functions \\(f : \mathbb{N} \to \mathbb{N} \\) - did anyone here talk about it yet? In this one we say \\(f \le g\\) if and only if

$$ \mathrm{limsup}_{n \to \infty} \frac{f(n)}{g(n)} \le 1 $$

> One partial order more, on the set of functions of the natural numbers on themselves. One function is bigger than another, when it is bigger in all natural numbers of the domain except at finite quantity at worst.

You mean except on a finite set of natural numbers? ("At finite quantity" isn't quite how English-speaking mathematicians talk.)

> The order is not total.

Right: if we take the characteristic function of the even numbers versus the characteristic function of the odd numbers, neither is \\(\le\\) the other.

> Plotkin has translated the German papers of Hausdorff on order theory in a beautiful book and this order appears there nontrivially.

Hmm, interesting! Gordon Plotkin? I know him, he's a great guy...but I have trouble imagining him translating the German papers of Hausdorff.

There's another great partial order on the set of functions \\(f : \mathbb{N} \to \mathbb{N} \\) - did anyone here talk about it yet? In this one we say \\(f \le g\\) if and only if

$$ \mathrm{limsup}_{n \to \infty} \frac{f(n)}{g(n)} \le 1 $$